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3 years ago

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How does work affect energy between objects so it can cause a change in the form of energy? Work transfers energy. Work changes

energy. Work increases energy. Work decreases energy.

2 answers:

Pavel [41]3 years ago

7 0

Answer: Work can transfer energy between objects and cause a change in the amount of total energy. Work can transfer energy between objects and cause a change in the form of energy. ... When a spring is compressed, the energy changes from kinetic to potential.

Aleksandr-060686 [28]3 years ago

6 0

Answer:

work can transfer energy

Explanation:

sense there is a ton of effort in work it can accumulate energy. physical and thermal energy. :)

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Which "spheres" are interacting when water evaporates from plants

Novay_Z [31]

The plant grows in the solid part of earth, the lithosphere. When water evaporates from the plant, it enters the hydrosphere, the portion if earth on kand and in the air that contains water. The atmosphere is part of the hydrosphere.

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3 years ago

Read 2 more answers

A major-league pitcher can throw a ball in excess of 40.1 m/s. If a ball is thrown horizontally at this speed, how much will it

mote1985 [20]

Answer:

The ball will drop 0.881 m by the time it reaches the catcher.

Explanation:

The position of the ball at time "t" is described by the position vector "r":

r = (x0 + v0x · t, y0 + v0y · t + 1/2 · g · t²)

Where:

x0 = initial horizontal position.

v0x = initial horizontal velocity.

t = time.

y0 = initial vertical position.

v0y = initial vertical velocity.

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

When the ball reaches the catcher, the position vector will be "r final" (see attached figure).

The x-component of the vector "r final", "rx final", will be 17.0 m. We have to find the y-component.

Using the equation of the x-component of the position vector, we can calculate the time it takes the ball to reach the catcher (notice that the frame of reference is located at the throwing point so that x0 and y0 = 0):

x = x0 + v0x · t

17.0 m = 0 m + 40.1 m/s · t

t = 17.0 m/ 40. 1 m/s = 0.424 s

With this time, we can calculate the y-component of the vector "r final", the drop of the ball:

y = y0 + v0y · t + 1/2 · g · t²

Initially, there is no vertical velocity, then, v0y = 0.

y = 1/2 · g · t²

y = -1/2 · 9.8 m/s² · (0.424 s)²

y = -0.881 m

The ball will drop 0.881 m by the time it reaches the catcher.

8 0

3 years ago

A pendulum of length L=36.1 cm and mass m=168 g is released from rest when the cord makes an angle of 65.4 degrees with the vert

pychu [463]

(a) -0.211 m

At the beginning the mass is displaced such that the length of the pendulum is L = 36.1 cm and the angle with the vertical is

$\theta=65.4^{\circ}$

The projection of the length of the pendulum along the vertical direction is

$L_y = L cos \theta = (36.1 cm)(cos 65.4^{\circ})=15.0 cm$

the full length of the pendulum when the mass is at the lowest position is

L = 36.1 cm

So the y-displacement of the mass is

$\Delta y = 15.0 cm - 36.1 cm = -21.1 cm = -0.211 m$

(b) 0.347 J

The work done by gravity is equal to the decrease in gravitational potential energy of the mass, which is equal to

$\Delta U = mg \Delta y$

where we have

m = 168 g = 0.168 kg is the mass of the pendulum

g = 9.8 m/s^2 is the acceleration due to gravity

$\Delta y = 0.211 m$ is the vertical displacement of the pendulum

So, the work done by gravity is

$W=(0.168 kg)(9.8 m/s^2)(0.211 m)=0.347 J$

And the sign is positive, since the force of gravity (downward) is in the same direction as the vertical displacement of the mass.

(c) Zero

The work done by a force is:

$W=Fd cos \theta$

where

F is the magnitude of the force

d is the displacement

$\theta$ is the angle between the direction of the force and the displacement

In this situation, the tension in the string always points in a radial direction (towards the pivot of the pendulum), while the displacement of the mass is tangential (it follows a circular trajectory): this means that the tension and the displacement are always perpendicular to each other, so in the formula

$\theta=90^{\circ}, cos \theta = 0$

and so the work done is zero.

5 0

3 years ago

A 500g object falls off a cliff and losers 100 J from its gravitational potential energy store. if the gravitational field stren

ludmilkaskok [199]

Answer : Height, h = 20.4 m

Explanation :

It is given that,

Mass of an object, m = 500 g = 0.5 kg

Gravitational potential energy, PE = 100 J

The Gravitational potential energy is the energy which is possessed due to the height and gravity of an object. It is given as :

PE = m g h

where,

h is the height of the cliff.

$100\ J=0.5\ kg\times 9.8\ m/s^2\times h$

h = 20.40 m

So, the height of the cliff is 20.4 m.

7 0

3 years ago

An object is 10 cm from thé mirror, its height is 1 cm and thé focal length is 5 cm. What is thé distance from thé mirror? S1= _

Viefleur [7K]

Note: I assume the mirror is concave, so that its focal length is positive (it is not specified in the text)

1a) We can use the mirror equation to find the distance of the image from the mirror:
$\frac{1}{f}= \frac{1}{p}+ \frac{1}{q}$
where
$f=5 cm$ is the focal length
$p=10 cm$ is the distance of the object from the mirror
$q$ is the distance of the image from the mirror.

Rearranging the equation, we find
$\frac{1}{q}= \frac{1}{f}- \frac{1}{p}= \frac{1}{5}- \frac{1}{10}= \frac{1}{10 cm}$
so, the distance of the image from the mirror is $q=10 cm$ .

1b) The image height is given by the magnification equation:
$\frac{h_i}{h_o}=- \frac{p}{q}$
where $h_i$ is the heigth of the image and $h_o=1 cm$ is the height of the object. By rearranging the equation and using p and q, we find
$h_i=-h_o \frac{p}{q}=-(1 cm) \frac{10 cm}{10 cm}=-1 cm$
and the negative sign means the image is inverted.

2) As before, we can find the distance of the image from the mirror by using the mirror equation:
$\frac{1}{f}= \frac{1}{p}+ \frac{1}{q}$
Rearranging it, we find
$\frac{1}{q}= \frac{1}{f}- \frac{1}{p}= \frac{1}{2}- \frac{1}{10}= \frac{4}{10 cm}$
so, the distance of the image from the mirror is
$q= \frac{10}{4}cm= 2.5 cm$

3) As before, we find the distance of the image from the mirror by using the mirror equation:
$\frac{1}{f}= \frac{1}{p}+ \frac{1}{q}$
Rearranging it, we find
$\frac{1}{q}= \frac{1}{f}- \frac{1}{p}= \frac{1}{2}- \frac{1}{10}= \frac{4}{10 cm}$
so, the distance of the image from the mirror is
$q= \frac{10}{4}cm= 2.5 cm$

And now we can use the magnification equation to find the image height:
$\frac{h_i}{h_o}=- \frac{p}{q}$
Rearranging it, we find
$h_i=-h_o \frac{p}{q}=-(3cm) \frac{10 cm}{2.5 cm}=-12 cm$
and the negative sign means the image is inverted.

5 0

3 years ago