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Leviafan [203]
3 years ago
6

An archer shoots an arrow with a mass of 45.0 grams from bow pulled

Physics
1 answer:
Sladkaya [172]3 years ago
3 0

Answer:

The force the archer need to pull in order to achieve the height is approximately 101.8 N

Explanation:

By energy conservation principle, puling an elastic bow with a force, for a given distance, performs work which is converted to the potential energy of the arrow at height

The given parameters are;

The mass of the arrow, m = 45.0 grams = 0.045 kg

The distance the elastic bow is pulled, d = 65.0 cm = 0.65 m

The height at which the arrow is reaches, h = 150.0 meters

Let 'F', represent the force the archer need to pull in order to achieve the height

Work done, W = Force × Distance moved in the direction of the force

Therefore;

The work done in pulling the arrow, W = F × d

By energy conservation, we have;

The work done in pulling the arrow, W = The potential energy gained by the arrow, P.E.

W = P.E.

The potential energy gained by the arrow, P.E. = m·g·h

Where;

m = The mass of the arrow

g = The acceleration due to gravity = 9.8 m/s²

h = The height the arrow reaches

∴ by plugging in the values, P.E. = 0.045 kg ×9.8 m/s² × 150 m = 66.15 J

W = F × d = F × 0.065 m

Also, W = P.E. = 66.15 J

∴ W = F × 0.065 m = 66.15 J

F × 0.065 m = 66.15 J

F = 66.15 J/(0.65 m) = 1323/13 N ≈ 101.8 N

The force the archer need to pull in order to achieve the height, F ≈ 101.8 N.

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A green truck is moving to the right. A red truck is moving to the left with a speed of 6 m/s. The mass of the red truck is 1,00
Contact [7]

Answer:

2 m/s

Explanation:

m_1 = Mass of red truck = 1000 kg

m_2 = Mass of green truck= 3000 kg

u_1 = Initial Velocity of red truck = 6 m/s

u_2 = Initial Velocity of green truck

v = Velocity with which they move together = 0

For elastic collision

m_1u_1 + m_2u_2 =(m_1 + m_2)v\\\Rightarrow m_1u_1 + m_2u_2 =0\\\Rightarrow m_1u_1 + m_2u_2 =(m_1 + m_2)v\\\Rightarrow m_1u_1 + m_2u_2 =0\\\Rightarrow u_2=-\frac{m_1u_1}{m_2}\\\Rightarrow u_2=-\frac{1000\times 6}{3000}\\\Rightarrow u_2=-2\ m/s

Velocity of the green truck is 2 m/s

8 0
3 years ago
Q11) If you were standing at the top of a building and you dropped a rock.
Dafna1 [17]

Answer:

Part A

The distance travel by the rock is approximately 132.496 m

Part B

The speed when the rock hits the ground is approximately 50.96 m/s

Explanation:

Part A

The question is focused on the kinetics equation of a free falling object

The given parameter is the time it takes the rock to hit the ground, t = 5.2 s

For an object in free fall, we have;

h = 1/2·g·t²

Where;

h = The height from which the object is dropped

g = The acceleration due to gravity ≈ 9.8 m/s²

t = The time taken to travel the distance, h = 5.2 s

∴ h = 1/2 × 9.8 m/s² × (5.2 s)² ≈ 132.496 m

The distance travel by the rock, h ≈ 132.496 m

Part B

The speed, 'v', when the rock hits the ground, is given by the following kinematic equation,

v = g·t

∴ v = 9.8 m/s² × 5.2 s = 50.96 m/s

The speed when the rock hits the ground, v ≈ 50.96 m/s.

8 0
2 years ago
A 544 g ball strikes a wall at 14.3 m/s and rebounds at 14.4 m/s. The ball is in contact with the wall for 0.042 s. What is the
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Answer:

F = 371.738\,N

Explanation:

Let assume that ball strikes a vertical wall in horizontal direction. The situation can be modelled by the appropriate use of the definition of Moment and Impulse Theorem, that is:

(0.544\,kg)\cdot (14.3\,\frac{m}{s})-F\cdot \Delta t = -(0.544\,kg)\cdot (14.4\,\frac{m}{s} )

F\cdot \Delta t = 15.613\,\frac{kg}{m\cdot s}

The average force acting on the ball during the collision is:

F = \frac{15.613\,\frac{kg}{m\cdot s} }{0.042\,s}

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3 years ago
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