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Zepler [3.9K]
2 years ago
11

The following graph shows the force exerted on and the displacement of object being pulled

Physics
1 answer:
Tomtit [17]2 years ago
3 0

The work done to pull the object 7.0 m is the total area under the graph from 0.0 m to 7.0 m, determined as 245 J.

<h3>Work done by the applied force</h3>

The area under force versus displacement graph is work done.

The total work done by pulling the object 7 m, can be grouped into two areas;

  • First area, A1 = area of triangle from 0 m to 2.0 m
  • Second area, A2 = area of trapezium, from 2.0 m to 7.0 m

A1 = ¹/₂ bh

A1 = ¹/₂ x (2) x (20)

A1 = 20 J

A2 = ¹/₂(large base + small base) x height

A2  = ¹/₂[(7 - 2) + (7-3)] x 50

A2 = ¹/₂(5 + 4) x 50

A2 = 225 J

<h3>Total work done </h3>

W = A1 + A2

W = 20 J + 225 J

W = 245 J

Learn more about work done here: brainly.com/question/8119756

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Answer:

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Explanation:

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Furthermore, a graph is made to see the production function.

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3 years ago
A 0.690 kg snowball is fired from a cliff 8.25 m high with an initial velocity of 10.3 m/s, directed 32.0° above the horizontal.
Lana71 [14]

Answer:

a) 16.4 m/s

b) 16.4 m/s

c) 16.4 m/s

Explanation:

a)

m = mass of the snowball = 0.690 kg

h = height of the cliff = 8.25 m

v₀ = initial speed of ball at the time of launch = 10.3 m/s

v = speed of the ball as it reach the ground

Using conservation of energy

initial kinetic energy + initial potential energy at the cliff = final kinetic energy just before reaching the ground

(0.5) m v₀² + mgh = (0.5) m v²

(0.5) v₀² + gh = (0.5) v²

(0.5) (10.3)² + (9.8 x 8.25) = (0.5) v²

v = 16.4 m/s

b)

As the launch angle is changed, the speed of the ball just before reaching the ground remain the same as the final speed does not depend on the angle of launch.

v = 16.4 m/s

c)

As the mass is changed, the speed of the ball just before reaching the ground remain the same as the final speed does not depend on the mass of the ball.

v = 16.4 m/s

7 0
4 years ago
A car accelerates from 21.0 m/s to 46.0 m/s in 2.5 seconds. What is the acceleration of the car.
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Answer:

10.0 m/s^2

Explanation:

The acceleration of an object is given by

a=\frac{v-u}{t}

where

v is the final velocity

u is the initial velocity

t is the time interval it takes for the velocity to change from u to v

For the car in this problem,

u = 21.0 m/s

v = 46.0 m/s

t = 2.5 s

Substituting, we find the acceleration

a=\frac{46.0-21.0}{2.5}=10.0 m/s^2

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Answer:

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