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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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A wheel accelerates from rest to 34.7 rad/s at a rate of 47.0 rad/s^2. Through what angle (in radians) did the wheel turn while
dem82 [27]

12.8 rad

Explanation:

The angular displacement \theta through which the wheel turned can be determined from the equation below:

\omega^2 = \omega_0^2 + 2\alpha\theta (1)

where

\omega_0 = 0

\omega = 34.7\:\text{rad/s}

\alpha = 47.0\:\text{rad/s}^2

Using these values, we can solve for \theta from Eqn(1) as follows:

2\alpha\theta = \omega^2 - \omega_0^2

or

\theta = \dfrac{\omega^2 - \omega_0^2}{2\alpha}

\:\:\:\:= \dfrac{(34.7\:\text{rad/s})^2 - 0}{2(47.0\:\text{rad/s}^2)}

\:\:\:\:= 12.8\:\text{rad}

7 0
2 years ago
In a physics lab, a 0.500-kg cart (Cart A) moving with a speed of 129 cm/s encounters a magnetic collision with a 1.50-kg cart (
omeli [17]

Answer:

58 cm/s

Explanation:

0.5×129=0.5×(-45)+1.5×V

V=58

7 0
3 years ago
An owl weighing 40N is sitting in a tree waiting to dive down to catch a mouse. If the owl's potential energy is 800 J with resp
Natali5045456 [20]

Answer:

<em>h = 20 m</em>

Explanation:

<u>Gravitational Potential Energy</u>

Gravitational potential energy (GPE) is the energy stored in an object due to its vertical position or height in a gravitational field.

It can be calculated with the equation:

U=m.g.h

Where m is the mass of the object, h is the height with respect to a fixed reference, and g is the acceleration of gravity or 9.8 m/s^2.

The weight of an object of mass m is:

W = m.g

Thus, the GPE is:

U=W.h

Solving for h:

\displaystyle h=\frac{U}{W}

The weight of the owl is W=40 N and its GPE is U=800 J.

\displaystyle h=\frac{800}{40}=20

h = 20 m

3 0
3 years ago
As a mercury atom absorbs a photon of
Allisa [31]
The energy absorbed by photon is 1.24 eV.
This is the perfect answer.
8 0
3 years ago
The acceleration of the object at time t = 0.7 s is most nearly equal to which of the following?
VladimirAG [237]

We have that for the Question "the acceleration of the object at time t = 0.7 s is most nearly equal to which of the following?"

  • it can be said that the acceleration of the object at time t = 0.7 s is most nearly equal to the slope of the line connecting the origin and the point where the graph where the graph crosses the 0.7s grid line

From the question we are told

the acceleration of the object at time t = 0.7 s is most nearly equal to which of the following?

Generally the equation for the Force  is mathematically given as

F=\frac{F}{dx}

Therefore

F=-kdx

k=600Nm^{-1}

now

K.E=0.5x ds^2

K.E=600*(-0.1^2)

K.E=3J

Therefore

the acceleration of the object at time t = 0.7 s is most nearly equal to the slope of the line connecting the origin and the point where the graph where the graph crosses the 0.7s grid line

For more information on this visit

brainly.com/question/23379286

6 0
3 years ago
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