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bekas [8.4K]
3 years ago
9

"a 10 kg rock is pushed off the edge of a bridge 50 meters above the ground. what was the kinetic energy of the rock at the midw

ay point of its fall?"
Physics
1 answer:
valentinak56 [21]3 years ago
8 0
Let's call h the initial height of the rock (h=50 m, the height of the bridge).

Initially, the rock has only gravitational potential energy, which is given by
U_i=mgh
where m=10 kg is the mass of the rock while g is the gravitational acceleration. So, the total mechanical energy of the rock at this point is
E_i = U_i = mgh=(10 kg)(9.81 m/s^2)(50 m)=4905 J

At midway point of its fall, its height is \frac{h}{2}, so its potential energy is
U_f = mg \frac{h}{2} = (10 kg)(9.81 m/s^2)(25 m)=2452.5 J
But now the rock is also moving by speed v, so it also has kinetic energy:
K_f =  \frac{1}{2}mv^2
So the total energy at the midway point of the fall is
E_f = U_f + K_f (1)

The mechanical energy must be conserved, so E_i = E_f, so we can rewrite (1) and solve it to find the kinetic energy of the rock at midway point of its fall:
E_i = U_f + K_f
K_f = E_i - U_f = 4905 J - 2452.5 J=2452.5 J
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irga5000 [103]
We are asked to solve and determine the magnitude of the current flowing through the first device. In order for us to have a better understanding of the problem, we can refer to the attached picture which contains electric circuit diagram. Since it the problem we are already given with an electromotive source or the voltage supply and since the two resistance is in parallel, it would clearly mean that the voltage drop in each resistance is just the same. The resistance 1 uses the 40 volts at the same time the resistance 2 uses 40 volts also. Solving further for the current, we can apply Ohm's law which V = IR where "V" represents the voltage, the "I" represents the current and "R" represents the resistance.

Such as the solution in obtaining current is shown below:
I = V / R, substitute values we have it
I = 40 volts / 1208 ohms
I = 0.0331 Amperes

Therefore, the current flowing in the first device is 0.033 Amperes or 33 milliAmperes.

6 0
3 years ago
What is the student's kinetic energy at the bottom of the hill if he is moving
soldi70 [24.7K]

Answer:

KE = 10530 J or 10.53 KJ

Explanation:

The formula for kinetic energy is KE = 1/2 mv^2

Let's apply the formula:

KE = 1/2 mv^2

KE = 1/2 (65kg) (18m/s)^2

KE = 10530 J or 10.53 KJ

5 0
3 years ago
A ball is thrown straight up with a speed of 30
konstantin123 [22]

Answer:

A ball is thrown straight up with a speed of 30

m/s. What is the maximum height reached by

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4 0
2 years ago
A mass-spring system has k = 56.8 N/m and m = 0.46 kg.
andriy [413]

Answer:

A. \omega=11.1121\ rad.s^{-1}

B. f=1.7685\ Hz

C. T=0.5654\ s

Explanation:

Given:

  • spring constant, k=56.8\ N.m^{-1}
  • mass attached, m=0.46\ kg

A)

for a spring-mass system the frequency is given as:

\omega=\sqrt{\frac{k}{m} }

\omega=\sqrt{\frac{56.8}{0.46}}

\omega=11.1121\ rad.s^{-1}

B)

frequency is given as:

f=\frac{\omega}{2\pi}

f=\frac{11.1121}{2\pi}

f=1.7685\ Hz

C)

Time period of a simple harmonic motion is given as:

T=\frac{1}{f}

T=0.5654\ s

7 0
3 years ago
Can a body have increasing velocity with decreasing acceleration?
Sergio039 [100]
Sure.  The acceleration may be decreasing, but as long as it stays
in the same direction as the velocity, the velocity increases.

I think you meant to ask whether the body can have increasing velocity
with negative acceleration.  That answer isn't simple either.

If the body's velocity is in the positive direction, then positive acceleration
means speeding up, and negative acceleration means slowing down.

BUT ... If the body's velocity is in the negative direction, then positive
acceleration means slowing down, and negative acceleration means
speeding up.

I know that's confusing. 

-- Take a piece of scratch paper, write a 'plus' sign at one edge and
a 'minus' sign at the other edge.  Those are the definitions of which
direction is positive and which direction is negative. 

-- Then sketch some cars ... one traveling in the positive direction, and
one driving in the negative direction.  Those are the directions of the
velocities.

-- Now, one car at a time:
. . . . . first push on the back of the car, in the direction it's moving;.
. . . . . then push on the front of the car, against its motion.
Each push causes the car to accelerate in the direction of the push.

When you see it on paper, all the positive and negative velocities
and accelerations will come clear for you.
3 0
3 years ago
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