All categories

3 years ago

Mechanical energy that has been ‘lost' to friction isn't really lost. It just is no longer in its mechanical form. True or False

Physics

1 answer:

Vadim26 [7]3 years ago

4 0

Answer:

True.

Explanation:

Energy can be defined as the ability (capacity) to do work. The two (2) main types of energy are;

a. Gravitational potential energy (GPE): it is an energy possessed by an object or body due to its position above the earth.

b. Kinetic energy (KE): it is an energy possessed by an object or body due to its motion.

Furthermore, the mechanical energy of a physical object or body is the sum of the potential energy and kinetic energy possessed by the object or body.

Mathematically, it is given by the formula;

Mechanical energy = G.P.E + K.E

Mechanical energy that has been ‘lost' to friction isn't really lost. It just is no longer in its mechanical form. This is ultimately in accordance with the law of conservation of energy, which states that energy cannot be destroyed but can only be converted or transformed from one form to another.

Hence, Mechanical energy that has been ‘lost' to friction isn't really lost but converted into heat energy.

You might be interested in

What is the current flow in amps in a circuit with an emf of 20v and

cluponka [151]

Answer:

voltage ÷ resistance; therefore

20 ÷ 4 = 5 amps

6 0

3 years ago

A girl (mass M) standing on the edge of a frictionless merry-go-round (radius R, rotational inertia I) that is not moving. She t

vladimir1956 [14]

a) $\omega=\frac{-mvR}{I+MR^2}$

b) $v=\frac{-mvR^2}{I+MR^2}$

Explanation:

Since there are no external torques acting on the system, the total angular momentum must remain constant.

At the beginning, the merry-go-round and the girl are at rest, so the initial angular momentum is zero:

$L_1=0$

Later, after the girl throws the rock, the angular momentum will be:

$L_2=(I_M+I_g)\omega +L_r$

where:

$I$ is the moment of inertia of the merry-go-round

$I_g=MR^2$ is the moment of inertia of the girl, where

M is the mass of the girl

R is the distance of the girl from the axis of rotation

$\omega$ is the angular speed of the merry-go-round and the girl

$L_r=mvR$ is the angular momentum of the rock, where

m is the mass of the rock

v is its velocity

Since the total angular momentum is conserved,

$L_1=L_2$

So we find:

$0=(I+I_g)\omega +mvR\\\omega=\frac{-mvR}{I+MR^2}$

And the negative sign indicates that the disk rotates in the direction opposite to the motion of the rock.

The linear speed of a body in rotational motion is given by

$v=\omega r$

where

$\omega$ is the angular speed

r is the distance of the body from the axis of rotation

In this problem, for the girl, we have:

$\omega=\frac{-mvR}{I+MR^2}$ is the angular speed

$r=R$ is the distance of the girl from the axis of rotation

Therefore, her linear speed is:

$v=\omega R=\frac{-mvR^2}{I+MR^2}$

5 0

3 years ago

Which one is the answer ?

ollegr [7]

It’s blurry the pic

But I think that is the second one

6 0

3 years ago

On the coast of Georgia, warm air usually picks up water vapor over the Atlantic Ocean. What will happen when the warm, moist ai

strojnjashka [21]

Answer:

C. The warm air will rise, causing water vapor to condense and form clouds.

Explanation:

8 0

3 years ago

10. A satellites is in a circular orbit around the earth at a height of 360 km above the earth’s surface. What is its time perio

Afina-wow [57]

Answer:

Orbital speed=8102.39m/s

Time period=2935.98seconds

Explanation:

For the satellite to be in a stable orbit at a height, h, its centripetal acceleration V2R+h must equal the acceleration due to gravity at that distance from the center of the earth g(R2(R+h)2)

V2R+h=g(R2(R+h)2)

V=√g(R2R+h)

V= sqrt(9.8 × (6371000)^2/(6371000+360000)

V= sqrt(9.8× (4.059×10^13/6731000)

V=sqrt(65648789.18)

V= 8102.39m/s

Time period ,T= sqrt(4× pi×R^3)/(G× Mcentral)

T= sqrt(4×3.142×(6.47×10^6)^3/(6.673×10^-11)×(5.98×10^24)

T=sqrt(3.40×10^21)/ (3.99×10^14)

T= sqrt(0.862×10^7)

T= 2935.98seconds

4 0

3 years ago