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Hatshy [7]
3 years ago
14

A student lifts a set of books off a table and places them in the upper shelf of a book case which is 2 meters above the table.

If she applies a force of 5 newtons to lift these books, what is the work done?
0 joules
2.5 joules
-2.5 joules
10 joules
-10 joules
Physics
2 answers:
Vilka [71]3 years ago
7 0

Answer:

D.10 joules

Explanation:

AfilCa [17]3 years ago
6 0
The work done is the product between the intensity of the force applied F, the amount of the displacement d of the book and the cosine of the angle \theta between the direction of the force and the direction of the displacement:
W=Fd \cos \theta
In our problem, the student is lifting the book, so he is applying a force directed upward, and the book is moving upward, so F and d are parallel and therefore the angle is zero, so \cos \theta = \cos 0=1
Therefore, the work done is
W=Fd=(5 N)(2 m)=10 J
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Answer:

Answer  for the question is given in the attachment.

Explanation:

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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

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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

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actual velocity is

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

5 0
3 years ago
Two long, straight wires are separated by a distance of 9.15 cm . One wire carries a current of 2.79 A , the other carries a cur
Dafna1 [17]

Answer:

The force is the same

Explanation:

The force per meter exerted between two wires carrying a current is given by the formula

\frac{F}{L}=\frac{\mu_0 I_1 I_2}{2\pi r}

where

\mu_0 is the vacuum permeability

I_1 is the current in the 1st wire

I_2 is the current in the 2nd wire

r is the separation between the wires

In this problem

I_1=2.79 A\\I_2=4.36 A\\r = 9.15 cm = 0.0915 m

Substituting, we find the force per unit length on the two wires:

\frac{F}{L}=\frac{(4\pi \cdot 10^{-7})(2.79)(4.36)}{2\pi (0.0915)}=2.66\cdot 10^{-5}N

However, the formula is the same for the two wires: this means that the force per meter exerted on the two wires is the same.

The same conclusion comes out  from Newton's third law of motion, which states that when an object A exerts a force on an object B, then object B exerts an equal and opposite force on object A (action-reaction). If we apply the law to this situation, we see that the force exerted by wire 1 on wire 2 is the same as the force exerted by wire 2 on wire 1 (however the direction is opposite).

3 0
3 years ago
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Gnoma [55]

Answer:

Option 5. 1 and 3

Solution:

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The ball after it left follows the path of trajectory and as it moves forward in the horizontal direction the force of the air acts on it.

In the whole projectile motion of the ball, the acceleration due to gravity acts on the ball thus the force of gravity acts on the ball in the downward direction  before it hits the ground.

6 0
3 years ago
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Kitty [74]

Explanation:

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3 0
3 years ago
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