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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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4 0
3 years ago
The pilot of an aircraft wishes to fly due west in a 33.9 km/h wind blowing toward the south. The speed of the aircraft in the a
diamong [38]

Answer:\theta =9.96^{\circ} North of west

Explanation:

Given

Plane wishes to fly in west

but wind with speed 33.9 km/h towards south obstructing its path

so plane must fly at an angle of \theta w.r.t west such that it final velocity is towards west

Plane absolute speed=195 km/h

To fly towards west velocity in Y direction should be zero

thus 195sin\theta =33.9

\theta =9.96^{\circ}

so Plane should head towards 9.96^{\circ} North of west in order to fly in west.

So plane

actual velocity is

v=-195cos9.96\hat{i}+195sin9.96\hat{j}

5 0
3 years ago
On a perfect fall day, you are hovering at low altitude in a hot-air balloon, accelerated neither upward nor downward. The total
Stella [2.4K]

Explanation:

Since the balloon is not accelerating means that the net force on the balloon is zero. This implies that the weight of balloon must be equal to the buoyant force on balloon.

Hence, the buoyant force equals the weight of air displaced by the balloon, also 20,000 N.

Weight of the air displaced = density of air × volume

The density of air at 1 atm pressure and 20º C is 1.2 kg/m³  

the volume V = 20,000/(1.2×9.8) =  1700 m³

3 0
3 years ago
A small 18 kilogram canoe is floating downriver at a speed of 1 m/s. What is the canoe's kinetic energy?
In-s [12.5K]

Kinetic energy = (1/2) (mass) (speed²).

A Physicist in the canoe, or on a raft floating downriver next to the canoe, will say that the canoe's kinetic energy is zero.

A Physicist on the riverbank, watching the canoe drift by at 1 m/s, will say that its kinetic energy is 9 Joules.

They're both correct.

8 0
3 years ago
A water balloon is thrown horizontally from a tower that is 45 m high. It strikes the shoes of an unsuspecting passerby who is 4
LekaFEV [45]

Answer:

14.85 m/s

Explanation:

From the question given above, the following data were obtained:

Height (h) of tower = 45 m

Horizontal distance (s) moved by the balloon = 45 m

Horizontal velocity (u) =?

Next, we shall determine the time taken for the balloon to hit the shoe of the passerby. This is illustrated below:

Height (h) of tower = 45 m

Acceleration due to gravity (g) = 9.8 m/s²

Time (t) =?

h = ½gt²

45 = ½ × 9.8 × t²

45 = 4.8 × t²

Divide both side by 4.9

t² = 45/4.9

Take the square root of both side

t = √(45/4.9)

t = 3.03 s

Finally, we shall determine the magnitude of the horizontal velocity of the balloon as shown below:

Horizontal distance (s) moved by the balloon = 45 m

Time (t) = 3.03 s

Horizontal velocity (u) =?

s = ut

45 = u × 3.03

Divide both side by 3.03

u = 45/3.03

u = 14.85 m/s

Thus, the magnitude of the horizontal velocity of the balloon was 14.85 m/s

4 0
3 years ago
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