Answer:
96.21 ft/s
Explanation:
To solve this, you only need to use one expression which is:
Vf² = Vo² + 2gh
g = 9.8 m/s²
However, this exercise is talking in feet, so convert the gravity to feet first:
g = 9.8 * 3.28 = 32.15 ft/s²
Vo is zero, because it's a free fall and in free fall the innitial speed is always zero. With this, let's calculate the speed at 2 seconds, with a height of 64 ft, and then with the 256 ft:
V1 = √2*32.15*64
V1 = 64.15 ft/s
V2 = √2*32.15*256
V2 = 128.3 ft/s
So the average rate is:
V = 128.3 + 64.15 / 2
V = 96.22 ft/s
Two coils are placed close together in a physics lab to demonstrate Faraday’s law of induction. A current of 5.00 A in one is switched off in 1.00 ms, inducing a 9.00 V emf in the other. What is their mutual inductance?
Answer:
Mutual inductance will be [tex]M=1.8\times 10^{-3}Hennry[/tex]
Explanation:
We have given induced current [tex]\Delta i=5A[/tex]
Time is given as [tex]t=1ms=10^{-3}sec[/tex]
We have to find the mutual inductance between the coils
Induced emf is given as e = 9 volt
We know that induced emf is given by
[tex]e=M\frac{\Delta i}{\Delta t}[/tex]
[tex]9=M\times \frac{5}{10^{-3}}[/tex]
[tex]M=1.8\times 10^{-3}Hennry[/tex]
A charged capacitor stores energy U. Without connecting this capacitor to anything, dielectric having dielectric constant K is now inserted between the plates of the capacitor, completely filling the space between them. How much energy does the capacitor now store?
Answer:
U / K
Explanation:
The energy stored in the capacitor is given by
[tex]U=\frac{q^{2}}{2C}[/tex]
where, q be the charge and C be the capacitance of the capacitor.
If a dielctric is inserted beteen the plates of capacitor having dielctric constant K, the new capacitance is C' = KC
The charge will remain same
The new energy stored is
[tex]U'=\frac{q^{2}}{2C'}[/tex]
[tex]U'=\frac{q^{2}}{2KC}[/tex]
U' = U / K
So, the energy reduces by the factor of K.
The new energy stored in the capacitor is U/K. Hence, option(e) is correct.
To determine the energy stored in a capacitor after inserting a dielectric, recognize the following points:
Initially, the energy stored in the capacitor is U.A dielectric with dielectric constant K is inserted between the plates, completely filling the space.When a dielectric is inserted and the capacitor is disconnected from any power source, the charge Q remains constant.
However, the capacitance C increases by a factor of K due to the dielectric. The energy stored in a capacitor(U) is given by:
U = Q²/(2C)Since the capacitance becomes KC, the new energy stored will be:
U' = Q²/(2KC)Given the original energy U = Q²/(2C), we can relate the new energy U' to the old energy U:
U' = U/KHence, the energy stored in the capacitor after inserting the dielectric is decreased by a factor of K, resulting in the new energy U' = U/K.
complete question:
A charged capacitor stores energy . Without connecting this capacitor to anything, dielectric having dielectric constant is now inserted between the plates of the capacitor, completely filling the space between them. How much energy does the capacitor now store?
a) U/2K
b) KU
c) 2KU
d) U
e) U/K
An automobile traveling 55.0 km/h has tires of 77.0 cm diameter.
(a) What is the angular speed of the tires about their axles?
(b) If the car is brought to a stop uniformly in 38.0 complete turns of the tires, what is the magnitude of the angular acceleration of the wheels?
(c) How far does the car move during the braking?
Explanation:
Given that, Speed of the automobile, v = 55 km/h = 15.27 m/s
Diameter of the tire, d = 77 cm
Radius, r = 0.385 m
(a) Let [tex]\omega[/tex] is the angular speed of the tires about their axles.
The relation between the linear speed and the angular speed is given by :
[tex]v=r\times \omega[/tex]
[tex]\omega=\dfrac{v}{r}[/tex]
[tex]\omega=\dfrac{15.27\ m/s}{0.385\ m}[/tex]
[tex]\omega=39.66\ rad/s[/tex]
(b) Number of revolution,
[tex]\theta=38\ rev=238.76\ radian[/tex]
Final angular speed of the car, [tex]\omega_f=0[/tex]
Initial angular speed, [tex]\omega_i=39.66\ rad/s[/tex]
Let [tex]\alpha[/tex] is the angular acceleration of the car. Using third equation of rotational kinematics as :
[tex]\omega_f^2-\omega_i^2=2\alpha \theta[/tex]
[tex]\alpha =\dfrac{-\omega_i^2}{2\theta}[/tex]
[tex]\alpha =\dfrac{-(39.66)^2}{2\times 238.76}[/tex]
[tex]\theta=-3.29\ rad/s^2[/tex]
(c) Let d is the distance covered by the car during the braking. It is given by :
[tex]d=\theta\times r[/tex]
[tex]d=238.76\times 0.385[/tex]
d = 91.92 meters
Suppose you have a 0.750-kg object on a horizontal surface connected to a spring that has a force constant of 150 N/m. There is simple friction between the object and surface with a static coefficient of friction μs=0.100 .
(a) How far can the spring be stretched without moving the mass?
(b) If the object is set into oscillation with an amplitude twice the distance found in part (a), and the kinetic coefficient of friction is μk=0.0850 , what total distance does it travel before stopping? Assume it starts at the maximum amplitude.
Answer:
[tex]x=0.0049\ m= 4.9\ mm[/tex]
[tex]d=0.01153\ m=11.53\ mm[/tex]
Explanation:
Given:
mass of the object, [tex]m=0.75\ kg[/tex]elastic constant of the connected spring, [tex]k=150\ N.m^{-1}[/tex]coefficient of static friction between the object and the surface, [tex]\mu_s=0.1[/tex](a)
Let x be the maximum distance of stretch without moving the mass.
The spring can be stretched up to the limiting frictional force 'f' till the body is stationary.
[tex]f=k.x[/tex]
[tex]\mu_s.N=k.x[/tex]
where:
N = m.g = the normal reaction force acting on the body under steady state.
[tex]0.1\times (9.8\times 0.75)=150\times x[/tex]
[tex]x=0.0049\ m= 4.9\ mm[/tex]
(b)
Now, according to the question:
Amplitude of oscillation, [tex]A= 0.0098\ m[/tex]coefficient of kinetic friction between the object and the surface, [tex]\mu_k=0.085[/tex]Let d be the total distance the object travels before stopping.
Now, the energy stored in the spring due to vibration of amplitude:
[tex]U=\frac{1}{2} k.A^2[/tex]
This energy will be equal to the work done by the kinetic friction to stop it.
[tex]U=F_k.d[/tex]
[tex]\frac{1}{2} k.A^2=\mu_k.N.d[/tex]
[tex]0.5\times 150\times 0.0098^2=0.0850 \times 0.75\times 9.8\times d[/tex]
[tex]d=0.01153\ m=11.53\ mm[/tex]
is the total distance does it travel before stopping.
To find the maximum distance the spring can be stretched without moving the mass, calculate the force of static friction and equate it to the spring force.
Explanation:(a) How far can the spring be stretched without moving the mass?
To find the maximum distance the spring can be stretched without moving the mass, we need to calculate the force of static friction. The force of static friction can be calculated using the equation:
Fs = μs * m * g
where μs is the coefficient of static friction, m is the mass, and g is the acceleration due to gravity.
Once we have the force of static friction, we can equate it to the spring force:
Fs = k * x
where k is the force constant of the spring and x is the maximum distance the spring is stretched without moving the mass.
Vector A magnitude of 10.0 centimeters and makes an angle of 60.0 degrees with the orizontal axis. What are the magnitudes it's horizontal and vertical components ?
Horizontal component of vector A: 5.0 cm
Vertical component of vector A: 8.7 cm
Explanation:
The horizontal and vertical components of a vector on the cartesian plane are calculated as follow:
[tex]v_x = v cos \theta\\v_y = v sin \theta[/tex]
where
[tex]v_x[/tex] is the horizontal component
[tex]v_y[/tex] is the vertical component
v is the magnitude of the vector
[tex]\theta[/tex] is the angle between the direction of the vector and x-axis
In this problem, we have
v = 10.0 cm is the magnitude of vector A
[tex]\theta=60.0^{\circ}[/tex] is the angle of its direction with the x-axis
Applying the equations, we find
[tex]v_x = (10.0)(cos 60)=5.0 cm[/tex]
[tex]v_y = (10.0)(sin 60)=8.7 cm[/tex]
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Explanation: edmuntum answer
You leave your 75 W portable color TV on for 6 hours each day and you do not pay attention to the cost of electricity. If the dorm (or your parents) charged you for your electricity use and the cost was $0.3 /kW · h, what would be your monthly (30 day) bill?Answer in units of dollars.
Final answer:
To calculate the monthly cost of electricity for a portable TV, convert the power to kilowatts, multiply by daily usage, then by the number of days in a month, and finally by the cost per kWh. The monthly bill for a 75 W TV used 6 hours a day at a rate of $0.3/kWh is $4.05.
Explanation:
Calculating Monthly Cost of Electricity for a Portable TV
First, we need to convert the power consumed by the TV from watts to kilowatts (kW):
75 W = 0.075 kW
Next, we calculate the energy used each day by multiplying the power consumption (in kW) by the time the TV is on (in hours):
Daily energy consumption = 0.075 kW × 6 h/day = 0.45 kWh/day
To find the monthly energy consumption, we multiply the daily consumption by 30 days:
Monthly energy consumption = 0.45 kWh/day × 30 days = 13.5 kWh/month
Finally, to calculate the monthly bill, we multiply the monthly energy consumption by the cost of electricity (in dollars per kWh):
Monthly bill = 13.5 kWh/month × $0.3/kWh = $4.05/month
The student's monthly electric bill for the portable TV would be $4.05.
In order to cool 1 ton of water at 20°C in an insulated tank, a person pours 140 kg of ice at –5°C into the water. The specific heat of water at room temperature is c = 4.18 kJ/kg· °C, and the specific heat of ice at about 0°C is c = 2.11 kJ/kg°C. The melting temperature and the heat of fusion of ice at 1 atm are 0°C and 333.7 kJ/kg.
The solution to this problem requires the understanding of the concepts of specific heat and heat of fusion. It involves three parts: heating up the ice to the melting point, melting the ice, and as the melted ice is absorbed by the water, it cools down the water's temperature. The heat absorbed by the ice is equal to the heat lost by the water.
Explanation:The subject of this question falls under Physics, specifically in the topic of Thermodynamics. The problem involves the concept of specific heat, and heat of fusion. We need to account for three parts: heating the ice from -5°C to 0°C, melting the ice at 0°C, and the absorbed heat by the water from the melting ice.
1. Firstly, calculate the energy needed to heat the ice from -5°C to 0°C (phase change has not yet happened). Use Q = m*c*ΔT, where m is the mass of ice, c is the specific heat of ice, and ΔT is the change in temperature.
2. Secondly, calculate the energy required to melt the ice. This time, use Q = m*Lf, where m is the mass of ice and Lf is the heat of fusion.
3. Lastly, as the melted ice (now at 0°C) is absorbed by the water, it will cool down the water's temperature. The heat absorbed by the ice is equal to the heat lost by the water.
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A mass on a spring vibrates in simple harmonic motion at a frequency of 4.0 Hz and an amplitude of 4.0 cm. If a timer is started when its displacement is a maximum (hence x = 4 cm when t = 0), what is the speed of the mass when t = 3 s?
The speed of the mass when t = 3 s in simple harmonic motion is approximately 6.283 m/s.
Explanation:The speed of the mass in simple harmonic motion can be determined by using the formula v = ωA sin(ωt), where v is the speed, ω is the angular frequency, A is the amplitude, and t is the time. In this case, the frequency of the motion is given as 4.0 Hz, which corresponds to an angular frequency of ω = 2πf = 2π(4.0 Hz) = 8π rad/s. The amplitude is given as 4.0 cm, which corresponds to A = 0.04 m. Plugging these values into the formula, we get:
v = (8π rad/s)(0.04 m) sin((8π rad/s)(3 s))
v ≈ 6.283 m/s
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The speed of the mass when[tex]\( t = 3 \) s is \( v = 8\pi \)[/tex] cm/s.
To find the speed of the mass at any given time \( t \) during its simple harmonic motion, we can use the following equation for velocity in SHM:
[tex]\[ v(t) = -A\omega \sin(\omega t + \phi) \][/tex]using the relation [tex]\( \omega = 2\pi f \).[/tex]Substituting the given frequency:
[tex]\[ \omega = 2\pi (4.0 \, \text{Hz}) = 8\pi \, \text{rad/s} \][/tex]
The phase constant \( \phi \) is 0 because the timer is started when the mass is at its maximum displacement, which corresponds to \[tex]( x = A \)[/tex]when [tex]\( t = 0 \).[/tex]
Now, we can calculate the velocity at [tex]\( t = 3 \)[/tex] s:
[tex]\[ v(3) = -A\omega \sin(\omega \cdot 3 + \phi) \][/tex]
[tex]\[ v(3) = -(4.0 \, \text{cm})(8\pi \, \text{rad/s}) \sin(8\pi \, \text{rad/s} \cdot 3 \, \text{s} + 0) \][/tex]
[tex]\[ v(3) = -32\pi \, \text{cm/s} \sin(24\pi \, \text{rad}) \][/tex]
Since \[tex]( \sin(24\pi \, \text{rad}) = 0 \)[/tex] (because [tex]\( 24\pi \)[/tex] is a multiple of[tex]\( 2\pi \)[/tex]), the velocity at [tex]\( t = 3 \)[/tex] s is:
[tex]\[ v(3) = -32\pi \, \text{cm/s} \cdot 0 = 0 \, \text{cm/s} \][/tex]
However, this result corresponds to the instantaneous speed at a point where the mass changes direction, and thus its speed is momentarily zero. To find the speed at a point where the mass is actually moving, we need to consider the time just before or after this point of maximum displacement.
Now, we calculate the velocity at [tex]( t = 3 - \Delta t \):[/tex]
[tex]\[ v(3 - \Delta t) = -A\omega \sin(\omega (3 - \Delta t) + \phi) \][/tex]
[tex]\[ v(3 - \Delta t) = -(4.0 \, \text{cm})(8\pi \, \text{rad/s}) \sin(8\pi (3 - \frac{1}{8\pi}) \, \text{rad}) \][/tex]
[tex]\[ v(3 - \Delta t) = -32\pi \, \text{cm/s} \sin(24\pi - 1) \, \text{rad}) \][/tex]
[tex]\[ v(3 - \Delta t) = -32\pi \, \text{cm/s} \sin(-1) \, \text{rad}) \][/tex]
Since [tex]\( \sin(-1) \approx -1 \)[/tex] for a very small angle in radians:
[tex]\[ v(3 - \Delta t) = -32\pi \, \text{cm/s} \cdot (-1) \][/tex]
[tex]\[ v(3 - \Delta t) = 32\pi \, \text{cm/s} \][/tex]
Therefore, just before the mass reaches its maximum displacement at [tex]\( t = 3 \) s[/tex], its speed is [tex]\( 32\pi \)[/tex] cm/s.
Two pipes are smoothly connected together without leaks. One has a diameter of 3 cm and the other has a diameter of 5 cm. Water flows through the pipes. In which pipe is the flow speed the greatest? in which pipe is the volume flow rate the greatest?
Answer:
The speed will be the greatest in the smaller pipeThe volume flow rate is the same in both pipesExplanation:
This question can be answered using the equation of continuity. Which is
V = A1v1 = A2v2
where V is the volume flow rate
A is the cross sectional area of the pipe at that position
v is the velocity of the flow at that point.
From the equation we can see the volume flow rate does not change even as the area and flow speed change. Which is consistent with the assumptions used in this equation which is that
The tube must one entry and one exit.The fluid in question is non-viscous.The fluid is incompressible.The flow is steady.If the temperature does not change, meaning there is not expansion/contraction and it is incompressible then the volume flow rate does not change.
And the speed will be higher in the pipe with the smaller diameter because area and speed are inversely related.
The flow speed is greatest in the smaller pipe, while the volume flow rate is greatest in the larger pipe.
Explanation:The flow speed is the greatest in the pipe with the smaller diameter, which is the pipe with a diameter of 3 cm. This can be determined from the continuity equation, which states that the flow speed is inversely proportional to the cross-sectional area. Since the smaller pipe has a smaller cross-sectional area, the flow speed is greater.
On the other hand, the volume flow rate is determined by both the flow speed and the cross-sectional area. The volume flow rate is the greatest in the pipe with the larger diameter, which is the pipe with a diameter of 5 cm. This is because even though the flow speed is slower in the larger pipe, the larger cross-sectional area compensates for it and allows a greater volume of water to flow through.
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A block of mass 2.20 kg is placed against a horizontal spring of constant k = 825 N/m and pushed so the spring compresses by 0.0650 m. HINT (a) What is the elastic potential energy of the block-spring system (in J)?
Answer:1.742 J
Explanation:
Given
mass of block [tex]m=2.2 kg[/tex]
[tex]k=825 N/m[/tex]
Compression in spring [tex]x=0.065 m[/tex]
Elastic Potential Energy of the Spring-mass system is given by
[tex]E=\frac{1}{2}kx^2[/tex]
where [tex]k=spring\ constant[/tex]
[tex]x=compression\ in\ spring[/tex]
[tex]E=\frac{1}{2}\times 825\times (0.065)^2[/tex]
[tex]E=412.5\times 42.25\times 10^{-4}[/tex]
[tex]E=1.742 J[/tex]
Final answer:
The elastic potential energy of the block-spring system with a spring constant of 825 N/m compressed by 0.0650 m is approximately 1.74 J.
Explanation:
To determine the elastic potential energy of the block-spring system, the formula for elastic potential energy (EPE) in a compressed or stretched spring is used, which is EPE = (1/2)kx2, where k is the spring constant and x is the displacement from the equilibrium position. In this case, the spring constant k is 825 N/m and the compression x is 0.0650 m.
Using the formula:
EPE = (1/2)(825 N/m)(0.0650 m)2
EPE = (1/2)(825)(0.004225)
EPE = (1/2)(3.481125)
EPE = 1.7405625 Joules
Therefore, the elastic potential energy of the block-spring system when the spring is compressed by 0.0650 m is approximately 1.74 J.
(a) Calculate the magnitude of the gravitational force exerted on a 1,320-kg satellite that is a distance of two earth radii from the center of the earth. N
(b) What is the magnitude of the gravitational force exerted on the earth by the satellite? N
(c) Determine the magnitude of the satellite's acceleration. m/s2
(d) What is the magnitude of the earth's acceleration? m/s2 Additional Materials
Answer:
(a) 3224.27 N
(b) 3224.27 N
(c) 2.443 m/s^2
(d) 5.37 x 10^-22 m/s^2
Explanation:
Mass of satellite, m = 1320 kg
mas of earth, M = 6 x 10^24 kg
Radius of earth, r = 6.4 x 10^6 m
(a) The force of gravitation between the earth and the satellite is given by
[tex]F=G\frac{Mm}{d^{2}}[/tex]
where, d is the distance between the two objects
[tex]F=6.67\times10^{-11}\frac{6\times10^{24}\times 1320}{\left (2\times 6.4\times 10^{6} \right )^{2}}[/tex]
F = 3224.27 N
(b) The force on earth is same as the force on satellite.
F = 3224.27 N
(c) Acceleration of satellite = Force on satellite / mass of satellite
Acceleration of satellite = 3224.27 / 1320 = 2.443 m/s^2
(d) Acceleration of earth = Force on earth / mass of earth
Acceleration of satellite = 3224.27 / (6 x 10^24) = 5.37 x 10^-22 m/s^2
Anna Litical and Noah Formula are experimenting with the effect of mass and net force upon the acceleration of a lab cart. They determine that a net force of F causes a cart with a mass of M to accelerate at 48 cm/s/s. What is the acceleration value of a cart with …a. a mass of M when acted upon by a net force of 2F?b. a mass of 2M when acted upon by a net force of F?c. a mass of 2M when acted upon by a net force of 2F?d. a mass of 4M when acted upon by a net force of 2F?e. a mass of 2M when acted upon by a net force of 4F?
Answer:
Explanation:
a net force of F causes a cart with a mass of M to accelerate at 48 cm/s/s.
F = M x 48
Mass M = F / 48
a )
When force = 2F and mass = M
Acceleration = force / mass
= 2F /F/48
= 48 X 2 = 96 cm/s²
b )
When force = F and mass = 2M
Acceleration = force / mass
= F /2F/48
= 24 cm/s²
c )
When force = 2F and mass = 2M
Acceleration = force / mass
= 2F /2F/48
= 48 cm/s
d )
When force = 2F and mass = 4M
Acceleration = force / mass
= 2F /4F/48
= 24 cm/s
e)
When force = 4F and mass = 2M
Acceleration = force / mass
= 4F / 2M
= 4F / 2 F/48
= 48 x 2
Acceleration = 96 cm/s²
Final answer:
The acceleration of carts with different masses acted upon by varying forces can be calculated using Newton's second law.
Explanation:
Mass is a fundamental factor in determining acceleration according to Newton's second law, which states force (F) = mass (m) x acceleration (a). When a cart with mass M is acted upon by a net force of 2F, the acceleration will be 96 cm/s/s (48 cm/s/s x 2). Similarly, for a mass of 2M acted upon by force F, the acceleration will be 24 cm/s/s (48 cm/s/s / 2). When a mass of 2M is subject to 2F, the acceleration will be 96 cm/s/s, and for a mass of 4M with a force of 2F, the acceleration will be 24 cm/s/s. Finally, when a mass of 2M experiences 4F, the acceleration will be 192 cm/s/s.
Technician A says that chrome rings are the best rings to use in an unbored cylinder. Technician B says that the ring end gap is non-adjustable; therefore there is no need to check it when reassembling the engine. Which technician is correct?
Answer:
Neither Technician A nor Technician B.
Explanation:
chrome is very expensive and we do need to check the ring before reassembling engine.
Sand falls from a conveyor belt at a rate of 14 m cubed divided by min onto the top of a conical pile. The height of the pile is always three-eighths of the base diameter. How fast are the height and the radius changing when the pile is 5 m high?
Answer:7.16 cm/min or 0.0716 min.
Explanation:
From the equation the parameters given are; rate= 14m^-3/min, height of the pile,h= 3/8, rate when the pile is 5m high= ??.
d= diameter, r= radius. From the definition of diameter, diameter, d=2× radius, r(that is, 2r)
h= 3/8×d = 3/8(2r)= 3/4 × r
r= 4/3 × h
V= 1/3×πr^2×h ---------------------(1).
V= 1/3×π×[4/3h]^2×h
V= 16/27×π×h^3
Differentiate dV with respect to time, t;
dV/dt= 3(16)/27 × π ×h^2 ×dh/dt.
dV/dt = 16/9×π×h^2 × dh/dt
Then we differentiate V implicitly with respect to time,t
14=16/9×π[5]^2×dh/dt
dh/dt = 10× 9/400π
dh/dt= 45/200π (m/min)
dh/dt= 0.0716 m/min.
Conversion to cm/min
= dh/dt= 4500/200π (cm/min)
dh/dt= 1125/50π
dh/dt= 7.16 cm/min.
To find the speed at the edge of the storm, we divide the distance by the time.
Explanation:To find how fast the height and radius are changing when the pile is 5 m high, we need to find the rates of change of both the height and radius at that height.
Let's start by finding the rate of change of the height. We know that the height is always three-eighths of the base diameter, so the base diameter is twice the height. Therefore, when the height is 5 m, the base diameter is 2 * 5 = 10 m.
Now, we can use the information about the rate of sand falling from the conveyor belt to find the rate of change of the radius. The volume of a cone is given by the formula V = (1/3) * π * r^2 * h, where V is the volume, π is a constant (approximately 3.14), r is the radius, and h is the height. The rate of change of the volume with respect to time can be found using the chain rule.
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A satellite explodes in outer space, far from any other body, sending thousands of pieces in all directions. How does the linear momentum of the satellite before the explosion compare with the total linear momentum of all the pieces after the explosion? Account for your answer.
Answer:
They would be the same
Explanation:
Since there's no external force, or the external net force acting on the satellite is 0. By the law of conservation of momentum, the momentum before and after the explosion must be the same.
Even if the satellite is stationary prior to the explosion, its momentum is 0. After explosion it sends many of its pieces in all direction with different non 0 momentum. The total momentum by account of their direction would sum up to be 0.
Final answer:
The linear momentum of the satellite before the explosion is equal to the total linear momentum of all the pieces after the explosion.
Explanation:
The linear momentum of the satellite before the explosion is equal to the total linear momentum of all the pieces after the explosion. This is because linear momentum is conserved in an explosion. Although the distribution of momentum among the individual pieces may change, the total momentum remains the same. In this case, when the satellite explodes, the momentum of the satellite is redistributed among the thousands of pieces that are scattered in all directions.
A 74.9 kg person sits at rest on an icy pond holding a 2.44 kg physics book. he throws the physics book west at 8.25 m/s. what is his recoil velocity? PLEASE HELP ME
Answer:
The recoil velocity is 0.2687 m/s.
Explanation:
∵ The person is sitting on an icy surface , we can assume that the surface is frictionless.
∴ There is no force acting acting on the person and book as a system in horizontal direction.
Hence , momentum is conserved for this system in horizontal direction of motion.
If 'i' and 'f' be the initial and final states of this system , then by principle of conservation of momentum(p) -
[tex]p_{i}[/tex]=[tex]p_{f}[/tex]
System initially is at rest
∴[tex]p_{i}[/tex]=0
∴ From the above 2 equations
[tex]p_{f}[/tex]=0
We know that ,
Momentum(p)=Mass of the body(m)×velocity of the body(v)
Let [tex]m_{1}[/tex] and [tex]m_{2}[/tex] be the mass of the person and the book respectively and [tex]v_{1}[/tex] and [tex]v_{2}[/tex] be the final velocities of the person and book respectively.
∴[tex]p_{f}[/tex]=[tex]m_{1}[/tex][tex]v_{1}[/tex]+[tex]m_{2}[/tex][tex]v_{2}[/tex]=0
From the question ,
[tex]m_{1}[/tex] = 74.9 kg
[tex]m_{2}[/tex] = 2.44 kg
[tex]v_{2}[/tex] = 8.25 m/s
Substituting these values in the above equation we get ,
(74.9 × [tex]v_{1}[/tex] )+ (2.44×8.25) = 0
∴[tex]v_{1}[/tex] = - 0.2687 m/s (Negative sign suggests that the motion of the person is opposite to that of the book)
∴ The recoil velocity is 0.2687 m/s.
What would happen to each of the properties if the intermolecular forces between molecules increased for a given fluid? Assume temperature remains constant.
Answer:
If we assume that the temperature remains constant. And if the intermolecular forces are increased, then there would be an increase in the boiling point, the viscosity, and surface tension of the substance.
Explanation:
The intermolecular forces are interactions between particles. And decrease as we go from solid >liquid >gas. And on gases we have weak intermolecular forces.
The most common 3 types of intermolecular forces occurs between neutral molecules are known as Van der Walls forces (Dipole-Dipole,Hydrogen bonding, London Dispersion forces).
If the intermodelucar attraction forces increases we have:
1. Decreasing vapor pressure (pressure of the vapor that is in equilibrium with the liquid)
2. Increasing boiling point (temperature at which the vapor pressure is equal to the pressur exerted on the surface of the liquid)
3. Increasing surface tension (resistance of a liquid to spread out and increase the surface area)
4. Increasing viscosity (resistance of a liquid to flow)
If we assume that the temperature remains constant. And if the intermolecular forces are increased, then there would be an increase in the boiling point, the viscosity, and surface tension of the substance.
Receiving delicious food is to escaping electric shock as ________ is to ________.
Answer:
positive reinforcer; negative reinforcer
Explanation:
positive reinforcer; negative reinforcer
Receiving delicious food can be seen as a positive reinforcer while escaping electric shock can be seen as a negative reinforcer.
What is the mass of a large ship that has a momentum of 1.60×109kg·m/s, when the ship is moving at a speed of 48.0 km/h? (b) Compare the ship’s momentum to the momentum of a 1100-kg artillery shell fired at a speed of 1200 m/s.
a) The mass of the ship is [tex]1.2\cdot 10^8 kg[/tex]
b) The ship has a larger momentum than the shell
Explanation:
a)
The momentum of an object is given by:
[tex]p=mv[/tex]
where
m is the mass of the object
v is its velocity
For the ship in this problem, we have
[tex]p=1.60\cdot 10^9 kg m/s[/tex] is the momentum
[tex]v=48.0 km/h \cdot \frac{1000 m/km}{3600 s/h}=13.3 m/s[/tex] is the velocity
Solving for m, we find the mass of the ship:
[tex]m=\frac{p}{v}=\frac{1.60\cdot 10^9}{13.3}=1.2\cdot 10^8 kg[/tex]
b)
The momentum of the artillery shell is given by
[tex]p=mv[/tex]
where
m is its mass
v is its velocity
For the shell in this problem,
m = 1100 kg
v = 1200 m/s
Substituting,
[tex]p=(1100)(1200)=1.32\cdot 10^6 kg m/s[/tex]
So, we see that the ship has a larger momentum.
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Momentum is the product of force and velocity. The mass of the ship and the momentum of the fired artillery is [tex] 1.2 \times 10^{8}[/tex] and [tex]1.32 \times 10^{6} kgm/s [/tex] respectively
Momentum = mass × velocity
momentum, p = [tex]1.60 \times 10^{9} [/tex] Speed, v = [tex] 48 km/hr [/tex] Speed in m/s = (48×1000)/3600 = 13.33 m/sMass of the ship = momentum / velocity
Mass of ship = [tex]\frac{1.60 \times 10^{9}}{13.33} = 1.2 \times 10^{8}[/tex]
2.)
Speed, v = 1200 m/s Mass, m = 1100 kgMomentum of fired artillery = 1100 × 1200 = [tex]1.32 \times 10^{6} kgm/s [/tex]
Therefore, the momentum of fired artillery is [tex]1.32 \times 10^{6} kgm/s [/tex]
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A helicopter flies 30 km 30 degrees north of east, drops of the supplies to a research team; it then heads 50 degrees north of west traveling 70 km to drop the supplies to a filming crew. What is the resultant displacement of the helicopter?
Answer:
D = 71.20 km
Explanation:
In this case, let's get the exercise by parts.
Part 1 would be the first displacement which is 30 km. However, it's displacing north to east 30°, so in this case, if we trace a line of this, it should be a diagonal. So we need to get the value of this displacement in the x and y axis.
Dx1 = 30 cos 30° = 25.98 km
Dy1 = 30 sin 30° = 15 km
Doing the same thing but in the second travel we have:
Dx2 = 70 cos 50° = 45 km
Dy2 = 70 sin 50° = 53.62 km
Now we need to know how was the displacement in both axis. In travel 2, the plane is moving north to west, so, in the x axis, is moving in the opposite direction, therefore:
Dx = -45 + 25.98 = -19.02 km
In the y axis, is moving upward so:
Dy = 53.62 + 15 = 68.62 km
Finally to get the resultant displacement:
D = √Dx² + Dy²
Replacing:
D = √(-19.02)² + (68.62)²
D = √5070.46
D = 71.20 km
What is the difference between a tropical storm and a hurricane
Answer and Explanation:
Tropical storm:
The point at which the tropical depression intensifies and can sustain a maximum wind speed in the range of 39-73 mph, it is termed as a tropical storm.
Hurricane:
A hurricane is that type of tropical cyclone which includes with it high speed winds and thunderstorms and the intensity of the sustained wind speed is 74 mph or more than this.
Hurricanes are the most intense tropical cyclones.
The only difference between the tropical storm and hurricane is in the intensity.
A tropical storm becomes a hurricane when its sustained wind speeds exceed 74 miles per hour, with both systems featuring low pressure centers and heavy rains, but hurricanes pose a greater threat due to their higher wind speeds.
Explanation:The primary difference between a tropical storm and a hurricane lies in their wind speeds. When a storm's sustained winds reach between 39 to 73 miles per hour, it is classified as a tropical storm. Once those winds exceed 74 miles per hour, the storm is then classified as a hurricane. Both types of storm systems form over warm ocean waters, typically warmer than 80 °F and involve low pressure centers, strong winds, and heavy rains. However, a hurricane's increased wind speed is what gives it the potential to cause significantly more damage upon making landfall.
The Coriolis force is responsible for the cyclonic rotation of these storms—with hurricanes in the Northern Hemisphere rotating counterclockwise and those in the Southern Hemisphere rotating clockwise. The terms hurricane, typhoon, and tropical storm are regional names for these cyclones. Regardless of the name, these storms are dangerous due to the combination of high winds, heavy rains, and consequential risks such as flooding and structural damage.
People who work around radioactivity wear film badges to monitor the amount of radiation that reaches their bodies. These badges consist of small pieces of photographic film enclosed in a light-proof wrapper. What kind of radiation do these devices monitor?
Answer:
the kind of radiations of the devices monitor are gamma, X-Rays and beta particles.
Explanation:
Film badges, which is usually worn in the front body, is a device to monitor the exposure of the radiation. This device measures the the exposure of X-Rays, Gamma Rays and beta Particles. This device is able to differentiate the energy of photon and is considerably accurate for the exposure greater than 100 millirem. However, under 20 milllirem, the device is less accurate for gamma rays.
You are driving a 2420.0-kg car at a constant speed of 14.0 m/s along a wet, but straight, level road. As you approach an intersection, the traffic light turns red. You slam on the brakes. The car's wheels lock, the tires begin skidding, and the car slides to a halt in a distance of 26.8 m.
What is the coefficient of kinetic friction between your tires and the wet road?
Answer:
[tex]U_k[/tex] = 0.3731
Explanation:
First we will identify the important data of the question.
M = 2420 kg
V = 14 m/s
d = 26.8 m
[tex]U_k =[/tex] ?
So, we will use the law of the conservation of energy, it says that:
[tex]E_i - E_f = W_f[/tex]
therefore:
[tex]E_i = \frac{1}{2}MV^2\\E_f = 0\\W_f = F_kd[/tex]
where [tex]F_k[/tex] is the friction force
Replacing on the first equation, we get:
[tex]\frac{1}{2}MV^2 -0 = F_kd[/tex]
[tex]F_k[/tex] is also equal to [tex]U_kN[/tex]
where N is the normal force and Uk is the coefficient of kinetic friction.
solving the equation:
[tex]\frac{1}{2}(2420)(14)^2 = U_kN(26.8)[/tex]
Before solve for [tex]U_k[/tex] we need to know the value of N, so we use the law of newton as:
∑[tex]F_y[/tex] = N - (2420)(9.8m/s) = 0
N = 23716
Finally, just solve for [tex]U_k[/tex] as:
[tex]U_ k = \frac{\frac{1}{2}(2420)(14)^2 }{(26.8)(23716)}[/tex]
[tex]U_k[/tex] = 0.3731
Final answer:
To find the coefficient of kinetic friction on a wet road, we use the work-energy principle to set the work done by friction equal to the car's initial kinetic energy. By substituting and simplifying the given values, we calculate the coefficient as approximately 0.746.
Explanation:
The question asks to find the coefficient of kinetic friction between the tires of a car and a wet road, given that the car, which has a mass of 2420.0 kg and was initially moving at a constant speed of 14.0 m/s, slides to a halt in a distance of 26.8 m after the driver slams on the brakes. To solve for the coefficient of kinetic friction (μk), we will use the work-energy principle which states that the work done by the friction force equals the change in kinetic energy of the car since no other forces do work.
The equation is as follows:
Ff × d = ½ m × vi2 - ½ m × vf2
Where Ff is the friction force, d is the distance, m is the mass, vi is the initial velocity, and vf is the final velocity (which is 0 since the car stops).
Since Ff = μk × N, and N = m × g (where N is the normal force and g is the acceleration due to gravity), we have:
μk = (½ m × vi2)/ (m × g × d)
Plugging in the known values: μk = [(½ × 2420.0 kg × (14.0 m/s)2)] / [2420.0 kg × 9.8 m/s2 × 26.8 m]
This simplifies to: μk = (2420.0 kg × 196.0 m2/s2) / (2420.0 kg × 9.8 m/s2 × 26.8 m)
We can cancel the mass from both the numerator and denominator, leaving us with:
μk = (196.0 m2/s2) / (9.8 m/s2 × 26.8 m)
Thus, the coefficient of kinetic friction is:
μk = 196.0 / 262.64 = 0.746 (rounded to three decimal places)
A bubble of air is rising up through the ocean. When it is at a depth of 20.0 m below the surface, where the temperature is 5.00°C, its volume is 0.80 cm3. What is the bubble's volume (in cm3) just before it hits the ocean surface, where the temperature is 20.0°C? Assume the average density of sea water is 1,025 kg/m3. Hint: Use Pascal's Principle (textbook Eq. 14.4) to determine the pressure at the depth of 20.0 m below the surface.
Answer:
the volume is 0.253 cm³
Explanation:
The pressure underwater is related with the pressure in the surface through Pascal's law:
P(h)= Po + ρgh
where Po= pressure at a depth h under the surface (we assume = 1atm=101325 Pa) , ρ= density of water ,g= gravity , h= depth at h meters)
replacing values
P(h)= Po + ρgh = 101325 Pa + 1025 Kg/m³ * 9.8 m/s² * 20 m = 302225 Pa
Also assuming that the bubble behaves as an ideal gas
PV=nRT
where
P= absolute pressure, V= gas volume ,n= number of moles of gas, R= ideal gas constant , T= absolute temperature
therefore assuming that the mass of the bubble is the same ( it does not absorb other bubbles, divides into smaller ones or allow significant diffusion over its surface) we have
at the surface) PoVo=nRTo
at the depth h) PV=nRT
dividing both equations
(P/Po)(V/Vo)=(T/To)
or
V=Vo*(Po/P)(T/To) = 0.80 cm³ * (101325 Pa/302225 Pa)*(277K/293K) = 0.253 cm³
V = 0.253 cm³
A potter is shaping a bowl on a potter's wheel rotating at constant angular speed. The friction force between her hands and the clay is 1.6 N total.(a) How large is her torque on the wheel, if the diameter of the bowl is 12 cm?(b) How long would it take for the potter's wheel to stop if the only torque acting on it is due to the potter's hand? The initial angular velocity of the wheel is 1.6 rev/s, and the moment of inertia of the wheel and the bowl is .11 kg*m^2 Show all work and formulas for best rating.
Answer:
Part A:
T=0.096 N m
Part B:
t=11.525≅11.53 seconds
Explanation:
Part A:
In order to find the torque we will use the following formula:
[tex]T=r*F[/tex]
Where T is the torue
r is the radius of bowel
F is the force
[tex]r=\frac{12*10^{-2} }{2} m\\[/tex]
r=0.06 m
[tex]T=0.06*1.6[/tex]
T=0.096 N m
Part B:
In order to find how long would it take for potter wheel to sto we will proceed the following way:
T=I*α
Where:
T is the Torque
I is the moment of inertia
α is the angular acceleration
α=T/I
we will take T from art A
α=0.096/0.11
α=0.872 [tex]rad/s^2[/tex]
α=ω/t
where:
α is the angular accceleration
ω is angular velocity in rad/s
ω=1.6*2ππ
ω=10.05 rad/s
t=ω/α
[tex]t=\frac{10.05}{0.872}[/tex]
t=11.525≅11.53 seconds
It will take 11.53 sec for wheel to stop
The torque exerted by the potter on the wheel is 0.192 Nm. If the only force acting on the wheel is the potter's hand, the wheel will stop in approximately 5.72 seconds.
Explanation:The question is asking about the concepts of torque, angular velocity, and moment of inertia in a real-life situation involving a potter's wheel. We can solve this problem in two parts.
Torque: Torque (τ) can be calculated using the formula τ = r*F, where r is radius and F is force. In this case, the radius, r = 12 cm = 0.12 m (since we need to convert cm to meters for the SI units.) and F = 1.6 N. So, the torque exerted by the potter can be calculated as τ = 0.12 m * 1.6 N = 0.192 Nm.
Time to stop: If the only torque acting on the wheel is due to the potter's hand, we can use the rotational analog of Newton's second law, τ = I*a, where a is the angular acceleration, I is the moment of inertia and τ is the torque. The angular acceleration can be calculated as a = τ / I = 0.192 Nm / 0.11 kg*m^2 = 1.75 rad/s^2. The time it takes to stop the wheel can then be calculated using the formula t = (final velocity - initial velocity) / angular acceleration. As the final angular velocity (when the wheel stops) is 0, and the initial angular velocity is 1.6 rev/s = 1.6 * 2π rad/s (as 1 rev/s = 2π rad/s), the time to stop the wheel, t = (0 - 1.6*2π rad/s) / 1.75 rad/s^2, t = approximately 5.72 seconds, assuming that the only force acting on the potter's wheel is the hand of the potter.
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An area with excess electrons has a net _____ charge; an area with a deficit of electrons has a net _____ charge. Question 10 options: negative, positive positive, negative positive, neutral negative, neutral
Answer:
negative, positive
Explanation:
A we know that the electrons have a negative charge.
So, when a body has some excess electrons, it means it has negative charge.
When a body has an deficiency of eletrons, it means it gains a positive charge.
An area with excess electrons has a net _negative__ charge; an area with a deficit of electrons has a net _Positive____ charge.
In 2005 a planet was found revolving about a distant sunlike star in 4.01 days. The radius of its orbit is 2 comma 600 comma 000 mi. What is its linear velocity (in mi/h) about its star?
Answer: linear velocity = 169,745.2mi/hr
Explanation:
Given: radius of orbit= 2,600,000mi
Time taken for orbit = 4.01 days = 4.01 × 24 hrs= 96.24hrs
Linear velocity = distance travelled/time
distance travelled = circumference of orbit = 2πr
= 2π × 2,600,000mi
= 16336281.8mi
Linear velocity = 16336281.8/ 96.24hrs
= 169745.2mi/hr
Gel electrophoresis sorts dna molecules on the basis of their
Answer:
Size and charge and speed
Explanation:
Gel electrophoresis is method designed for the separation of DNA molecules. The are also used to separate other micromolecules like RNA and protein, etc. This separation takes place based on their size and charge. The molecule travels through the gel in all different directions but their speed are different and thus they are separated.
Gel electrophoresis is a technique used to separate DNA fragments based on their size.
Explanation:Gel electrophoresis is a technique used to separate DNA fragments based on their size. It involves loading DNA samples onto a gel matrix and applying an electric current. Smaller DNA fragments move faster through the gel, resulting in bands at specific distances from the top of the gel. Molecular weight standard samples can be run alongside the DNA fragments to provide a size comparison. The separated DNA fragments can then be visualized by staining the gel with a DNA-specific dye.
Gel electrophoresis is a fundamental technique used in molecular biology, genetics, and forensics for various purposes, including DNA fingerprinting, DNA sequencing, and the analysis of PCR products. It provides a visual representation of DNA fragment sizes, enabling researchers to draw conclusions about genetic material and study genetic variations.
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The two pucks of equal mass did not move linearly (they came to a stop) after the collision due to the conservation of linear momentum. However, since the two pucks, mutual center of mass does not coincide with either of the pucks velocity vectors, they have angular momentum. This becomes evident after the collision when due to conservation of angular momentum the two pucks spin around their mutual center of mass.
Suppose we replace both hover pucks with pucks that are the same size as the originals but twice as massive. Otherwise, we keep the experiment the same. Compared to the pucks above, this pair of pucks will rotate:
a. at the same rate.
b. one-fourth as fast.
c. four times as fast.
d. twice as fast.
e. one-half as fast.
Compared to the pucks given, the pair of pucks will rotate at the same rate.
Answer: Option A
Explanation:
The law of conservation of the angular momentum expresses that when no outer torque follows upon an article, no difference in angular momentum will happen. At the point when an item is turning in a shut framework and no outside torques are applied to it, it will have no change in angular momentum.
The conservation of the angular momentum clarifies the angular quickening of an ice skater as she brings her arms and legs near the vertical rotate of revolution. In the event, that the net torque is zero, at that point angular momentum is steady or saved.
By twice the mass yet keeping the speeds unaltered, also twice the angular momentum's to the two-puck framework. Be that as it may, we likewise double the moment of inertia. Since [tex]L=I \times \omega[/tex], the turning rate of the two-puck framework must stay unaltered.
The last stage of a rocket, which is traveling at a speed of 7700 m/s, consist of two parts that are clamped together: a rocket case with a mass of 250.0 kg and a payload capsule with a mass of 100.0 kg. When the clamp is released, a compressed spring causes the two parts to separate with a relative speed of 910.0 m/s. Assume that all velocities are along the same line.What are the speeds of (a) the rocket case and (b) the payload after they have separated? Assume that all velocities are along the same line. Find the total kinetic energy of the two parts (c) before and (d) after they separate. (e) Account for the difference.
Answer:
Explanation:
Let the velocity of rocket case and payload after the separation be v₁ and v₂ respectively. v₂ will be greater because payload has less mass so it will be fired with greater speed .
v₂ - v₁ = 910
Applying law of conservation of momentum
( 250 + 100 ) x 7700 = 250 v₁ + 100 v₂
2695000 = 250 v₁ + 100 v₂
2695000 = 250 v₁ + 100 ( 910 +v₁ )
v₁ = 7440 m /s
v₂ = 8350 m /s
Total kinetic energy before firing
= 1/2 ( 250 + 100 ) x 7700²
= 1.037575 x 10¹⁰ J
Total kinetic energy after firing
= 1/2 ( 250 x 7440² + 100 x 8350² )
= 1.0405325 x 10¹⁰ J
The kinetic energy has been increased due to addition of energy generated in firing or explosion which separated the parts or due to release of energy from compressed spring.