Answer:
magnitude : 0.6 m/s²
Direction : Left
Explanation:
m = mass of the cart = 90 kg
Taking force in right direction as positive and force in left direction as negative
F₁ = Force applied by man 1 = 140 N
F₂ = Force applied by other man = - 195 N
a = acceleration of the cart
Force equation for the motion of the cart is given as
F₁ + F₂ = ma
140 + (- 195) = 90 a
a = - 0.6 m/s²
magnitude of acceleration is 0.6 m/s²
The negative sign indicates the direction of acceleration towards left
Final answer:
The magnitude of the cart's acceleration is approximately 0.611 m/s², and the direction is to the left.
Explanation:
When calculating the acceleration of the cart, we need to consider the net force acting on it and its mass. The net force is found by subtracting the smaller force from the larger force, taking into account their directions. With forces of 140 N to the right and 195 N to the left, the net force is the difference, which is 195 N - 140 N = 55 N, directed to the left since the larger force is in that direction. Using Newton's second law, acceleration (a) is the net force (Fnet) divided by the mass (m).
To find the magnitude of the acceleration, calculate a = Fnet / m. So, a = 55 N / 90 kg which equals approximately 0.611 m/s2. Since the larger force was to the left, the direction of the acceleration is also to the left.
An air-track cart is attached to a spring and completes one oscillation every 5.67 s in simple harmonic motion. At time t = 0.00 s the cart is released at the position x = +0.250 m. What is the position of the cart when t = 29.6 s?
Answer:
0.2447 m
Explanation:
Amplitude, A = 0.25 m, T = 5.67 s, t = 29.6 s, y = ?
The general equation of SHM is given by
y = A Sin wt
y = 0.25 Sin (2 x 3.14 t /5.67)
Put t = 29.6 s
y = 0.25 Sin (2 x 3.14 x 29.6 / 5.67)
y = 0.2447 m
The sound intensity at the ear of passenger in a car with a damaged muffler is 8.0 × 10-3 W/m2. What is the intensity level of this sound in decibels? Use the threshold of hearing (1.0 × 10-12 W/m2) as the refere
Answer:
99 dB
Explanation:
We have given that sound intensity with a damaged muffler =8× [tex]10^{-3}[/tex]
we have to find the intensity level of the this sound in decibels
for calculating in decibels we have to use the formula
β=[tex]10log\frac{I}{I_0}[/tex]
=[tex]10log\frac{.008}{10^{-12}}[/tex]
=10 log8+10 log[tex]10^{9}[/tex]
=10 log8+90 log10
=10×0.9030+90
=99 dB
Final answer:
To calculate the intensity level of a sound in decibels, use the formula β = 10 log10(I/I0). Given the intensity of [tex]8.0 \times 10^{-3[/tex] threshold of hearing of [tex]10^{-12[/tex] the sound intensity level is approximately 99 decibels.
Explanation:
The student asks about calculating the intensity level of sound in decibels, given the sound intensity at the ear of a passenger in a car with a damaged muffler (8.0 × 10-3 W/m2). To find the intensity level in decibels (dB), we use the formula:
β = 10 log10(I/I0)
where:
β is the sound intensity level in decibels (dB)I is the sound intensity in watts per meter squared (W/m2)I0 = 10-12 W/m2 is the reference intensity, which is the threshold of hearingPlugging in the values:
β = 10 log10(8.0 × 10-3 / 10-12)
β = 10 log10(8.0 × 109)
β = 10 (log10(8) + log10(109))
β = 10 (0.903 + 9)
β = 10 × 9.903
β = 99.03 dB
The intensity level of the sound in the car with the damaged muffler is approximately 99 decibels.
An archer pulls her bowstring back 0.376 m by exerting a force that increases uniformly from zero to 251 N. (a) What is the equivalent spring constant of the bow? N/m (b) How much work does the archer do on the string in drawing the bow? J
Answer:
(A) 667.5 N/m
(B)
Explanation:
(A) Let the spring constant be k.
Using the formula F = kx
k = 251 / 0.376
K = 667.5 N/m
(B)
Work done
W = 0.5 × kx^2
W = 0.5 × 667.5 × 0.376 × 0.376
W = 47.2 J
Consider two copper wires. One has twice the length and twice the cross-sectional area of the other. How do the resistances of these two wires compare? A) Both wires have the same resistance. B) The shorter wire has twice the resistance of the longer wire. The longer wire has twice the resistance of the shorter wire. D) The longer wire has four times the resistance of the shorter wire. E) The shorter wire has four times the resistance of the longer wire.
Final answer:
The longer wire will have twice the resistance of the shorter wire.
Explanation:
The resistance of a wire depends on both its length and cross-sectional area. In this case, the longer wire has twice the length and twice the cross-sectional area of the shorter wire. The resistance of a wire is directly proportional to its length and inversely proportional to its cross-sectional area. Therefore, the longer wire will have twice the resistance of the shorter wire.
Velocity of a wave is the. (a) Wavelength x frequency (b) Wave number x frequency (c) Time period x phase (d) None
Answer:
option (a)
Explanation:
Wavelength is defined as the distance traveled by the wave in one complete oscillation.
The number of oscillations completed in one second is called frequency.
The relation for the wave velocity is given by
wave velocity = frequency x wavelength
You're driving your pickup truck around a curve that has a radius of 22 m.How fast can you drive around this curve before a steel toolbox slides on the steel bed of the truck?
The maximum speed at which the truck can briefly move around the curve without making the toolbox slide can be found by setting the sliding acceleration equal to the centripetal acceleration (v²/r) and solving for v. Factors like tire friction and flat terrain are also taken into account.
Explanation:To determine how fast the truck can go before the toolbox starts to slide, we need to calculate the maximum static friction (the force that prevents sliding).
The formula to identify the maximum speed that didn't result in sliding, we would use centripetal acceleration, which is the product of tangential speed squared divided by the curve's radius.
The acceleration of the crate on the truck bed must equal centripetal acceleration to prevent sliding. Given that the sliding acceleration is 2.06 m/s², setting this value to equal centripetal acceleration (v²/r where r is 22 m) and solving for v (velocity/speed), would provide the maximum speed at which the truck can move without the toolbox sliding.Tire friction also plays a role, as it allows vehicles to move at higher speeds without sliding off the road.
This also assumes that the road is relatively flat and the truck moves uniformly, similar to the situation illustrated in Figure 24.7 with a motorist moving in a straight direction.
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The speed at which a toolbox in a truck will start to slide when the truck is turning around a curve depends on the balance of the centripetal force and static friction force. These depend on the truck's speed, the curve's radius, the toolbox's mass and the friction coefficient. Without specific values, a numerical answer can't be given.
Explanation:The question is asking about the speed at which the toolbox in the truck would start sliding due to the forces acting upon it when the truck turns around a curve. This is related to centripetal force and frictional force.
When the truck turns, the toolbox experiences a centripetal force which pushes it towards the center of the curve. At the same time, friction between the toolbox and the truck bed is resisting this pushing force, keeping the box in place. At some point, if the truck is going too fast, the centripetal force will overcome friction, and the toolbox will start to slide.
We can use the formula for centripetal force, which is F = mv²/r, where m is the mass of the toolbox, v is the velocity of the truck and r is the radius of the curve. And we know that the maximum static friction force (F_max) is f_s * m * g, where f_s is the static friction coefficient and g is the acceleration due to gravity.
The toolbox starts to slide when F = F_max, so we can find the maximum speed (v_max) before sliding happens by making these two equations equal to each other and solving for v. Without the specific values for the mass of the toolbox and the static friction coefficient, we cannot calculate a numerical answer.
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Calculate the mass (in SI units) of (a) a 160 lb human being; (b) a 1.9 lb cockatoo. Calculate the weight (in English units) of (c) a 2300 kg rhinoceros; (d) a 22 g song sparrow.
Explanation:
1. Force applied on an object is given by :
F = W = mg
(a) A 160 lb human being, F = 160 lb
g = acceleration due to gravity, g = 32 ft/s²
[tex]m=\dfrac{F}{g}[/tex]
[tex]m=\dfrac{160\ lb}{32\ ft/s^2}[/tex]
m = 5 kg
(b) A 1.9 lb cockatoo, F = 1.9 lb
[tex]m=\dfrac{F}{g}[/tex]
[tex]m=\dfrac{1.9\ lb}{32\ ft/s^2}[/tex]
m = 0.059 kg
2. (a) A 2300 kg rhinoceros, m = 2300 kg
[tex]W=2300\ kg\times 32\ ft/s^2=73600\ lb[/tex]
(b) A 22 g song sparrow, m = 22 g = 0.022 kg
[tex]W=0.022\ kg\times 32\ ft/s^2=0.704\ lb[/tex]
Hence, this is the required solution.
Final answer:
The mass of a 160 lb human being is 72.576 kg, and the mass of a 1.9 lb cockatoo is 0.86184 kg. A 2300 kg rhinoceros weighs 5060 lbs in English units, while a 22 g song sparrow weighs 0.0484 lbs.
Explanation:
To calculate the mass (in SI units) of a 160 lb human being, we use the conversion factor 1 lb = 0.4536 kg. So, the calculation is:160 lb x 0.4536 kg/lb = 72.576 kg
(a) The mass of a 160 lb human being in SI units is 72.576 kg.
For a 1.9 lb cockatoo, the conversion to kilograms is:1.9 lb x 0.4536 kg/lb
= 0.86184 kg
(b) The mass of a 1.9 lb cockatoo in SI units is 0.86184 kg
To convert the mass of a 2300 kg rhinoceros to weight in English units, knowing that 1 kg = 2.2 lbs (where weight in pounds is considered the gravitational force on the mass), the weight is calculated as:
2300 kg x 2.2 lbs/kg= 5060 lbs
(c) The weight of a 2300 kg rhinoceros in English units is 5060 lbs.
Finally, for converting a 22 g song sparrow to weight in English units:
22 g x (1 kg/1000 g) x (2.2 lbs/kg)= 0.0484 lbs
(d) The weight of a 22 g song sparrow in English units is 0.0484 lbs.
Two infinite parallel surfaces carry uniform charge densities of 0.20 nC/m2 and -0.60 nC/m2. What is the magnitude of the electric field at a point between the two surfaces?
Answer:
Electric field, E = 45.19 N/C
Explanation:
It is given that,
Surface charge density of first surface, [tex]\sigma_1=0.2\ nC/m^2=0.2\times 10^{-9}\ C/m^2[/tex]
Surface charge density of second surface, [tex]\sigma=-0.6\ nC/m^2=-0.6\times 10^{-9}\ C/m^2[/tex]
The electric field at a point between the two surfaces is given by :
[tex]E=\dfrac{\sigma}{2\epsilon_o}[/tex]
[tex]E=\dfrac{\sigma1-\sigma_2}{2\epsilon_o}[/tex]
[tex]E=\dfrac{0.2\times 10^{-9}-(-0.6\times 10^{-9})}{2\times 8.85\times 10^{-12}}[/tex]
E = 45.19 N/C
So, the magnitude of the electric field at a point between the two surfaces is 45.19 N/C. Hence, this is the required solution.
A circular coil of wire of radius 8 cm and 19 turns is placed with its plane perpendicular to a 5.6 T magnetic field, produced by an electromagnet. The current to the electromagnet is then switched off, causing the magnetic field to collapse to zero in a time of 744 ms. What is the emf generated in the coil while the magnetic field is collapsing, in units of volts?
Answer:
2.87 Volt
Explanation:
N = 19
change in magnetic field = 5.6 T
Time, dt = 744 ms = 0.744 s
Induced emf,
e = N × change in flux / time
e = 19 × 3.14 × 0.08 × 0.08 ×5.6/0.744
e = 2.87 Volt
A large in-falling fragment could be tracked using radar. Explain how distance, speed, and the direction of motion, of the fragment could be determined. (15 points)
Answer:
speed, distance and direction of motion of the object can be determined by analyzing the radio wave.
Explanation:
We know that radar operates by transmitting radio waves to a destination and these waves are comes back to the receiver station. By Considering this transmission and receiver process, we can measure the distance, velocity and path of an object's movement.
Distance can be assessed by taking following consideration, the velocity of the waves is V. It can help to assess the time made for the waves to be emitted by the radar and felt by the receiver, let the time be t.
Therefore distance can be determine as D= v*t/2,
here 2 signifies that the distance travelled by the wave in either direction ( from transmitter to receiver and vice verse)
Using the source wave frequency, speed can be computed. In a specific frequency, the radar starts sending out the frequencies and the reflected wave will have a distinct frequency. The velocity can be determine by
[tex]v= (\Delta f/f)(c/2),[/tex]
where[tex]\Delta f[/tex]is the change in frequency and
c is the speed of light (the wave).
Direction can be determine by applying above principle. change in frequency is used to determine the direction in the following way:
When the frequency transition is very low, the object moves away from the radar and vice verse.
In a head-on collision, an alpha particle (Z = 2) of energy 8.80 MeV bounces straight back from a nucleus of charge 82.0 e. How close were the centers of the objects at closest approach?
The close were the centers of the objects at closest approach 2.7 x 10^-14 m
E =8.8 MeV = 8.8 x 1.6 x 10^-13 J
q = 2 e = 2 x 1.6 x 10^-19 C
Q = 82 e = 82 x 1.6 x 10^-19 C
Let d be the distance of closest approach
E = k Q q / d
Where, K = 9 x 10^9 Nm^2 / C^2
d = k Q q / E
d = (9 x 10^9 x 82 x 1.6 x 10^-19 x 2 x 1.6 x 10^-19) / (8.8 x 1.6 x 0^-13)
d = 2.7 x 10^-14 m
In jumping to block a shot, a volleyball player with a weight of 600 N generates an average vertical ground reaction force of 900 N for 0.37 seconds. What is the net vertical impulse that causes her velocity to increase in the upward direction?
Answer:
111 N
Explanation:
weight (downwards) = 600 N
reaction force (upwards) = 900 N
t = 0.37 s
Net force
F = reaction force - weight = 900 - 600 = 300 N
Impulse = force x time = 300 x 0.37 = 111 N
A capacitor has a charge of 4.6 μC. An E-field of 1.8 kV/mm is desired between the plates. There's no dielectric. What must be the area of each plate?
Answer:
[tex]A = 0.2875 m^2[/tex]
Explanation:
As we know that
[tex]Q = 4.6 \mu C[/tex]
E = 1.8 kV/mm
now we know that electric field between the plated of capacitor is given as
[tex]E = \frac{\sigma}{\epsilon_0}[/tex]
now we will have
[tex]1.8 \times 10^6 = \frac{\sigma}{\epsilon_0}[/tex]
[tex]\sigma = (1.8 \times 10^6)(8.85 \times 10^{-12})[/tex]
[tex]\sigma = 1.6 \times 10^{-5} C/m^2[/tex]
now we have
[tex]\sigma = \frac{Q}{A}[/tex]
now we have area of the plates of capacitor
[tex]A = \frac{Q}{\sigma}[/tex]
[tex]A = \frac{4.6 \times 10^{-6}}{1.6 \times 10^{-5}}[/tex]
[tex]A = 0.2875 m^2[/tex]
A 2.4-m-diameter merry-go-round with a mass of 270 kg is spinning at 20 rpm. John runs around the merry-go-round at 5.0 m/s, in the same direction that it is turning, and jumps onto the outer edge. John’s mass is 34 kg . Part A Part complete What is the merry-go-round's angular speed, in rpm, after John jumps on?
Answer:
23.98 rpm
Explanation:
d = diameter of merry-go-round = 2.4 m
r = radius of merry-go-round = (0.5) d = (0.5) (2.4) = 1.2 m
m = mass of merry-go-round = 270 kg
I = moment of inertia of merry-go-round
Moment of inertia of merry-go-round is given as
I = (0.5) m r² = (0.5) (270) (1.2)² = 194.4 kgm²
M = mass of john = 34 kg
Moment of inertia of merry-go-round and john together after jump is given as
I' = (0.5) m r² + M r² = 194.4 + (34) (1.2)² = 243.36 kgm²
w = final angular speed
w₀ = initial angular speed of merry-go-round = 20 rpm = 2.093 rad/s
v = speed of john before jump
using conservation of angular momentum
Mvr + I w₀ = I' w
(34) (5) (1.2) + (194.4) (2.093) = (243.36) w
w = 2.51 rad/s
w = 23.98 rpm
When two resistors are wired in series with a 12 V battery, the current through the battery is 0.33 A. When they are wired in parallel with the same battery, the current is 1.60 A. Part A What are the values of the two resistors?
Answer:
If R₂=25.78 ohm, then R₁=10.58 ohm
If R₂=10.57 then R₁=25.79 ohm
Explanation:
R₁ = Resistance of first resistor
R₂ = Resistance of second resistor
V = Voltage of battery = 12 V
I = Current = 0.33 A (series)
I = Current = 1.6 A (parallel)
In series
[tex]\text{Equivalent resistance}=R_{eq}=R_1+R_2\\\text {From Ohm's law}\\V=IR_{eq}\\\Rightarrow R_{eq}=\frac{12}{0.33}\\\Rightarrow R_1+R_2=36.36\\ Also\ R_1=36.36-R_2[/tex]
In parallel
[tex]\text{Equivalent resistance}=\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}\\\Rightarrow {R_{eq}=\frac{R_1R_2}{R_1+R_2}[/tex]
[tex]\text {From Ohm's law}\\V=IR_{eq}\\\Rightarrow R_{eq}=\frac{12}{1.6}\\\Rightarrow \frac{R_1R_2}{R_1+R_2}=7.5\\\Rightarrow \frac{R_1R_2}{36.36}=7.5\\\Rightarrow R_1R_2=272.72\\\Rightarrow(36.36-R_2)R_2=272.72\\\Rightarrow R_2^2-36.36R_2+272.72=0[/tex]
Solving the above quadratic equation
[tex]\Rightarrow R_2=\frac{36.36\pm \sqrt{36.36^2-4\times 272.72}}{2}[/tex]
[tex]\Rightarrow R_2=25.78\ or\ 10.57\\ If\ R_2=25.78\ then\ R_1=36.36-25.78=10.58\ \Omega\\ If\ R_2=10.57\ then\ R_1=36.36-10.57=25.79\Omega[/tex]
∴ If R₂=25.78 ohm, then R₁=10.58 ohm
If R₂=10.57 then R₁=25.79 ohm
The two resistors in question, when wired in a series, have a combined resistance of about 36.36 Ohms. When wired in parallel, they share a resistance of roughly 7.5 Ohms. The two resistor values would then most likely be around 28.86 Ohms and 7.5 Ohms.
Explanation:The nature of your question indicates a focus on the properties and calculations associated with electrical resistors in a circuit. These resistors can either be configured in a series or parallel connection, which will drastically change their behavior and derived readings.
From the context of your question, it is evident that a 12V battery is being used in conjunction with two resistors. In a series connection, the sum of the voltages across each resistor will equal the total voltage, effectively segmenting the 12V battery's output. In a parallel connection, each resistor would experience the full 12V impact, leading to larger current readings. Your values indicate a 0.33A current in a series scenario and a 1.60A current in a parallel situation.
Knowing this, we can apply Ohm's Law, which states Voltage (V) equals Current (I) times Resistance (R). In the series connection, the total resistance can be calculated as 12V divided by 0.33A equals approximately 36.36 Ohms. For the parallel connection, the total resistance would be 12V divided by 1.60A equals approximately 7.5 Ohms. Subtracting these two values, we can find that one of the resistors is around 28.86 Ohms while the other is roughly 7.5 Ohms. This would satisfy both the series and parallel conditions outlined in your question.
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A ball having a mass of 200 g is released from rest at a height of 400 mm above a very large fixed metal surface. If the ball rebounds to a height of 325 mm above the surface, determine the coefficient of restitution between the ball and the surface.
Answer:
0.9
Explanation:
h = 400 mm, h' = 325 mm
Let the coefficient of restitution be e.
h' = e^2 x h
325 = e^2 x 400
e^2 = 0.8125
e = 0.9
The coefficient of restitution between the ball and the surface is 0.9.
The coefficient of restitution (e) between the ball and the surface can be determined using the formula:
[tex]\[ e = \sqrt{\frac{\text{height after collision}}{\text{height before collision}}} \][/tex]
First, we need to convert the heights from millimeters to meters for consistency, since the standard units for height in physics are meters.
The initial height (h_i) before the collision is 400 mm, which is equivalent to 0.4 m (since 1 m = 1000 mm).
The final height (h_f) after the collision is 325 mm, which is equivalent to 0.325 m.
Now, we can plug these values into the formula for the coefficient of restitution:
[tex]\[ e = \sqrt{\frac{0.325 \text{ m}}{0.4 \text{ m}}} \][/tex]
[tex]\[ e = \sqrt{\frac{325}{400}} \][/tex]
[tex]\[ e = \sqrt{0.8125} \][/tex]
[tex]\[ e = 0.9 \][/tex]
Therefore, the coefficient of restitution between the ball and the surface is 0.9. This means that the collision is relatively elastic, with the ball retaining a significant portion of its initial kinetic energy after the rebound.
A square loop of wire consisting of a single turn is perpendicular to a uniform magnetic field. The square loop is then re-formed into a circular loop, which also consists of a single turn and is also perpendicular to the same magnetic field. The magnetic flux that passes through the square loop is 4.5 x 10 -3 Wb. What is the flux that passes through the circular loop?
a) How fast must a meter stick be moving if its length is observed to shrink to 0.6 m? b) With what speed will a clock have to be moving in order to run at a rate that is one-half the rate of a clock at rest?
Answer:
A)
0.8 c
B)
0.87 c
Explanation:
A)
L₀ = Original length of the meter stick = 1 m
L = Length observed = 0.6 m
[tex]v[/tex] = speed of the meter stick
Using the equation
[tex]L = L_{o} \sqrt{1 - \left ( \frac{v}{c} \right )^{2}}[/tex]
[tex]0.6 = 1 \sqrt{1 - \left ( \frac{v}{c} \right )^{2}}[/tex]
[tex]0.36 = 1 - \left ( \frac{v}{c} \right )^{2}[/tex]
[tex]v[/tex] = 0.8 c
B)
T₀ = Time of the clock at rest = t
T = Time of the clock at motion = (0.5) t
[tex]v[/tex] = speed of the clock
Using the equation
[tex]T = T_{o} \sqrt{1 - \left ( \frac{v}{c} \right )^{2}}[/tex]
[tex]0.5 t = t \sqrt{1 - \left ( \frac{v}{c} \right )^{2}}[/tex]
[tex]0.5 = \sqrt{1 - \left ( \frac{v}{c} \right )^{2}}[/tex]
v = 0.87 c
A batter hits a baseball into the air. The formula y=-16x^2+64x+5models the baseball’s height above the ground, y, in feet, x seconds after it is hit.When does the baseball reach its maximum height? What is that height?
Explanation:
A batter hits a baseball into the air. The formula that models the baseball’s height above the ground, y, in feet, x seconds after it is hit is given by :
[tex]y=-16x^2+64x+5[/tex].........(1)
(a) We need to find the maximum height reached by the baseball. The maximum height reached is calculated by differentiating equation (1) wrt x as :
[tex]\dfrac{dy}{dx}=0[/tex]
[tex]\dfrac{d(-16x^2+64x+5)}{dx}=0[/tex]
[tex]-32x+64=0[/tex]
x = 2 seconds
(b) The height of the ball is given by equation (1) and putting the value of x in it as :
[tex]y=-16(2)^2+64(2)+5[/tex]
y = 69 feets
So, at 2 seconds the baseball reaches its maximum height and its maximum height is 69 feets.
Answer:
wow
Explanation:
1) A fan is to accelerate quiescent air to a velocity of 8 m/s at a rate of 9 m3/s. Determine the minimum power that must be supplied to the fan. Take the density of air to be 1.18 kg/m3.
Answer:
[tex]\dot{W} = 339.84 W[/tex]
Explanation:
given data:
flow Q = 9 m^{3}/s
velocity = 8 m/s
density of air = 1.18 kg/m^{3}
minimum power required to supplied to the fan is equal to the POWER POTENTIAL of the kinetic energy and it is given as
[tex]\dot{W} =\dot{m}\frac{V^{2}}{2}[/tex]
here [tex]\dot{m}[/tex]is mass flow rate and given as
[tex]\dot{m} = \rho*Q[/tex]
[tex]\dot{W} =\rho*Q\frac{V^{2}}{2}[/tex]
Putting all value to get minimum power
[tex]\dot{W} =1.18*9*\frac{8^{2}}{2}[/tex]
[tex]\dot{W} = 339.84 W[/tex]
The minimum power required to accelerate air to a velocity of 8 m/s at a rate of 9 m^3/s with a density of 1.18 kg/m^3 is roughly 339.84 Watts.
Explanation:
The subject of this question is physics, specifically the topic of power in mechanical systems. To find the minimum power required, we use the formula Power = 0.5 * density * volume flow rate * velocity^2. Plugging in the given values: Power = 0.5 * 1.18 kg/m^3 * 9 m^3/s * (8 m/s)^2, we get approximately 339.84 W. Therefore, the minimum power that must be supplied to the fan is roughly 339.84 Watts.
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A 61 kg skier starts from rest at the top of a 1200 m long trail which drops a total of 227 m from top to bottom. At the bottom, the skier is moving 11 m/s. How much energy was dissipated by friction?
Answer:
energy dissipated = 132 kJ
Explanation:
mass = 61 kg
height drop = 227 m
velocity = 11 m/s
potential energy due to height drop from top to bottom
P.E. = m g h
P.E. = 61× 9.8× 227
P.E. = 135,700 J
kinetic energy = [tex]\frac{1}{2}mv^2[/tex]
= [tex]\frac{1}{2}\times 61 \times 11^2[/tex]
= 3690.5 J
energy dissipated = P.E - K.E.
= 135,700 J -3690.5 J
=132,009.5 J = 132 kJ
A steel wire of length 4.7 m and cross section 3 x 103 m2 stretches by the same amount as a copper wire of length 3.5 m and cross section 4 x 10-5 m2 under a given load. What is the ratio of the Young's modulus of steel to that of copper? (a) 3.83 x 103 (b) 1.46 x 10-2 (d) 5.85 x 10-3 (c) 1.79 x 10-2 2.
Answer:
The ratio of the young's modulus of steel and copper is [tex]1.79\times10^{-2}[/tex]
(c) is correct option.
Explanation:
Given that,
Length of steel wire = 4.7 m
Cross section[tex]A = 3\times10^{-3}\ m^2[/tex]
Length of copper wire = 3.5 m
Cross section[tex]A = 4\times10^{-5}\ m^2[/tex]
We need to calculate the ratio of young's modulus of steel and copper
Using formula of young's modulus for steel wire
[tex]Y=\dfrac{\dfrac{F}{A}}{\dfrac{\Delta l}{l}}[/tex]
[tex]Y_{s}=\dfrac{Fl_{s}}{A_{s}\Delta l}[/tex]....(I)
The young's modulus for copper wire
[tex]Y_{c}=\dfrac{Fl_{c}}{A_{c}\Delta l}[/tex]....(II)
From equation (I) and (II)
The ratio of the young's modulus of steel and copper
[tex]\dfrac{Y_{s}}{Y_{c}}=\dfrac{\dfrac{Fl_{s}}{A_{s}\Delta l}}{\dfrac{Fl_{c}}{A_{c}\Delta l}}[/tex]
[tex]\dfrac{Y_{s}}{Y_{c}}=\dfrac{A_{c}\times l_{s}}{A_{s}\times l_{c}}[/tex]
[tex]\dfrac{Y_{s}}{Y_{c}}=\dfrac{4\times10^{-5}\times4.7}{3\times10^{-3}\times3.5}[/tex]
[tex]\dfrac{Y_{s}}{Y_{c}}=1.79\times10^{-2}[/tex]
Hence, The ratio of the young's modulus of steel and copper is [tex]1.79\times10^{-2}[/tex]
An electron enters a magnetic field of 0.43 T with a velocity perpendicular to the direction of the field. At what frequency does the electron traverse a circular path? ( m el = 9.11 × 10-31 kg, e = 1.60 × 10-19 C)
Answer:
Frequency, [tex]f=1.2\times 10^{10}\ Hz[/tex]
Explanation:
It is given that,
Magnetic field, B = 0.43 T
We need to find the frequency the electron traverse a circular path. It is also known as cyclotron frequency. It is given by :
[tex]f=\dfrac{qB}{2\pi m}[/tex]
[tex]f=\dfrac{1.6\times 10^{-19}\ C\times 0.43\ T}{2\pi \times 9.11\times 10^{-31}\ Kg}[/tex]
[tex]f=1.2\times 10^{10}\ Hz[/tex]
Hence, this is the required solution.
Nitroglycerin flows through a pipe of diameter 3.0 cm at 2.0 m/s. If the diameter narrows to 0.5 cm, what will the velocity be?
Answer:
72 m/s
Explanation:
D1 = 3 cm, v1 = 2 m/s
D2 = 0.5 cm,
Let the velocity at narrow end be v2.
By use of equation of continuity
A1 v1 = A2 v2
3.14 × 3 × 3 × 2 = 3.14 × 0.5 ×0.5 × v2
v2 = 72 m/s
A 1210 W radiant heater is constructed to operate at 115 V. (a) What is the current in the heater when the unit is operating? (b) What is the resistance of the heating coil? (c) How much thermal energy is produced in 4.07 h?
Answer:
(a) 10.52 A
(b) 11 ohm
(c) 1.77 x 10^7 J
Explanation:
P = 1210 W, V = 115 V,
Let i be the current.
(a) Use the formula of power
P = V i
1210 = 115 x i
i = 10.52 A
(b) Let the resistance of heating coil is R.
Use the formula given below
V = i x R
R = V / i = 115 / 10.52 = 11 ohm
(c) Use the formula for energy
E = V x i x t
E = 115 x 10.52 x 4.07 x 60 x 60 = 1.77 x 10^7 J
What is the wavelength of an electron that has a kinetic energy of 0.50 MeV (relativistic)?
Answer:
The wavelength of electron is [tex]6.99\times 10^{-22}\ m[/tex]
Explanation:
The kinetic energy of the electron is, [tex]E=0.5\ MeV=0.5\times 10^6\ eV[/tex]
We need to find the wavelength of this electron. It can be calculated using the concept of DE-broglie wavelength as :
[tex]\lambda=\dfrac{h}{\sqrt{2mE} }[/tex]
h is Plank's constant
m is the mass of electron
[tex]\lambda=\dfrac{6.67\times 10^{-34}\ J-s}{\sqrt{2\times 9.1\times 10^{-31}\ kg\times 0.5\times 10^6\ eV} }[/tex]
[tex]\lambda=6.99\times 10^{-22}\ m[/tex]
So, the wavelength of electron is [tex]6.99\times 10^{-22}\ m[/tex]. Hence, this is the required solution.
The wavelength of an electron with a kinetic energy of 0.50 MeV (relativistic) is approximately 7.28 x 10^-12 m.
Explanation:The wavelength of an electron with a kinetic energy of 0.50 MeV can be calculated using the relativistic de Broglie equation:
λ = h/(m*c)
Where λ is the wavelength, h is Planck's constant (6.63 x 10^-34 Js), m is the mass of the electron (9.11 x 10^-31 kg), and c is the speed of light (3.00 x 10^8 m/s).
Substituting the values:
λ = (6.63 x 10^-34 Js)/((9.11 x 10^-31 kg)*(3.00 x 10^8 m/s))
λ ≈ 7.28 x 10^-12 m
Therefore, the wavelength of the electron is approximately 7.28 x 10^-12 m.
Four objects are situated along the y axis as follows: a 2.00-kg object is at 13.00 m, a 3.00-kg object is at 12.50 m, a 2.50-kg object is at the origin, and a 4.00-kg object is at 20.500 m. Where is the center of mass of these objects?
The center of mass of these objects is located at approximately 12.652 meters on the y-axis.
We can find the center of mass (center of gravity) of these objects by calculating the weighted average of their positions along the y-axis. Here's how:
Consider Each Object:
We have four objects with masses m1, m2, m3, and m4 at positions y1, y2, y3, and y4 respectively.
Center of Mass Formula:
The center of mass coordinate (y_cm) is calculated as:
y_cm = (Σ(mi * yi)) / Σ(mi)
Σ (sigma) represents summation over all objects (i = 1 to 4 in this case).
mi is the mass of the i-th object.
yi is the y-axis position of the i-th object.
Apply the formula to our case:
y_cm = [(2.00 kg * 13.00 m) + (3.00 kg * 12.50 m) + (2.50 kg * 0.00 m) + (4.00 kg * 20.500 m)] / (2.00 kg + 3.00 kg + 2.50 kg + 4.00 kg)
Calculate:
y_cm = [26.00 kgm + 37.50 kgm + 0.00 kgm + 82.00 kgm] / 11.50 kg
y_cm ≈ 145.50 kg*m / 11.50 kg ≈ 12.652 m (rounded to four decimal places)
Therefore, the center of mass of these objects is located at approximately 12.652 meters on the y-axis.
Calculate the force required to double the length of a steel wire of area of cross section 5 x 10-5 m2? (Y=2x 1011N m2) (a) 10-5 N (b) 10-7 N n's D (c) 107 N (d) 105 N
Answer:
Option C is the correct answer.
Explanation:
We equation for elongation
[tex]\Delta L=\frac{PL}{AE}[/tex]
Here we need to find load required,
We need to double the wire, that is ΔL = 2L - L = L
A = 5 x 10⁻⁵ m²
E = 2 x 10¹¹ N/m²
Substituting
[tex]L=\frac{PL}{5\times 10^{-5}\times 2\times 10^{11}}\\\\P=10^7N[/tex]
Option C is the correct answer.
A 120-V rms voltage at 1000 Hz is applied to an inductor, a 2.00-μF capacitor and a 100-Ω resistor, all in series. If the rms value of the current in this circuit is 0.680 A, what is the inductance of the inductor?
Answer:
The inductance of the inductor is 35.8 mH
Explanation:
Given that,
Voltage = 120-V
Frequency = 1000 Hz
Capacitor [tex]C= 2.00\mu F[/tex]
Current = 0.680 A
We need to calculate the inductance of the inductor
Using formula of current
[tex]I = \dfrac{V}{Z}[/tex]
[tex]Z=\sqrt{R^2+(L\omega-\dfrac{1}{C\omega})^2}[/tex]
Put the value of Z into the formula
[tex]I=\dfrac{V}{\sqrt{R^2+(L\omega-\dfrac{1}{C\omega})^2}}[/tex]
Put the value into the formula
[tex]0.680=\dfrac{120}{\sqrt{(100)^2+(L\times2\pi\times1000-\dfrac{1}{2\times10^{-6}\times2\pi\times1000})^2}}[/tex]
[tex]L=35.8\ mH[/tex]
Hence, The inductance of the inductor is 35.8 mH
Answer:
Inductance,L:
"The property of the conductor or the solenoid to generate the electromotive force,emf due to the flow of current,I."
Unit: henry,H as it is equivalent to, kg.m².sec⁻².A⁻².
Explanation:
Data:
Voltage,v=120 v-rms,Frequency,f=1000 Hz,Capacitor, C=2.00 μF,Current,I=0.680 A,Solution:
We need to calculate the inductance, L of the solenoid inside a circuit,
I=v/z,Z=√R²+(Lω-1/Cω)²,putting the values I=V/√R²+(Lω-1/Cω)²,0.680=120/√(100)²+(L×2π×1000-1/2×10⁻⁶×2π×1000)²,L=35.8×10⁻³H, or L=35.8 mH.⇒AnswerA mass of 100 kg is pulled by a 392 N force in the +X direction along a rough surface (uk=0.4) with uniform velocity v=20 m/s. What is the TOTAL work done by ALL forces on the object after s=10 m?
Answer:
The total work done will be zero.
Explanation:
Given that,
Mass = 100 kg
Force = 392 N
Velocity = 20 m/s
Distance s= 10 m
We need to calculate the work done
Using balance equation
The net force will be
[tex]F'=F-\mu mg[/tex]
[tex]F'=392-0.4\times100\times9.8[/tex]
[tex]F'=0[/tex]
The net force is zero.
Hence, The total work done will be zero by all forces on the object.