A large crate sits on the floor of a warehouse. Paul and Bob apply constant horizontal forces to the crate. The force applied by
Paul has magnitude 48.0 N and direction 61.0∘ south of west. How much work does Paul’s force do during a displacement of the crate that is 12.0 m in the direction 22.0∘ east of north?
1 answer:
Answer:
-570.4 J
Explanation:
Parameters given:
Force = 48N, 61° SW
Displacement = 12m, 22.0 NE
Work done is given by:
W = F*d*cosθ
To find θ:
θ = 61 + 90 + 22 = 172°
=> W = 48 * 12 * cos172
W = - 570.4 J
Hence, the work to be done is - 570.4J.
This value is negative because the applied force is opposite the direction of the displacement.
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The force of attraction between two objects can be illustrated using Newton's Law of Universal Gravitation.
The relation between the different parameters is shown in the attached image.
Now, from the relation, we can deduce that the force between the two objects is directly proportional to the masses of the two objects.
This means that, if the mass of one object is doubled, then the force between the two objects will also be doubled.
The relation between the different parameters is shown in the attached image.
Now, from the relation, we can deduce that the force between the two objects is directly proportional to the masses of the two objects.
This means that, if the mass of one object is doubled, then the force between the two objects will also be doubled.
Answer:
a) y₂ = 49.1 m , t = 1.02 s , b) y = 49.1 m , t= 1.02 s
Explanation:
a) We will solve this problem with the missile launch kinematic equations, to find the maximum height, at this point the vertical speed is zero
² =
² - 2 g (y –yo)
The origin of the coordinate system is on the floor and the ball is thrown from a height
y-yo = =
- g t
t = / g
t = 10 / 9.8
t = 1.02 s
b) the maximum height
y- 44.0 = ² / 2 g
y - 44.0 = 5.1
y = 5.1 +44.0
y = 49.1 m
The time is the same because it does not depend on the initial height
t = 1.02 s