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Sidana [21]
3 years ago
10

A girl drops a pebble from a high cliff into a lake far below. She sees the splash of the pebble hitting the water 2.00s later.

How fast was the pebble going when it hit the water?
Physics
1 answer:
Natali5045456 [20]3 years ago
3 0

Answer:

19.62 ms

Explanation:

t = Time taken = 2 s

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s² (downward direction is taken as positive)

Equation of motion

v=u+at\\\Rightarrow v=0+9.81\times 2\\\Rightarrow v=19.62\ m/s

The speed of the pebble when it hit the water is 19.62 ms

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The net force is 0 N.

To find the total net force, all you need to do is add the forces.

So 100 + -100 = 0

Most of these questions are trick questions, but if you know Newton's laws of forces, then you should know that all forces should cancel out with each other.

Hope this helps!!!

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3 years ago
300N effort is applied to lift the load of 900N by using a first class lever. If the distance between the load and fulcrum is 20
Klio2033 [76]

Answer:

Effort=300N

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7 0
2 years ago
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How do contact and non-contact forces affect the design of different modes of transportation?
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Contact and non contact forces are big factors to consider in designing of different modes of transportation because this factors are resistances for the mode of transportation. These contact and non- contact forces should be minimized in order for the energy requirement to be also minimum. But not the extent of risking the safety, for example a non contact force is the wieght, the should be optimum, safety of the design should not be compromised just to reduce weight, just like by removing essential parts, support just to remove weight is not good. 

7 0
4 years ago
A college student is working on her physics homework in her dorm room. her room contains a total of 6.0×1026 gas molecules. as s
ella [17]
<span>6.6 degrees C Let's model the student as a 125 w furnace that's been operating for 11 minutes. So 125 w * 11 min = 125 kg*m^2/s^3 * 11 min * 60 s/min = 82500 kg*m^2/s^2 = 82500 Joule So the average kinetic energy increase of each gas molecule is 82500 J / 6.0x10^26 = 1.38x10^-22 J Now the equation that relates kinetic energy to temperature is: E = (3/2)Kb*Tk E = average kinetic energy of the gas particles Kb = Boltzmann constant (1.3806504Ă—10^-23 J/K) Tk = Kinetic temperature in Kelvins Notice the the energy level of the gas particles is linear with respect to temperature. So we don't care what the original temperature is, we just need to know by how much the average energy of the gas particles has increased by. So let's substitute the known values and solve for Tk E = (3/2)Kb*Tk 1.38x10^-22 J = (3/2)1.3806504Ă—10^-23 J/K * Tk 1.38x10^-22 J = 2.0709756x10^-23 J/K * Tk 6.64 K = Tk Rounding to 2 significant digits gives 6.6K. So the temperature in the room will increase by 6.6 degrees K or 6.6 degrees C, or 11.9 degrees F.</span>
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3 years ago
_______ is energy that is transferred due to a difference in temperatures.
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Answer:

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