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Montano1993 [528]
3 years ago
12

State a situation in which force is applied on a body, but no work is done​

Physics
2 answers:
LenaWriter [7]3 years ago
8 0
If you press on your arm force is applied work done is if it moves.
One answer could be if I was to press my hand on a table.
Have a great day!
Yanka [14]3 years ago
3 0
If a force is applied but the object doesn't move, no work is done; if a force is applied and the object moves a distance d in a direction other than the direction of the force, less work is done than if the object moves a distance d in the direction of the applied force.
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HELLLP CORRECT ANSWER IS BRAINIEST!!!!!
HACTEHA [7]

Answer: the answer is D

Explanation:

When we look at an object and see its color, we are seeing all of the light that reflects off of that object. Red objects reflect red light, green objects reflect green light, and so on. But what happens to the rest of the colors that hit that object? They get absorbed!

4 0
3 years ago
Read 2 more answers
A 62.0-kg skier is moving at 6.10 m/s on a frictionless, horizontal, snow-covered plateau when she encounters a rough patch 4.10
7nadin3 [17]

b) 747.3 J

a) 7.88 m/s

Explanation:

b)

We start by solving part b) first.

The internal energy generated when crossing the rough patch is equal (in magnitude) to the work done by the friction force on the skier.

The magnitude of the friction force is:

F_f=\mu mg

where:

\mu=0.300 is the coefficient of friction

m = 62.0 kg is the mass of the skier

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

Therefore, the work done by friction is:

W=-F_f d =-\mu mg d

where

d = 4.10 m is the length of the rough patch

The negative sign is due to the fact that the friction force is opposite to the direction of motion of the skier

Substituting,

W=-(0.300)(62.0)(9.8)(4.10)=-747.3 J

So, the internal energy generated in crossing the rough patch is 747.3 J.

a)

If we take the bottom of the hill as reference level, the initial mechanical energy of the skier is sum of his kinetic energy + potential energy:

E=K_i +U_i = \frac{1}{2}mu^2 + mgh

where

m = 62.0 kg is the mass

u = 6.10 m/s is the initial speed

h = 2.50 m is the height of the hill

After crossing the rough patch, the new mechanical energy is

E'=E+W

where

W = -747.3 J is the work done by friction

At the bottom of the hill, the final energy is just kinetic energy,

E' = K_f = \frac{1}{2}mv^2

where v is the final speed.

According to the law of conservation of energy, we can write:

E+W=E'

So we find v:

\frac{1}{2}mu^2 +mgh+W = \frac{1}{2}mv^2\\v=\sqrt{u^2+2gh+\frac{2W}{m}}=\sqrt{6.10^2+2(9.8)(2.50)+\frac{2(-747.3)}{62.0}}=7.88 m/s

8 0
3 years ago
Jennifer and Jamie have a class assignment to identify an example of climate change and an example of a change in weather. They
viva [34]

Answer:

II only

Explanation:

Hope this helped?

8 0
3 years ago
Two in-phase loudspeakers that emit sound with the same frequency are placed along a wall and are separated by a distance of 5.0
e-lub [12.9K]

Answer:

f = 421.8 Hz

Explanation:

When she moved a distance of 1 m from mid point she observe first destructive interference due to two speakers

so we can say that path difference of sound due to two speakers will be equal to half of the wavelength

so path difference is given as

\Delta L = {3.5^2 + 12^2}^{0.5} - {1.5^2 + 12^2}^{0.5}

so it will be

\Delta L = 12.5 - 12.093

\Delta L = 0.4066

now we know that

\frac{\lambda}{2} = 0.4066

\lambda = 0.813

now frequency of sound is given as

f = \frac{v}{\lambda}

f = \frac{343}{0.813}

f = 421.8 Hz

4 0
3 years ago
In an electron dot diagram, the symbol for an element is used to represent
SIZIF [17.4K]
A the nucleus bruh it can't be anything else
7 0
3 years ago
Read 2 more answers
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