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Vadim26 [7]
3 years ago
13

A 0.5 kg object, initially at rest, is pulled to the right along a frictionless horizontal surface by a constant horizontal forc

e of 25 N for a
distance of 0.5m.
a. What is the work done by the force?
b. What is the change in the kinetic energy of the block?
c. What is the speed of the block after the force is removed?
Help asap plz
Physics
1 answer:
sergij07 [2.7K]3 years ago
6 0

Answer:

A.) 12.5 J

B.) 12.5 J

C.) 7.1 m/s

Explanation:

Given that a 0.5 kg object, initially at rest, is pulled to the right along a frictionless horizontal surface by a constant horizontal force of 25 N for a distance of 0.5m.

a. What is the work done by the force?

Work done = force × distance

Work done = 25 × 0.5

Work done = 12.5 J

b. What is the change in the kinetic energy of the block?

Work done = energy

Change in Kinetic energy = work done

Change in kinetic energy = 12.5 J

c. What is the speed of the block after the force is removed?

Kinetic energy = 1/2mV^2

12.5 = 1/2 × 0.5 × V^2

25 = 0.5V^2

V^2 = 25/0.5

V^2 = 50

V = 7.1 m/s

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An airplane is moving at 350 km/hr. If a bomb is
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Answers:

a) -171.402 m/s

b) 17.49 s

c) 1700.99 m

Explanation:

We can solve this problem with the following equations:

y=y_{o}+V_{oy}t-\frac{1}{2}gt^{2} (1)

x=V_{ox}t (2)

V_{f}=V_{oy}-gt (3)

Where:

y=0 m is the bomb's final jeight

y_{o}=1.5 km \frac{1000 m}{1 km}=1500 m is the bomb'e initial height

V_{oy}=0 m/s is the bomb's initial vertical velocity, since the airplane was moving horizontally

t is the time

g=9.8 m/s^{2} is the acceleration due gravity

x is the bomb's range

V_{ox}=350 \frac{km}{h} \frac{1000 m}{1 km} \frac{1 h}{3600 s}=97.22 m/s is the bomb's initial horizontal velocity

V_{f} is the bomb's fina velocity

Knowing this, let's begin with the answers:

<h3>b) Time</h3>

With the conditions given above, equation (1) is now written as:

y_{o}=\frac{1}{2}gt^{2} (4)

Isolating t:

t=\sqrt{\frac{2 y_{o}}{g}} (5)

t=\sqrt{\frac{2 (1500 m)}{9.8 m/s^{2}}} (6)

t=17.49 s (7)

<h3>a) Final velocity</h3>

Since V_{oy}=0 m/s, equation (3) is written as:

V_{f}=-gt (8)

V_{f}=-(97.22)(17.49 s) (9)

V_{f}=-171.402 m/s (10) The negative sign ony indicates the direction is downwards

<h3>c) Range</h3>

Substituting (7) in (2):

x=(97.22 m/s)(17.49 s) (11)

x=1700.99 m (12)

5 0
3 years ago
hello friends,i need your help my home work now in physics,topic:motion.40 marks +brainliest if correct .​
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Answer:

See below

Explanation:

Vertical position is given by

df = do + vo t - 1/2 a t^2      df = final position = 0 (on the ground)

                                           do =original position = 2 m

                                            vo = original <u>VERTICAL</u> velocity = 0

                                            a = acceleration of gravity = 9.81 m/s^2

THIS BECOMES

0 = 2 + 0 * t  - 1/2 ( 9.81)t^2

  to show t =<u> .639 seconds to hit the ground </u>

During this .639 seconds it flies horizontally at 10 m/s for a distance of

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An overhead electric power line carries a maximum current of 125 A. What is the magnitude of the maximum magnetic field at a poi
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Answer:

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Explanation:

B= μoI/(2πr)

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r: distance from the wire

I : current in the wire

B= (4π×10^(-7)×125)/(2π×4.5)

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Answer:

each resistor draws 1/3 of an amp or 0.33333 amps

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