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2 years ago

How much of the total energy in Problem 3 and 4 has been transformed to kinetic energy?

Physics

1 answer:

ale4655 [162]2 years ago

6 0

I am only a HS Freshman so I might be wrong

Answer:

3. Its potential energy and the formula for PE = mgh so you would use the weight given (Weight = mg) and mg is in the (Potential Energy) forumula so we multiply 3 by 30 and get 90 Joules

4. We would do the same thing with this: PE = mgh; PE = 10*3 = 30 Joules

Once again I am only a Freshman and I have taken Private AP Physics classes so just maybe wait for someone who is more confident.

This is just what I would answer

btw, for problem 5, the fromula for kinetic energy is 1/2*m*v^2

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Why is the law of buoyancy considered a law?

solong [7]

Answer:

Explanation:

because it is a function of physics that will always be true, hope this helps :)

4 0

2 years ago

Two objects are dropped from rest from the same height. Object A falls through a distance during a time t, and object B falls th

UNO [17]

Answer:

Distance covered by B is 4 times distance covered by A

Explanation:

For an object in free fall starting from rest, the distance covered by the object in a time t is

$s=\frac{1}{2}gt^2$

where

s is the distance covered

g is the acceleration due to gravity

t is the time elapsed

In this problem:

- Object A falls through a distance $s_A$ during a time t, so the distance covered by object A is

$s_A=\frac{1}{2}gt^2$

- Object B falls through a distance $s_B$ during a time 2t, so the distance covered by object B is

$s_B=\frac{1}{2}g(2t)^2 = 4(\frac{1}{2}gt^2)=4s_A$

So, the distance covered by object B is 4 times the distance covered by object A.

5 0

2 years ago

A 58.0-kg projectile is fired at an angle of 30.0° above the horizontal with an initial speed of 140 m/s from the top of a cliff

strojnjashka [21]

(a) $6.43\cdot 10^5 J$

The total mechanical energy of the projectile at the beginning is the sum of the initial kinetic energy (K) and potential energy (U):

$E=K+U$

The initial kinetic energy is:

$K=\frac{1}{2}mv^2$

where m = 58.0 kg is the mass of the projectile and $v=140 m/s$ is the initial speed. Substituting,

$K=\frac{1}{2}(58 kg)(140 m/s)^2=5.68\cdot 10^5 J$

The initial potential energy is given by

$U=mgh$

where g=9.8 m/s^2 is the gravitational acceleration and h=132 m is the height of the cliff. Substituting,

$U=(58.0 kg)(9.8 m/s^2)(132 m)=7.5\cdot 10^4 J$

So, the initial mechanical energy is

$E=K+U=5.68\cdot 10^5 J+7.5\cdot 10^4 J=6.43\cdot 10^5 J$

(b) $-1.67 \cdot 10^5 J$

We need to calculate the total mechanical energy of the projectile when it reaches its maximum height of y=336 m, where it is travelling at a speed of v=99.2 m/s.

The kinetic energy is

$K=\frac{1}{2}(58 kg)(99.2 m/s)^2=2.85\cdot 10^5 J$

while the potential energy is

$U=(58.0 kg)(9.8 m/s^2)(336 m)=1.91\cdot 10^5 J$

So, the mechanical energy is

$E=K+U=2.85\cdot 10^5 J+1.91 \cdot 10^5 J=4.76\cdot 10^5 J$

And the work done by friction is equal to the difference between the initial mechanical energy of the projectile, and the new mechanical energy:

$W=E_f-E_i=4.76\cdot 10^5 J-6.43\cdot 10^5 J=-1.67 \cdot 10^5 J$

And the work is negative because air friction is opposite to the direction of motion of the projectile.

(c) 88.1 m/s

The work done by air friction when the projectile goes down is one and a half times (which means 1.5 times) the work done when it is going up, so:

$W=(1.5)(-1.67\cdot 10^5 J)=-2.51\cdot 10^5 J$

When the projectile hits the ground, its potential energy is zero, because the heigth is zero: h=0, U=0. So, the projectile has only kinetic energy:

E = K

The final mechanical energy of the projectile will be the mechanical energy at the point of maximum height plus the work done by friction:

$E_f = E_h + W=4.76\cdot 10^5 J +(-2.51\cdot 10^5 J)=2.25\cdot 10^5 J$

And this is only kinetic energy:

$E=K=\frac{1}{2}mv^2$

So, we can solve to find the final speed:

$v=\sqrt{\frac{2E}{m}}=\sqrt{\frac{2(2.25\cdot 10^5 J)}{58 kg}}=88.1 m/s$

4 0

3 years ago

If car A passes car B, then car A must be

Tcecarenko [31]

Answer:

B. moving faster than car B, but not necessarily accelerating

Explanation:

Velocity is the speed of something. So car A's velocity is greater than car B but does not mean car A is accelerating.

4 0

2 years ago

Each of the following statements is arguably true of thermometers. Which of them is most helpful to keep in mind if you are cond

Lera25 [3.4K]

Answer:

The temperature reported by a thermometer is never precisely the same as its surroundings

Explanation:

In this experiment to determine the specific heat of a material the theory explains that when a heat interchange takes place between two bodies that were having different temperatures at the start, the quantity of heat the warmer body looses is equal to that gained by the cooler body to reach the equilibrium temperature. <u>This is true only if no heat is lost or gained from the surrounding.</u> If heat is gained or lost from the surrounding environment, the temperature readings by the thermometer will be incorrect. The experimenter should therefore keep in mind that for accurate results, the temperature recorded by the thermometer is similar to that of the surrounding at the start of the experiment and if it differs then note that there is either heat gained or lost to the environment.

3 0

3 years ago