The angular momentum of the ice skater spinning at 6.00 rev/s is 15.82 kg · m²/s. When the skater reduces their rate of spin to 1.95 rev/s, the moment of inertia is 1.29 kg · m². If the skater keeps their arms in and allows friction with the ice to slow them to 3.00 rev/s in 23.0 seconds, the average torque exerted is -0.343 N · m (indicating the direction of the torque).
Explanation:(a) The angular momentum of the ice skater can be calculated by multiplying the angular velocity (in radians per second) by the moment of inertia (in kg · m²). First, we need to convert the angular velocity from rev/s to rad/s. Since 1 rev = 2π rad, the angular velocity in rad/s is 6.00 rev/s * 2π rad/rev = 37.70 rad/s. Now, we can calculate the angular momentum: Angular momentum = angular velocity * moment of inertia = 37.70 rad/s * 0.420 kg · m² = 15.82 kg · m²/s.
(b) To find the new moment of inertia when the angular velocity decreases to 1.95 rev/s, we can rearrange the formula for angular momentum to solve for moment of inertia. Angular momentum = angular velocity * moment of inertia. Rearranging the equation: Moment of inertia = angular momentum / angular velocity = 15.82 kg · m²/s / (1.95 rev/s * 2π rad/rev) = 1.29 kg · m².
(c) To calculate the average torque exerted when the skater slows to 3.00 rev/s, we can use the equation torque = change in angular momentum / time. First, we need to calculate the change in angular momentum by subtracting the initial angular momentum from the final angular momentum. The initial angular momentum is 6.00 rev/s * 2π rad/rev * 0.420 kg · m² = 15.82 kg · m²/s, and the final angular momentum is 3.00 rev/s * 2π rad/rev * 0.420 kg · m² = 7.91 kg · m²/s. The change in angular momentum is 7.91 kg · m²/s - 15.82 kg · m²/s = -7.91 kg · m²/s. Finally, we can calculate the average torque: Torque = change in angular momentum / time = -7.91 kg · m²/s / 23.0 s = -0.343 N · m (negative sign indicates the direction of the torque).
What is the maximum speed that a car can maneuver over a circular turn of radius r = 75.0 m without sliding if the coefficient of static friction is µs = 0.780? (b) What is the maximum speed that a car can maneuver over a circular turn of radius r = 25.0 m without sliding if the coefficient of static friction is µs = 0.120?
Answer:
Explanation:
Given
Radius of Track [tex]r_1=75 m[/tex]
coefficient of Static Friction [tex]\mu _s=0.78[/tex]
Here centripetal Force is Balanced by Friction Force
thus
[tex]\frac{mv^2}{r}=\mu _sg[/tex]
[tex]\frac{v^2}{r}=\mu _sg[/tex]
[tex]v=\sqrt{\mu _srg}[/tex]
[tex]v=\sqrt{0.78\times 75\times 9.8}[/tex]
[tex]v=23.94 m/s[/tex]
(b)For [tex]r_2=25 m[/tex]
[tex]\mu _s=0.12[/tex]
[tex]v=\sqrt{\mu _sr_2g}[/tex]
[tex]v=\sqrt{0.12\times 25\times 9.8}[/tex]
[tex]v=5.42 m/s[/tex]
Final answer:
The maximum speed that a car can maneuver over a circular turn without sliding depends on the coefficient of static friction and the radius of the turn. This can be calculated using the formula v = sqrt(μs * g * r). For the given scenarios, the maximum speeds are 30.5 m/s and 7.36 m/s respectively.
Explanation:
The maximum speed that a car can maneuver over a circular turn without sliding depends on the coefficient of static friction and the radius of the turn. The maximum speed can be calculated using the formula:
v = sqrt(μs * g * r)
where v is the maximum speed, μs is the coefficient of static friction, g is the acceleration due to gravity (9.8 m/s²), and r is the radius of the turn.
For the first scenario with r = 75.0 m and μs = 0.780, the maximum speed is:
v = sqrt(0.780 * 9.8 * 75.0) = 30.5 m/s
For the second scenario with r = 25.0 m and μs = 0.120, the maximum speed is:
v = sqrt(0.120 * 9.8 * 25.0) = 7.36 m/s
Problem 6.056 Air enters a compressor operating at steady state at 15 lbf/in.2, 80°F and exits at 275°F. Stray heat transfer and kinetic and potential energy effects are negligible. Assuming the ideal gas model for the air, determine the maximum theoretical pressure at the exit, in lbf/in.2 p2,max = lbf/in.2
To solve this process it is necessary to consider the concepts related to the relations between pressure and temperature in an adiabatic process.
By definition the relationship between pressure and temperature is given by
[tex](\frac{P_2}{P_1})=(\frac{T_2}{T_1})^{(\frac{\gamma}{\gamma-1})}[/tex]
Here
P = Pressure
T = Temperature
[tex]\gamma =[/tex]The ratio of specific heats. For air normally is 1.4.
Our values are given as,
[tex]P_1 = 15lb/in^2\\T_1= 80\°F = 299.817K\\T_2 =400\°F = 408.15K[/tex]
Therefore replacing we have,
[tex](\frac{P_2}{P_1})=(\frac{T_2}{T_1})^{(\frac{\gamma}{\gamma-1})}[/tex]
[tex](\frac{P_2}{15})=(\frac{408.15}{299.817})^{(\frac{1.4}{1.4-1})}[/tex]
Solving for [tex]P_2,[/tex]
[tex]P_2 = 15*(\frac{408.15}{299.817})^{(\frac{1.4}{1.4-1})}[/tex]
[tex]P_2 = 44.15Lbf/in^2[/tex]
Therefore the maximum theoretical pressure at the exit is [tex]44.15Lbf/in^2[/tex]
To determine the maximum theoretical pressure at the exit of the compressor, we can use the ideal gas law. Given the initial pressure and temperature, as well as the final temperature, we can calculate the maximum theoretical pressure using the equation p2,max = (p1 * T2) / T1.
Explanation:To determine the maximum theoretical pressure at the exit of the compressor, we can use the ideal gas law. The ideal gas law is given by the equation pV = nRT, where p is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin. We can rearrange this equation to solve for the pressure:
p = (nRT) / V
Since the volume and the number of moles are constant, we can rewrite the equation as:
p1 / p2 = T1 / T2
where p1 and T1 are the initial pressure and temperature, and p2 and T2 are the final pressure and temperature.
Substituting the given values, we have:
p2,max = (p1 * T2) / T1
Plugging in the values from the problem statement, we have:
p2,max = (15 lbf/in.2 * 275°F) / 80°F
p2,max = 51.56 lbf/in.2
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A typical tornado can be envisioned as a cylinder of height 640 m, with a diameter of about 230 m (note: the visible funnel cloud is usually somewhat less than the actual diameter). The rotation in the tornado is approximately a `solid body rotation', meaning that it rotates very much like a solid cylinder would, even though air is a gas. If the outer edge of the cylinder has a speed of 51 m/s, and the air has a density of 0.97 kg/m3, determine the kinetic energy contained by the tornado.
Answer:
16771720740.20324 J
Explanation:
[tex]\rh0[/tex] = Density = 0.97 kg/m³
V = Volume = [tex]\pi r^2h[/tex]
d = Diameter of cylinder = 230 m
r = Radius = [tex]\frac{d}{2}=\frac{230}{2}=115\ m[/tex]
h = Height of the cylinder = 640 m
v = Velocity of cylinder = 51 m/s
Mass of object is given by
[tex]m=\rho V\\\Rightarrow m=0.97\times \pi 115^2\times 640\\\Rightarrow m=25792727.013\ kg[/tex]
Moment of inertia of a cylinder
[tex]I=\frac{1}{2}mr^2\\\Rightarrow I=\frac{1}{2}\times 25792727.013\times 115^2\\\Rightarrow I=170554407373.4625\ kgm^2[/tex]
Angular speed
[tex]\omega=\frac{v}{r}\\\Rightarrow \omega=\frac{51}{115}\ rad/s[/tex]
Kinetic energy is given by
[tex]K=\frac{1}{2}I\omega^2\\\Rightarrow K=\frac{1}{2}\times 170554407373.4625\times \left(\frac{51}{115}\right)^2\\\Rightarrow K=16771720740.20324\ J[/tex]
The kinetic energy contained by the tornado is 16771720740.20324 J
The Sun delivers an average power of 1150 W/m2 to the top of the Earth’s atmosphere. The permeability of free space is 4π × 10−7 T · N/A and the speed of light is 2.99792 × 108 m/s. Find the magnitude of Em for the electromagnetic waves at the top of the atmosphere. Answer in units of N/C.
Answer:
E=930.84 N/C
Explanation:
Given that
I = 1150 W/m²
μ = 4Π x 10⁻⁷
C = 2.999 x 10⁸ m/s
E= C B
C=speed of light
B=Magnetic filed ,E=Electric filed
Power P = I A
A=Area=4πr² ,I=Intensity
[tex]I=\dfrac{CB^2}{2\mu_0}[/tex]
[tex]I=\dfrac{CE^2}{2\mu_0 C^2}[/tex]
[tex]E=\sqrt{{2I\mu_0 C}}[/tex]
[tex]E=\sqrt{{2\times 1150\times 4\pi \times 10^{-7}(2.99792\times 10^8)}}[/tex]
E=930.84 N/C
Therefore answer is 930.84 N/C
To find the magnitude Em of the electromagnetic waves at the top of the earth's atmosphere, we use the intensity of electromagnetic wave and solving the equation Em = sqrt(2Icμo), we can find the magnitude of Em in units of N/C.
Explanation:To find the magnitude Em of the electromagnetic waves at the top of the Earth's atmosphere, we use the fact that the power received per unit area is the intensity I of the electromagnetic wave. According to the given information, this intensity is 1150 W/m2. The relationship between the intensity and electromagnetic fields is given by the equation I = 0.5 * E²/c * μo. Solving for Em, we get Em = sqrt(2Icμo), where μo = 4π × 10-7 T N/A² is the permeability of free space and c = 2.99792 × 10⁸ m/s is the speed of light.
Subbing in the given values, we can compute Em as:
Em = sqrt[2 * 1150 W/m² * 2.99792 × 10⁸ m/s * 4π × 10-7 T N/A²]
This computation will give the strength of the electric field at the top of the earth’s atmosphere in units of N/C.
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Two billiard balls of equal mass move at right angles and meet at the origin of an xy coordinate system. Initially ball A is moving upward along the y axis at 2.0 m/s and ball B is moving to the right along the x axis with speed 3.7 m/s. After the collision ball B is moving along the positive y axis. a. What is the speed of ball A and B after the collision? b. What is the direction of motion of ball A after the collision? c. What is the total momentum and kinetic energy of the two balls after the collision?
The momentum conservation allows to find the results for the questions about the collision of the balls are:
A) The velocities after the collision are: vₐ = 3.7 i m/s and [tex]v_b[/tex] = 2 j m/s
B) The direction of ball A is: x-axis
C) The total moment is: [tex]p_{total}[/tex] = 4.2m and the direction is θ = 28.4º
The total kinetic energy is: [tex]K_{total}[/tex] = 8.84m J
Given parameters
Ball A initially moves y-axis at [tex]v_{oay}[/tex] = 2.0 m / s Ball B moves end x axis [tex]v_[obx}[/tex]vobx = 3.7 m / s Ball B moves on the y-axis after collisionTo find
A) The speeds after the collision.
B) direction of the ball A.
C) The moment and energy after the crash.
Momentum is defined by the product of mass and velocity. When a system is isolated, there is no external force, the momentum is conserved. This is an important conservation law of physics.
p = m v
Where the bold letters indicate vectors, m is the mass and v the velocity.
A) The system is formed by the two balls, therefore it is isolated and the momentum is conserved. Since the momentum is a vector quantity we write the components on each axis, see attached for a schematic.
Initial instant. Before the crash.
They indicate that ball B moves in the axis and after the collision.
x-axis
p₀ₓ = m [tex]v_{obx}[/tex]vobx
y-axis
[tex]p_{oy} = m v_{oay}[/tex]
Final moment. After the crash.
x-axis
[tex]p_{fx} = m v_{fax}[/tex]
y- axis
[tex]p_{fy} = m v_{fay} + m v_{fby}[/tex]
The momentum is preserved
x- axis
p₀ₓ = [tex]p_{fx}[/tex]
m [tex]v_{obx}[/tex] = m [tex]v_{fax}[/tex]
[tex]v_{fax} = v _{obx}\\v_{fax} = 3.7 m/s[/tex]
y- axis
[tex]p_{oy} = p_[fy} \\m v_{oay} = m v_{fay} + m v_[fby}\\v_{oay} = v_{fay} + v_{fby}[/tex]
2 = [tex]v_{fay} + v_fby}[/tex]
Ball A moves in the x-axis therefore it has no final velocity in the y-axis.
[tex]v_{fay}=0[/tex]
[tex]v_{fby}[/tex] = 2 m / s
B) Ball A moves on the x-axis.
C) the total momentum is the sum of the momentum of each particle
[tex]p_{total} = p_a+p_b[/tex]
To find the module let's use the Pythagorean theorem.
[tex]p_{total} = m \sqrt{2^2 + 3.7^2}[/tex]
p = 4.2m
Let's use trigonometry for the direction.
tan θ = [tex]\frac{p_{fb}}{p_{fa}}[/tex]
θ = tan⁻¹ [tex]\frac{p_{fb}x}{p_{fa}}[/tex]
θ = tam⁻¹ [tex]\frac{2}{3.7}[/tex]
θ = 28.4º
The total kinetic energy is
[tex]K_{total}= K_1+K_2[/tex]
[tex]K_{total}[/tex] = ½ m [tex]v_a^2[/tex] + ½ m [tex]v_b^2[/tex]
[tex]K_{total}[/tex] = [tex]\frac{m}{2}[/tex] (3.7² + 2²)
[tex]K_{total}[/tex] = 8.845 m
In conclusion using momentum conservation we can find the results for the questions about the collision of the balls are:
A) The velocities after the collision [tex]v_a[/tex] = 3.7 i m/s and [tex]v_b[/tex] = 2 j m/s
B) The direction of ball A is: x-axis
C) The total momentum is: [tex]p_{total}[/tex] = 4.2m and the direction is θ = 28.4º
The total kinetic energy is: K_ {total} = 8.84m J
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illustrates an Atwood's machine. Let the masses of blocks A and B be 6.00 kg and 3.00 kg , respectively, the moment of inertia of the wheel about its axis be 0.220 kg⋅m2, and the radius of the wheel be 0.120 m. There is no slipping between the cord and the surface of the wheel.
Answer:
The magnitude of the linear acceleration of block A is 1.21 m/s² in downward.
Explanation:
Given that,
Mass of block A = 6.00 kg
Mass of block B = 3.00 kg
Moment of inertia = 0.220 kg.m²
Radius = 0.120 m
Suppose we need to find the the magnitude of the linear acceleration of block A
Let a is the acceleration of the blocks.
Let [tex]T_{a}[/tex] and [tex]T_{b}[/tex] are the tension in the A and B cord.
According to figure,
We need to calculate the magnitude of the linear acceleration of block A
Net force acting on block A,
[tex]F_{A}=m_{A}g-T_{A}[/tex]
[tex]m_{A} a=m_{A}g-T_{A}[/tex]
[tex]T_{A}=m_{A}g-m_{A}a[/tex]...(I)
Net force acting on block B,
[tex]F_{B}=T_{B}-m_{B}g[/tex]
[tex]m_{B}a=T_{B}-m_{B}g[/tex]
[tex]T_{B}=m_{B}a+m_{B}g[/tex]...(II)
Net torque acting on pulley
[tex]T_{net}=I\times\alpha[/tex]
[tex]T_{A}r-T_{B}r=I\times \dfrac{a}{r}[/tex]
[tex]T_{A}-T_{B}=I\times\dfrac{a}{r^2}[/tex]
[tex]m_{A}g-m_{A}a-(m_{B}g+m_{B}a)=I\times\dfrac{a}{r^2}[/tex]
[tex]g(m_{A}-m_{B})-a(m_{A}+m_{B})=I\times\dfrac{a}{r^2}[/tex]
[tex]g(m_{A}-m_{B})=I\times\dfrac{a}{r^2}+a(m_{A}+m_{B})[/tex]
[tex]g(m_{A}-m_{B})=a(\dfrac{I}{r^2}+(m_{A}+m_{B}))[/tex]
[tex]a=\dfrac{g(m_{A}-m_{B})}{(\dfrac{I}{r^2}+(m_{A}+m_{B}))}[/tex]
Put the value into the formula
[tex]a=\dfrac{9.8\times(6.00-3.00)}{\dfrac{0.220}{(0.120)^2}+(6.00+3.00)}[/tex]
[tex]a=1.21\ m/s^2[/tex]
Hence, The magnitude of the linear acceleration of block A is 1.21 m/s² in downward.
The potential difference V(t) between the parallel plates of a capacitor is instantaneously increasing at a rate of 107 V/s. What is the displacement current (in mA) between the plates if the separation of the plates is 1.06 cm and they have an area of 0.174 m2?
To solve this problem it is necessary to apply the related concepts to the scalar value of displacement current, which can be expressed in terms of electric flux as
[tex]I_d = \epsilon_0 \frac{d\Phi_E}{dt}[/tex]
Where,
[tex]\epsilon_0[/tex] = Permitibitty of free space constant
[tex]\Phi_E[/tex] = Magnetic flux
t = time
We know as well that the Flux can be expressed as
[tex]\Phi = EA[/tex]
Here
A= Cross-sectional area
E = Electric Potential in a Uniform electric field
At the same time the electric potential is expressed in terms of Voltage and distance, that is
[tex]E = \frac{V}{d}[/tex]
Using this equation we have then that
[tex]I_d = \epsilon_0 \frac{d\Phi_E}{dt}[/tex]
[tex]I_d = \epsilon_0 \frac{d(EA)}{dt}[/tex]
[tex]I_d = \epsilon_0*A (\frac{d(E)}{dt})[/tex]
[tex]I_d = \epsilon_0*A (\frac{d(V)}{dt*d})[/tex]
[tex]I_d = \frac{\epsilon_0*A}{d} (\frac{d(V)}{dt})[/tex]
According to our values we have that
[tex]\frac{dV}{dt} = 107V/s[/tex]
[tex]A = 0.174m^2[/tex]
[tex]d = 1.06*10^{-3}m[/tex]
[tex]\epsilon = 8.85418^{-12} m^{-3}kg^{-1}s^4A^2[/tex]
Replacing,
[tex]I_d = \frac{\epsilon_0*A}{d} (\frac{d(V)}{dt})[/tex]
[tex]I_d = \frac{(8.85418^{-12})*(0.174)}{1.06*10^{-3}} (107)[/tex]
[tex]I_d = 7.565*10^{-8}A[/tex]
Therefore the displacement current is [tex]7.565*10^{-5}mA[/tex]
The displacement current (in mA) between these plates is equal to [tex]1.56 \times 10^{-5}\;mA[/tex]
Given the following data:
Rate of increment = 107 V/s.Distance = 1.06 cm to m = [tex]1.06 \times 10^{-2}\;m[/tex]Area = 0.174 [tex]m^2[/tex].Scientific data:
Permittivity of free space = [tex]8.854 \times 10^{-12}[/tex]How to calculate the displacement current.Mathematically, the displacement current (in mA) between the plates in an electric field is given by this formula:
[tex]I_d=\frac{\epsilon _o A}{d} (\frac{d(v)}{dt} )\\\\[/tex]
Where:
[tex]\epsilon_o[/tex] is the permittivity of free space.d is the distance.A is the area.[tex]\frac{d(v)}{dt}[/tex] is the rate of change.Substituting the given parameters into the formula, we have;
[tex]I_d=\frac{8.854 \times 10^{-12} \times 0.174 \times 107}{1.06 \times 10^{-2}} \\\\I_d=1.56 \times 10^{-8}\\\\I_d=1.56 \times 10^{-5}\;mA[/tex]
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The y component of the electric field of an electromagnetic wave traveling in the +x direction through vacuum obeys the equation Ey = (375 N/C) cos [kx - (2.20 × 1014 rad/s)t]. What is the wavelength of this electromagnetic wave?
Answer: 8.6 µm
Explanation:
At a long distance from the source, the components (the electric and magnetic fields) of the electromagnetic waves, behave like plane waves, so the equation for the y component of the electric field obeys an equation like this one:
Ey =Emax cos (kx-ωt)
So, we can write the following equality:
ω= 2.2 1014 rad/sec
The angular frequency and the linear frequency are related as follows:
f = ω/ 2π= 2.2 1014 / 2π (rad/sec) / rad = 0.35 1014 1/sec
In an electromagnetic wave propagating through vacuum, the speed of the wave is just the speed of light, c.
The wavelength, speed and frequency, are related by this equation:
λ = c/f
λ = 3.108 m/s / 0.35. 1014 1/s = 8.6 µm.
The wavelength of the electromagnetic wave is 8.57 micrometers.
The wavelength of an electromagnetic wave can be calculated using the following equation:
λ = c / f
where:
λ is the wavelength
c is the speed of light in vacuum (3.0 × 10^8 m/s)
f is the frequency
The frequency of the electromagnetic wave can be calculated from the angular frequency ω using the following equation:
f = ω / 2π
where:
ω is the angular frequency (2.20 × 10^14 rad/s)
Plugging in the values for ω and c, we get:
f = (2.20 × 10^14 rad/s) / (2π)
f = 3.50 × 10^13 Hz
Now we can calculate the wavelength:
λ = (3.0 × 10^8 m/s) / (3.50 × 10^13 Hz)
λ = 8.57 × 10^-6 m
Therefore, the wavelength of the electromagnetic wave is 8.57 micrometers.
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A produce distributor uses 800 packing crates a month, which it purchases at a cost of $10 each. The manager has assigned an annual carrying cost of 35 percent of the purchase price per crate. Ordering costs are $28. Currently the manager orders once a month.
Answer:
$364.29
Explanation:
given,
Packing of crates per month (u)= 800
annual carrying cost of 35 percent of the purchase price per crate.
Ordering cost(S) = $ 28
D = 800 x 12 = 9600 crates/year
H = 0.35 P
H = 0.35 x $10
H = $3.50/crate per yr.
Present Total cost
= [tex]\dfrac{800}{2}\times 3.50 + \dfrac{9600}{800}\times 28[/tex]
= 1400 + 336
= $ 1,736
[tex]Q_0 = \sqrt{\dfrac{2DS}{H}}[/tex]
[tex]Q_0 = \sqrt{\dfrac{2\times 9600 \times 28}{3.50}}[/tex]
[tex]Q_0 =\$ 391.92[/tex]
Total cost at EOQ
= [tex]\dfrac{391.92}{2}\times 3.50 + \dfrac{9600}{391.92}\times 28[/tex]
= 685.86 + 685.85
= $ 1,371.71
the firm save annually in ordering and carrying costs by using the EOQ
= $ 1,736 - $ 1,371.71
= $364.29
The question focuses on Economic Order Quantity (EOQ) in inventory management for a produce distributor's packing crates. EOQ helps determine the optimal number of units a company should add to its inventory to minimize total holding and setup costs.
Explanation:The subject of this question is
Economic Order Quantity (EOQ)
, which is an operational efficiency measure used in inventory management. It refers to the number of units that a company should add to its inventory at a time to minimize total holding and setup costs.
In this case, the produce distributor uses 800 crates a month, purchased at $10 each. The annual carrying cost is 35% of the purchase price per crate, and the ordering cost is $28. Currently, the distributor orders once a month. Hence, to determine the optimal number of orders, or the EOQ, various factors such as ordering cost, holding cost, and demand rate need to be considered.
This cost minimization issue can be mathematically represented and solved using the EOQ formula: EOQ = √((2DS)/H) where D is the demand rate, S is the setup (or order) cost, and H is the holding (or carrying) cost.
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Two skiers travel two different frictionless paths. The two paths start at the same place and finish at the same place. Skier A is heavier than Skier B. With no more information, which of the skiers has the greater speed at the finish? Ignore air resistance.
A. Skier A has the same speed as Skier B at the finish.
B. Skier A has the greater speed at the finish.
C. Skier B has the greater speed at the finish.
D. There is no way to know which skier has the greater speed at the finish.
Answer:
D. There is no way to know which skier has the greater speed at the finish.
Explanation::
Total energy is conserved because there is no friction
(Et)₀ = (Et)f
(Et)₀ = Initial total energy (J)
(Et)f = Final total energy (J)
(Et)₀= K₀ + U₀
(Et)f= Kf + Uf
K₀ : Initial kinetic energy
U₀ : Initial potential energy
Kf : final kinetic energy
Uf : final potential energy
The formulas to calculate the kinetic energy (K) and potential energy (U) are:
K = ( 1/2)m*v²
U = m* g*h
m : mass (kg)
v: speed ( m/s)
h: hight ( m)
Problem development
Skier A
(Et)₀ = (Et)f
K₀ + U₀ = Kf + Uf
( 1/2)mA*(v₀A)² + mA*g*h₀ = ( 1/2)mA*(vfA)² + mA*g*hf ,
We divide by mA on both sides of the equation
( 1/2)*(v₀A)²+ g*h₀ = ( 1/2)(vfA)² + g*hf
( 1/2)*(v₀A)²+ g*h₀ - g*hf = ( 1/2)(vfA)²
We multiply by 2 both sides of the equation
(v₀A)²+2g(h₀ -hf) = (vfA)²
(vfA)² = (v₀A)²+2g(h₀ -hf) Equation (1)
Skier B
(Et)₀ = (Et)f
K₀ + U₀ = Kf + Uf
(1/2)mB*(v₀B)² + mB*g*h₀ = ( 1/2)mB*(vfB)² + mB*g*hf
We perform the same procedure above:
(vfB)² = (v₀B)²+2g(h₀ -hf) Equation (2)
Comparison of equation (1) with equation (2)
The term 2g (h₀ -hf) is the same in both equations because the paths of the two skiers start in the same place and end in the same place.
The final speed (vf) of skiers depends on their initial speed (v₀).
Because the initial speed of the skiers is unknown, it cannot be determined which has the highest final speed
Two kids create a makeshift seesaw by setting a 4-m long uniform plank on a saw horse. The saw horse is 0.5 m to the left of the center of mass of the plank. The child of mass m1 = 42 kg sits at the left end of the plank. The child of mass m2 = 31 kg sits 1 m to the right of the center of mass of the plank. What is the mass of the plank?
Answer:
The mass of the plank is 33 kg.
Explanation:
Given that,
Length of plank = 4 m
Distance in left from the center= 0.5 m
Mass of child = 42 kg
Mass of other child = 31 kg
Distance in right from the center= 1 m
We can sum torques around the saw horse; each force generates a torque equal to force x distance from saw horse
We need to calculate the torque for first child
Using formula of torque
[tex]\tau =mg\times d[/tex]
Put the value into the formula
[tex]\tau=42\times9.8\times1.5[/tex]
[tex]\tau=617.4\ N-m[/tex]
We need to calculate the torque for second child
The child is 1.5 m from the saw horse
Using formula of torque
[tex]\tau' =mg\times d[/tex]
Put the value into the formula
[tex]\tau'=31\times9.8\times1.5[/tex]
[tex]\tau'=455.7\ N-m[/tex]
The weight of the plank produces a clockwise torque
[tex]\tau''=mg\times d[/tex]
[tex]\tau''=mg\times 0.5[/tex]
We need to calculate the mass of the plank
Using balance equation of torque
[tex]\tau=\tau''+\tau'[/tex]
Put the value into the formula
[tex]617.4=mg\times0.5+455.7[/tex]
[tex]m=\dfrac{617.4-455.7}{0.5\times9.8}[/tex]
[tex]m=33\ kg[/tex]
Hence, The mass of the plank is 33 kg.
The mass of the plank is found to be 29 kg.
To solve for the mass of the plank, we need to use the principles of static equilibrium. The seesaw must be balanced, meaning the sum of moments around the pivot (saw horse) must be zero.
Given -
Length of the plank = 4 m
Pivot position (0.5 m to the left of the center) = 2-0.5 m from the left end = 1.5 m
Mass of the first child, m1 = 42 kg (at the left end, 1.5 m from pivot)
Mass of the second child, m2 = 31 kg (1 m to the right of the center, 1.5+1 m = 2.5 m from the pivot)
Equation of moments about the pivot (taking counterclockwise as positive):
Moment of m1 = 42 kg * 1.5 m (clockwise) = -42 * 1.5 Nm
Moment of m2 = 31 kg * 2.5 m (counterclockwise) = 31 * 2.5 Nm
Moment of the plank's weight (assuming it acts at its center, which is 0.5 m from the pivot):
Moment of plank = M * 0.5 m (clockwise) = -M * 0.5 Nm
Setting the sum of moments to zero for equilibrium:
31 * 2.5 - 42 * 1.5 - M * 0.5 = 0
Solving for M:
31 * 2.5 - 42 * 1.5 = M * 0.5
77.5 - 63 = 0.5M
14.5 = 0.5M
M = 14.5 / 0.5 = 29 kg
Therefore, the mass of the plank is 29 kg.
A solid uniform sphere and a thin-walled, hollow sphere have the same mass M and radius R. If they roll without slipping up a ramp that is inclined at an angle β above the horizontal and if both have the same vcm before they start up the incline, which object reaches the greater height, or do both of them reach the same height?
Answer:Hollow sphere
Explanation:
Given
same mass for solid and hollow sphere
same [tex]v_{cm}[/tex] before they start up incline
Moment of inertia of solid Sphere
[tex]I_1=\frac{2}{5}Mr^2[/tex]
Moment of inertia of hollow sphere
[tex]I_2=\frac{2}{3}Mr^2[/tex]
Conserving Energy at bottom and top point for solid sphere
kinetic energy +Rotational Energy=Potential energy
[tex]\frac{1}{2}Mv_{cm}^2+\frac{1}{2}I\omega ^2=mgh_1[/tex]
for pure rolling [tex]v_{cm}=\omega r[/tex]
[tex]\frac{1}{2}Mv_{cm}^2+\frac{1}{2}\times \frac{2}{5}Mr^2=Mgh_1[/tex]
[tex]\frac{7}{10}Mv_{cm}^2=Mgh_1[/tex]
[tex]h_1=\frac{7v_{cm}^2}{10g}[/tex]
For hollow sphere
[tex]\frac{1}{2}Mv_{cm}^2+\frac{1}{2}\times \frac{2}{3}Mr^2=Mgh_2[/tex]
[tex]h_2=\frac{5v_{cm}^2}{6g}[/tex]
therefore height gained by hollow sphere is more
The hollow sphere reached the greatest height.
Height reached by each object
The height reached by each object is determined by applying the principle of conservation of energy as show below;
K.E = P.E
¹/₂mv² + ¹/₂Iω² = mgh
where;
I moment of inertia ω is angular speed = v/rHeight reached by solid sphereI = ²/₅mr²
¹/₂mv² + ¹/₂Iω² = mgh
¹/₂mv² + ¹/₂(²/₅mr²)(v/r)² = mgh
¹/₂v² + ¹/₅v² = gh
7v² = 10gh
h = 7v²/10g
h = 0.7(v²/g)
Height reached by hollow sphereI = ²/₃mr²
¹/₂mv² + ¹/₂Iω² = mgh
¹/₂mv² + ¹/₂(²/₃mr²)(v/r)² = mgh
¹/₂v² + ¹/₃v² = gh
5v² = 6gh
h = 5v²/6g
h = 0.83(v²/g)
Thus, the hollow sphere reached the greatest height.
Learn more about conservation of energy here: https://brainly.com/question/166559
A 66-kg base runner begins his slide into second base when he is moving at a speed of 3.4 m/s. The coefficient of friction between his clothes and Earth is 0.70. He slides so that his speed is zero just as he reaches the base(a) How much mechanical energy is lost due to friction acting on the runner. (b) How far does he slide?
Answer:
a ) = 381.48 J
b )= 84.25 cm
Explanation:
Kinetic energy of the runner
= 1/2 m v²
= .5 x 66 x 3.4²
= 381.48 J
The final kinetic energy of the runner is zero .
Loss of mechanical energy
= 381.48 J
This loss in mechanical energy is due to action of frictional force .
b )
Let s be the distance of slide
deceleration due to frictional force
= μmg/m
.7 x 66 x 9.8 / 66
a = - 6.86 m s⁻¹
v² = u² - 2 a s
0 = 3.4² - 2x6.86 s
s = 3.4² / 2x6.86
= .8425 m
84.25 cm
A spring with an 8-kg mass is kept stretched 0.4 m beyond its natural length by a force of 32 N. The spring starts at its equilibrium position and is given an initial velocity of 4 m/s. Find the position of the mass at any time t. (Use x for the displacement in meters from the equilibrium position.)
Answer:
x = 1.26 sin 3.16 t
Explanation:
Assume that the general equation of the displacement given as
x = A sinω t
A=Amplitude ,t=time ,ω=natural frequency
We know that speed V
[tex]V=\dfrac{dx}{dt}[/tex]
V= A ω cosωt
Maximum velocity
V(max)= Aω
Given that F= 32 N
F = K Δ
K=Spring constant
Δ = 0.4 m
32 =0.4 K
K = 80 N/m
We know that ω²m = K
8 ω² = 80
ω = 3.16 s⁻¹
Given that V(max)= Aω = 4 m/s
3.16 A = 4
A= 1.26 m
Therefore the general equation of displacement
x = 1.26 sin 3.16 t
Answer:
Explanation:
mass, m = 8 kg
extension, y = 0.4 m
force, F = 32 N
maximum velocity, v = 4 m/s
maximum velocity , v = ω A
where, ω be the angular velocity and A be the amplitude
4 = ω x 0.4
ω = 10 rad/s
position
x = A Sin ωt
x = 0.4 Sin 10 t
One end of a thin rod is attached to a pivot, about which it can rotate without friction. Air resistance is absent. The rod has a length of 0.80 m and is uniform. It is hanging vertically straight downward. The end of the rod nearest the floor is given a linear speed v0, so that the rod begins to rotate upward about the pivot. What must be the value of v0, such that the rod comes to a momentary halt in a straight-up orientation, exactly opposite to its initial orientation?
Answer:
6.86 m/s
Explanation:
This problem can be solved by doing the total energy balance, i.e:
initial (KE + PE) = final (KE + PE). { KE = Kinetic Energy and PE = Potential Energy}
Since the rod comes to a halt at the topmost position, the KE final is 0. Therefore, all the KE initial is changed to PE, i.e, ΔKE = ΔPE.
Now, at the initial position (the rod hanging vertically down), the bottom-most end is given a velocity of v0. The initial angular velocity(ω) of the rod is given by ω = v/r , where v is the velocity of a particle on the rod and r is the distance of this particle from the axis.
Now, taking v = v0 and r = length of the rod(L), we get ω = v0/ 0.8 rad/s
The rotational KE of the rod is given by KE = 0.5Iω², where I is the moment of inertia of the rod about the axis of rotation and this is given by I = 1/3mL², where L is the length of the rod. Therefore, KE = 1/2ω²1/3mL² = 1/6ω²mL². Also, ω = v0/L, hence KE = 1/6m(v0)²
This KE is equal to the change in PE of the rod. Since the rod is uniform, the center of mass of the rod is at its center and is therefore at a distane of L/2 from the axis of rotation in the downward direction and at the final position, it is at a distance of L/2 in the upward direction. Hence ΔPE = mgL/2 + mgL/2 = mgL. (g = 9.8 m/s²)
Now, 1/6m(v0)² = mgL ⇒ v0 = [tex]\sqrt{6gL}[/tex]
Hence, v0 = 6.86 m/s
To find the value of v0 that will cause the rod to come to a momentary halt in a straight-up orientation, we need to consider conservation of energy and angular momentum. The equation (3/2)gh = 4(v0)^2 can then be used to solve for v0. The value of v0 is equal to the square root of (3/8)gh.
Explanation:To find the value of v0 that will cause the rod to come to a momentary halt in a straight-up orientation, we need to consider conservation of energy and angular momentum. When the rod is released, its potential energy is converted to kinetic energy and rotational kinetic energy. At the moment the rod comes to a halt, all of its initial kinetic energy will be converted back into potential energy. Since the rod is initially hanging vertically downward, we can equate the initial potential energy to the potential energy at the moment of momentary halt:
mgh = (1/2)Iω2
Where m is the mass of the rod, g is the acceleration due to gravity, h is the height of the rod, I is the moment of inertia of the rod about the pivot point, and ω is the angular velocity of the rod at the moment of momentary halt. The moment of inertia of a rod rotating about one end is given by I = (1/3)mL2, where L is the length of the rod. Therefore, we can rewrite the equation as:
mgh = (1/2)(1/3)mL2ω2
Simplifying the equation, we have:
(3/2)gh = ω2L2
To find ω, we need to relate it to the linear speed v0. Since the rod is rotating about a pivot point, the linear speed of a point on the rod is related to the angular velocity by the equation v = ωr, where r is the distance of the point from the pivot. In this case, the distance r is equal to half the length of the rod, so r = L/2. Substituting this into the equation, we have:
v0 = ω(L/2)
From this equation, we can solve for ω:
ω = (2v0)/L
Substituting this into the previous equation, we get:
(3/2)gh = ((2v0)/L)2L2
Simplifying further, we have:
(3/2)gh = 4(v0)2
Finally, solving for v0, we get:
v0 = √((3/8)gh)
A student standing on a cliff that is a vertical height d = 8.0 m above the level ground throws a stone with velocity v0 = 24 m/s at an angle θ = 21 ° below horizontal. The stone moves without air resistance; use a Cartesian coordinate system with the origin at the stone's initial position.
Part (a) With what speed, vf in meters per second, does the stone strike the ground 50%
Part (b) If the stone had been thrown from the clifftop with the same initial speed and the same angle, but above the horizontal would its impact velocity be different? YesNo Grade Summary 0% 100% Potential
Answer:
a) Vf = 27.13 m/s
b) It would have been the same
Explanation:
On the y-axis:
[tex]Y=-Vo*sin\theta*t-1/2*g*t^2[/tex]
[tex]-8=-24*sin(21)*t-1/2*10*t^2[/tex]
Solving for t:
t1 = 0.67s t2= -2.4s
Discarding the negative value and using the positive one to calculate the velocity:
[tex]Vf_y = -Vo*sin\theta-g*t[/tex]
[tex]Vf_y = -15.3m/s[/tex]
So, the module of the velocity will be:
[tex]Vf=\sqrt{(-15.3)^2+(24*cos(21))^2}[/tex]
[tex]Vf=27.13m/s[/tex]
If you throw it above horizontal, it would go up first, and when it reached the initial height, the velocity would be the same at the throwing instant. And starting then, the movement will be the same.
To determine the speed at which the stone strikes the ground, we can break the initial velocity into its horizontal and vertical components and calculate the time and horizontal distance traveled by the stone. By using the equations for vertical and horizontal motion and the initial conditions provided, we can calculate the speed at which the stone strikes the ground. If the stone had been thrown from a greater height above the horizontal, the impact velocity would be different as the horizontal component of the velocity would be greater.
Explanation:To determine the speed at which the stone strikes the ground, we need to analyze the motion of the stone both horizontally and vertically. Since the stone is thrown with an initial velocity of 24 m/s at an angle of 21° below the horizontal, we can break the initial velocity into its horizontal and vertical components. The horizontal component, Vx, can be calculated using Vx = v0 * cos(θ), where v0 is the initial velocity and θ is the angle. The vertical component, Vy, can be calculated using Vy = v0 * sin(θ).
In this case, Vx = 24 m/s * cos(21°) = 22.4 m/s and Vy = 24 m/s * sin(21°) = 8.7 m/s. The time it takes for the stone to hit the ground can be calculated using the equation y = y0 + Vy0t + 0.5gt^2, where y0 is the initial vertical position, Vy0 is the initial vertical velocity, g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time. In this case, y0 = 8.0 m, Vy0 = 8.7 m/s, and g = 9.8 m/s^2. We can solve for t using the quadratic equation.
Once we have the time, we can calculate the horizontal distance traveled by the stone using x = Vx0t. Since Vx0 is constant and equal to Vx, we can substitute Vx into the equation. In this case, x = 22.4 m/s * t.
Therefore, the speed vf at which the stone strikes the ground can be calculated using vf = √(Vx^2 + Vy^2). We can substitute the values we calculated for Vx and Vy to find the answer.
For part (b) of the question, if the stone had been thrown from the clifftop with the same initial speed and the same angle but above the horizontal, the impact velocity will be different. The vertical component of the velocity will be the same, as it only depends on the initial speed and angle of projection. However, the horizontal component will be different. The horizontal component of the velocity depends on the height of the cliff. When thrown from a higher height, the horizontal component will be greater, resulting in a higher impact velocity.
A long steel I-beam used in bridge construction has a length of 15.0 m. The beam may be exposed to temperatures as low as −23°C in the winter and as high as 32°C in the summer. What is the difference in length of the beam between these two temperature extremes? (Give the absolute value of the difference. Enter your answer in mm.)
To solve this problem it is necessary to apply the concepts of thermal expansion. Thermal expansion can be expressed in longitudinal terms such as
[tex]\Delta L = L_0 \alpha \Delta T[/tex]
Where,
[tex]\alpha =[/tex] Thermal Expanssion Coefficient
[tex]L_0 =[/tex] Initial Length
[tex]\Delta T =[/tex] Change in Temperature
Our values are given as
[tex]\alpha = 11*10^{-6}/\°C \rightarrow[/tex] from Steel
[tex]L_0 = 15m[/tex]
[tex]T_1 = -23\°C[/tex]
[tex]T_2 = 32\°C[/tex]
Replacing we have that,
[tex]\Delta L = L_0 \alpha \Delta T[/tex]
[tex]\Delta L = (17)(11*10^{-6})(32-(-21))[/tex]
[tex]\Delta L = 0.009911m[/tex]
[tex]\Delta L = 9.911mm[/tex]
Therefore the difference in length of the beam between these two temperature extremes is 9.911mm
A baseball is pitched horizontally toward home plate with a velocity of 110 km/h. In which of the following scenarios does the baseball have the largest change in momentum?
a. The baseball is hit straight back to the pitcher at a speed of 110 km/h.
b. The catcher catches the ball.
c. The ball is popped straight up at a speed of 110 km/h.
d. The catcher catches the ball, and the ball is popped straight up at a speed of 110 km/h have the same change in momentum.
Answer:
option A
Explanation:
given,
velocity to hit home plate = 110 Km/h
a) When the baseball is hit straight back
Assuming the momentum before hitting be P and after hitting the bal is also equal to P.
change in momentum = P - (-P) = 2P
b) When the catcher catches the ball the change in momentum is equal to zero.
c) when the ball is popped up the change in momentum
p_x = P and P_y = P
resultant momentum = [tex]\sqrt{P^2+ P^2} = \sqrt{2}\ P[/tex]
the maximum change in momentum will be in case A
The correct answer is option A
Final answer:
The scenario in which the baseball has the largest change in momentum is when it is hit straight back to the pitcher at a speed of 110 km/h.
Explanation:
The scenario in which the baseball has the largest change in momentum is when it is hit straight back to the pitcher at a speed of 110 km/h. This is because when the baseball is hit straight back, it completely reverses its direction of motion, resulting in a large change in momentum. The other scenarios, such as the baseball being caught by the catcher or popped straight up, do not involve a complete reversal of direction, so the change in momentum is smaller.
The magnetic field around the head has been measured to be approximately 3.00×10−8 gauss . Although the currents that cause this field are quite complicated, we can get a rough estimate of their size by modeling them as a single circular current loop 16.0 cm (the width of a typical head) in diameter.What is the current needed to produce such a field at the center of the loop? Answer in A.
Answer:
[tex]3.81972\times 10^{-7}\ A[/tex]
Explanation:
B = Magnetic field = [tex]3\times 10^{-8}\ G[/tex]
d = Diameter of loop = 16 cm
r = Radius = [tex]\frac{d}{2}=\frac{16}{2}=8\ cm[/tex]
i = Current
[tex]\mu_0[/tex] = Vacuum permeability = [tex]4\pi \times 10^{-7}\ H/m[/tex]
The magnetic field of a loop is given by
[tex]B=\frac{\mu_0i}{2r}\\\Rightarrow i=\frac{B2r}{\mu_0}\\\Rightarrow i=\frac{3\times 10^{-8}\times 10^{-4}\times 2\times 0.08}{4\pi\times 10^{-7}}\\\Rightarrow i=3.81972\times 10^{-7}\ A[/tex]
The current needed to produce such a field at the center of the loop is [tex]3.81972\times 10^{-7}\ A[/tex]
The Americium nucleus, 241 95 Am, decays to a Neptunium nucleus, 237 93 Np, by emitting an alpha particle of mass 4.00260 u and kinetic energy 5.5 MeV. Estimate the mass of the Neptunium nucleus, ignoring its recoil, given that the Americium mass is 241.05682 u
Answer:
The mass of Neptunium is 237.054 u.
Explanation:
Given that,
Mass of Americium = 241.05682 u
Mass of alpha particle = 4.00260 u
The equation is,
[tex]_{95}^{241}Am\rightarrow _{93}^{237}NP+\alpha\ particle+5.5 MeV[/tex]
Let the mass of Neptunium is m.
Since the mass remain same.
We need to calculate the mass of Neptunium
Using formula of mass
Mass of Neptunium = Mass of Americium -Mass of alpha particle
Put the value into the formula
[tex]m=241.05682 -4.00260[/tex]
[tex]m=237.054\ u[/tex]
Hence, The mass of Neptunium is 237.054 u.
The mass of the Neptunium nucleus is estimated to be 237.05422 u.
The Americium nucleus has a given mass of 241.05682 u and decays by emitting an alpha particle with a mass of 4.00260 u and kinetic energy of 5.5 MeV. To estimate the mass of the Neptunium nucleus, ignoring its recoil, we initially consider the conservation of mass-energy principle.
Mass of Neptunium = Mass of Americium -Mass of alpha particle
The mass of the Neptunium nucleus can be calculated simply by subtracting the mass of the alpha particle from the original Americium nucleus, which results in -:
m = 241.05682 u - 4.00260 u = 237.05422 u
A person with mass mp = 71 kg stands on a spinning platform disk with a radius of R = 1.62 m and mass md = 193 kg. The disk is initially spinning at ω = 1.8 rad/s. The person then walks 2/3 of the way toward the center of the disk (ending 0.54 m from the center). 1)What is the total moment of inertia of the system about the center of the disk when the person stands on the rim of the disk?
Answer:
439.587 kgm²
Explanation:
[tex]m_p[/tex] = Mass of person = 71 kg
R = Radius of platform = 1.62 m
[tex]m_d[/tex] = Mass of disc = 193 kg
The moment of inertia of the system is given by
[tex]I=m_pR^2+\frac{1}{2}m_dR^2\\\Rightarrow I=71\times 1.62^2+\frac{1}{2}\times 193\times 1.62^2\\\Rightarrow I=439.587\ kgm^2[/tex]
The total moment of inertia of the system about the center of the disk when the person stands on the rim of the disk is 439.587 kgm²
Suppose that the strings on a violin are stretched with the same tension and each has the same length between its two fixed ends. The musical notes and corresponding fundamental frequencies of two of these strings are G (196 Hz) and E (659.3 Hz). The linear density of the E string is 3.40 × 10-4 kg/m. What is the linear density of the G string?
Answer:
0.00384 kg/m
Explanation:
The fundamental frequency of string waves is given by
[tex]f=\frac{1}{2L}\sqrt{\frac{F}{\mu}}[/tex]
For some tension (F) and length (L)
[tex]f\propto\frac{1}{\mu}[/tex]
Fundamental frequency of G string
[tex]f_G=196\ Hz[/tex]
Fundamental frequency of E string
[tex]f_E=659.3\ Hz[/tex]
Linear mass density of E string is
[tex]\mu_E=3.4\times 10^{-4}\ kg/m[/tex]
So,
[tex]\frac{F_G}{F_E}=\sqrt{\frac{\mu_E}{\mu_G}}\\\Rightarrow \frac{F_G^2}{F_E^2}=\frac{\mu_E}{\mu_G}\\\Rightarrow \mu_G=3.4\times 10^{-4}\times \frac{659.3^2}{196^2}\\\Rightarrow \mu_G=0.00384\ kg/m[/tex]
The linear density of the G string is 0.00384 kg/m
A block with mass m = 0.250 kg is attached to one end of an ideal spring and moves on a horizontal frictionless surface. The other end of the spring is attached to a wall. When the block is at x = +0.240 m, its acceleration is ax = -13.0 m/s2 and its velocity is vx = +4.00 m/s.1. What is the spring's force constant k?2. What is the amplitude of the motion?3. What is the maximum speed of the block during its motion?4. What is the maximum magnitude of the block's acceleration during its motion?
Answer:
13.54 N/m
0.6 m
4.37 m/s
32.496 m/s²
Explanation:
m = Mass of block = 0.25 kg
k = Spring constant
A = Amplitude
x = Compression of spring = 0.24 m
a = Acceleration = -13 m/s²
v = Velocity = 4 m/s
The weight of the block and force on spring is equal
[tex]ma=-kx\\\Rightarrow k=-\frac{ma}{x}\\\Rightarrow k=-\frac{0.25\times -13}{0.24}\\\Rightarrow k=13.54\ N/m[/tex]
The spring's force constant is 13.54 N/m
Total energy of the system is given by
[tex]E=\frac{1}{2}(mv^2+kx^2)\\\Rightarrow E=\frac{1}{2}(0.25\times 4^2+13.54\times 0.24^2)\\\Rightarrow E=2.39\ J[/tex]
At maximum displacement v = 0
[tex]E=\frac{1}{2}(mv^2+kA^2)\\\Rightarrow E=\frac{1}{2}(0+kA^2)[/tex]
[tex]\\\Rightarrow E=\frac{1}{2}kA^2\\\Rightarrow A=\sqrt{\frac{E2}{k}}\\\Rightarrow A=\sqrt{\frac{2\times 2.39}{13.54}}\\\Rightarrow A=0.6\ m[/tex]
The amplitude of the motion is 0.6 m
Speed of the block
[tex]E=\frac{1}{2}mv_m^2+0\\\Rightarrow v_m=\sqrt{\frac{2E}{m}}\\\Rightarrow v_m=\sqrt{\frac{2\times 2.39}{0.25}}\\\Rightarrow v_m=4.37\ m/s[/tex]
The maximum speed of the block during its motion is 4.37 m/s
Forces in the spring
[tex]ma_m=kA\\\Rightarrow a_m=\frac{kA}{m}\\\Rightarrow a_m=\frac{13.54\times 0.6}{0.25}\\\Rightarrow a_m=32.496\ m/s^2[/tex]
Maximum magnitude of the block's acceleration during its motion is 32.496 m/s²
The force constant of the spring is 13.54 N/m. The amplitude of the oscillatory motion is 0.240 m. The maximum velocity and acceleration are 4.72 m/s and 13 m/s² respectively.
Explanation:First, we can calculate the spring's force constant k using Newton's second law (F = ma) and Hooke's Law (F = kx), by equating the forces F (Note that the negative acceleration is due to the force acting opposite to displacement). Hence, having the force due to spring F = m * a = 0.250 kg * -13.0 m/s² = -3.25 N and applying force due to spring as per Hooke's law as -k * x. Equating these forces gives k = 13.54 N/m.
Secondly, the amplitude A of the motion can be identified as |x| = 0.240 m, since for a simple harmonic oscillator, the amplitude is the maximum displacement from equilibrium (x = 0).
For the third part, the maximum speed Vmax during its motion can be calculated using the energy conservation principle, as given by the formula, Vmax = sqrt(k/m * A²), which gives Vmax = 4.72 m/s.
Finally, the maximum acceleration of the block Amax happens at the points of maximum displacement, i.e., at the amplitude points, which is Amax = k/m * A = 13m/s².
Learn more about Simple Harmonic Motion here:https://brainly.com/question/28208332
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A 259-kV power transmission line carrying 429 A is hung from grounded metal towers by ceramic insulators, each having a 0.71×109 Ω resistance. What is the resistance to ground of 95 of these insulators? Give your answer in MΩ. (Please note that the ceramic insulators are like resistors that are connected in parallel.)
Answer:
The resistance is [tex]7.47\times10^{6}\ \Omega[/tex].
Explanation:
Given that,
Power = 259 kV
Current = 429 A
Resistance [tex]R=0.71\times10^{9}\ Omega[/tex]
We need to calculate the current in each insulator
Using formula of current
[tex]I=\dfrac{P}{R}[/tex]
Put the value into the formula
[tex]I=\dfrac{259\times10^{3}}{0.71\times10^{9}}[/tex]
[tex]I=3.64\times10^{-4}\ A[/tex]
So all 95 insulators are in parallel
We need to calculate the resistance
Using formula of resistance
[tex]\dfrac{1}{R}=\sum_{i=1}^{95}\dfrac{1}{R_{i}}[/tex]
Put the value into the formula
[tex]\dfrac{1}{R}=\dfrac{95}{0.71\times10^{9}}[/tex]
[tex]\dfrac{1}{R}=1.338\times10^{-7}[/tex]
[tex]R=\dfrac{1}{1.338\times10^{-7}}[/tex]
[tex]R=7473841.5=7.47\times10^{6}\ \Omega[/tex]
Hence, The resistance is [tex]7.47\times10^{6}\ \Omega[/tex].
Water flows through a valve at the rate of 1000 lbm/s. The pressure just upstream of the valve is 90 psi and the pressure drop across the valve is 50 psi. The inside diameters of the valve inlet and outlet pipes are 12 and 24 in. The flow occurs in a horizontal plane. Determine the loss in available energy units across the valve.
Final answer:
The loss in available energy units across the valve is equal to the pressure drop, which is 344,738 N/m².
Explanation:
To determine the loss in available energy units across the valve, we need to calculate the change in kinetic energy and potential energy before and after the valve. The available energy is given by the equation:
Available energy = pressure drop + change in kinetic energy + change in potential energy
Given that the pressure drop across the valve is 50 psi, we can convert it to SI units (N/m²) by multiplying by 6894.76:
Pressure drop = 50 psi × 6894.76 N/m²/psi = 344,738 N/m²
The change in kinetic energy can be calculated using the following formulas:
Change in kinetic energy = (1/2) × mass flow rate × (outlet velocity² - inlet velocity²)
Change in potential energy = mass flow rate × g × (outlet height - inlet height)
Since the flow occurs in a horizontal plane, there is no change in potential energy. Therefore, the loss in available energy units across the valve is equal to the pressure drop:
Loss in available energy units = 344,738 N/m²
A grocery cart with a mass of 20 kg is pushed at constant speed along an aisle by a force F = 12 N. The applied force acts at a 20° angle to the horizontal.
Find the work done by each of the forces on the cart if the aisle is 18 m long.
Answer:
Explanation:
Given
mass of cart m=20 kg
Force applied F=12 N at angle of [tex]\theta =20^{\circ}[/tex] with horizontal
Force can be resolved in two components i.e. [tex]F\cos \theta [/tex]and [tex]F\sin \theta [/tex]
[tex]F\sin \theta [/tex]is will act Perpendicular to the motion of cart and [tex]F\cos \theta [/tex]will help to move the cart .
Since cart is moving with constant velocity therefore net force is zero on the cart as friction is opposing the [tex]F\cos \theta [/tex]
Work done by [tex]F\cos \theta [/tex]is
[tex]W_1=F\cos \theta \cdot L[/tex]
[tex]W_1=12\cos (20)\cdot 18=202.978 J[/tex]
Work done by Friction will be same but in Opposite Direction
[tex]W_f=-202.978 J[/tex]
Work done by [tex]F\sin \theta [/tex] is
[tex]W_2=0[/tex] because direction of force and motion is Perpendicular to each other.
A liquid of density 1230 kg/m3 flows steadily through a pipe of varying diameter and height. At Location 1 along the pipe, the flow speed is 9.39 m/s and the pipe diameter d1 is 11.5 cm. At Location 2, the pipe diameter d2 is 15.3 cm. At Location 1, the pipe is Δy=9.59 m higher than it is at Location 2. Ignoring viscosity, calculate the difference ΔP between the fluid pressure at Location 2 and the fluid pressure at Location 1.
To develop this problem it is necessary to use the continuity equations and Bernoullie's theorem.
It is known from Bernoullie's theorem that
[tex]P_1 + \rho gh_1+\frac{1}{2} \rho v_2^2 =P_2 + \rho gh_2+\frac{1}{2} \rho v_2^2[/tex]
Where
P = Pressure
g = Gravity
h= Height
v = Velocity
[tex]\rho[/tex] = Density
On the other hand we have that the continuity equation is given by
[tex]A_1v_1 = A_2 v_2[/tex]
Where A is the Cross-sectional area and v the velocity.
For our values we know that
[tex]A_1v_1 = A_2 v_2[/tex]
[tex](\frac{\pi d_1^2}{4})v_1 =(\frac{\pi d_2^2}{4})v_2[/tex]
[tex]d_1^2v_1=d_2^2v_2[/tex]
[tex](11.5cm)^2(9.39)=(15.3)^2v_2[/tex]
[tex]v_2 = 5.305m/s[/tex]
Using Bernoulli's expression we can now find the pressure difference,
[tex]P_1 + \rho gh_1+\frac{1}{2} \rho v_2^2 =P_2 + \rho gh_2+\frac{1}{2} \rho v_2^2[/tex]
[tex]P_1-P_2=-\rho gh_1-\frac{1}{2}\rho v_2^2 +\rho gh_2+\frac{1}{2} \rho v_2^2[/tex]
[tex]P_1-P_2 = \rho g (h_1-h_2)+\frac{1}{2}\rho(v_1^2-v_2^2)[/tex]
[tex]P_1-P_2 = (1.3*10^3)(9.8)(9.59)+\frac{1}{2}(1.3*10^3)((9.39)^2-(5.305)^2)[/tex]
[tex]P_1-P_2 = 1.612*10^5Pa[/tex]
A vertical spring has a length of 0.25 m when a 0.175 kg mass hangs from it, and a length of 0.775 m when a 2.075 kg mass hangs from it. gA. What is the force constant of the spring, in newtons per meter?
B. What is the unloaded length of the spring, in centimeters?
Answer:
A) 35.5N/m b) 20.1cm
Explanation:
Using Hooke's law;
F = Ke where F is the weight of the object = mass of the object in kg * acceleration due to gravity in m/s^2 and k if the force constant of the spring in N/m and e is the extension of the spring which original length of the spring - new length after extension in meters
For the first body, m*g = K * (0.25- li)
Where li is the initial length of the spring
0.175*9.81 = k(0.25-li)
1.72 = k(0.25-li) as equation 1
For the second body, m *g = K* ( 0.775-li)
2.075*9.81 = k (0.775-li) equation 2
20.36 = k(0.775-li)
Make li subject of the formula;
li = 0.775 - 20.36/k
Substitute for li in equation 1
1.72 = k(0.25- (0.775 - 20.36/k))
1.72 = k ( 0.25 - 0.775 + 20.36/k)
Open the bracket with k
1.72 = 0.25k - 0.775k + 20.36 (since k cancel k)
Collect the like terms:
1.72 - 20.36 = - 0.525k
- 18.64 = -0.525k
Divide both side by -0.525
-18.64/-0.525 = -0.525/-0.525k
K = 35.5N/m
B) substitute for k in using
li = 0.775 - 20.36/k
li = 0.775 - 20.36/35.5
li = 0.775 - 0.574
li = 0.201 in meters
li = 0.201 * 100 centimeters = 20.1cm
(A) The force constant of the spring is obtained as 34.66 N/m
(B) The unloaded length of the spring is 20 cm.
Hooke's Law(A) According to Hooke's law;
[tex]F = kx[/tex]
Where 'F' is the force needed to extend or compress the spring, 'k' is the force constant and 'x' is the extended length or compressed length.
Here, the gravitational force of the mass provides the necessary force for the spring to elongate.
Therefore, we can say that in the case of first mass;
[tex]k\,(0.25-l_0) = m_1g = (0.175\,kg \times 9.8\,m/s^2)[/tex]
Where [tex]l_0[/tex] is the unloaded length of the spring.
[tex]\implies k\,(0.25-l_0)=1.715\,N[/tex]
In the case of the second mass, we can write;
[tex]k\,(0.775\,m-l_0) = m_2g = (2.075\,kg \times 9.8\,m/s^2)[/tex]
[tex]\implies k\,(0.775\,m-l_0) = 20.335\,N[/tex]
From both these equations, we can write;
[tex]k\,(0.775\,m-l_0-0.25\,m+l_0)=20.335\,N- 1.715\,N[/tex]
[tex]\implies0.525\, k=18.2\,N[/tex]
[tex]\therefore k=\frac{18.2\,N}{0.525\,m}=34.66\,N/m[/tex]
(B) Applying the value of 'k' in any of the equations for force of given masses, we get;
[tex]k\,(0.25-l_0)=1.715\,N[/tex]
[tex]\implies (34.66\,N/m)\times\,(0.25-l_0)=1.715\,N[/tex]
[tex]\implies 8.665N-34.66\,l_0=1.715\,N[/tex]
[tex]\implies \,l_0=\frac{-6.95N}{-34.66} =0.2\,m=20\,cm[/tex]
Learn more about Hooke's law here:
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Arrange the following kinds of electromagnetic radiation in order of increasing wavelength: infrared, green light, red light, radio waves, X rays, ultraviolet light. Rank from shortest to longest. To rank items as equivalent, overlap them.
Answer:
X rays, ultraviolet light, green light, red light, infrared radio waves
Explanation:
The electromagnetic spectrum can be thought of as being arranged either in decreasing frequency or increasing wavelength. The full spectrum is as follows, from the shortest wavelength to the longest:
Gamma rays, X rays, ultraviolet light, violet light, indigo light, blue light, green light, yellow light, orange light, red light, infrared, radio waves, radar waves, microwaves, television waves, radio waves.
The electromagnetic radiation types can be arranged from shortest to longest wavelength as follows: X rays, ultraviolet light, green light, red light, infrared, and radio waves.
Explanation:The electromagnetic radiation types provided can be arranged in order of increasing wavelength as follows: X rays, ultraviolet light, green light, red light, infrared, radio waves. This order represents the electromagnetic spectrum from shortest to longest wavelengths. In simpler terms, X rays have the shortest wavelength, followed by ultraviolet light, green light, red light, infrared, and finally radio waves with the longest wavelength. Wavelength and frequency are inversely related - the shorter the wavelength, the higher the frequency.
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Find the maximum number of lines per centimeter a diffraction grating can have and produce a first-order maximum for the largest wavelength of visible light. (Assume the wavelengths of visible light range from 380 nm to 760 nm in a vacuum.)
To solve this problem it is necessary to apply the concepts related to constructive interference for multiple split.
The precaution is given by,
[tex]dsin\theta = m\lambda[/tex]
Where,
d = Distance between the slits
[tex]\theta =[/tex] Angle between the path and a line from the slits to the screen
m = Any integer, representing the number of repetition of the spectrum.
[tex]\lambda =[/tex]Wavelength
For first order equation we have that m = 1 then
[tex]d sin\theta = \lambda[/tex]
As the maximum number of lines corresponds to the smallest d values, we have that [tex]\theta = 90[/tex]
[tex]d sin90=\lambda[/tex]
[tex]d = 760nm[/tex]
Therefore the maximum numbers of lines per centimeter would be
[tex]N = \frac{10^{-2}m}{d}[/tex]
[tex]N = \frac{10^{-2}m}{760*10^{-9}m}[/tex]
[tex]N = 13157.89[/tex]
The maximum numbers of lines per centimeter is 13158