The longer the lever, the greater the

what is the transfer of heat energy by direct contact. A . convection b. radiation c. thermal or heat and d. conduction

katrin2010 [14]

Answer:

d. conduction

Explanation:

Conduction involves the transfer of electric charge or thermal energy due to the movement of particles. When the conduction relates to electric charge, it is known as electrical conduction while when it relates to thermal energy, it is known as heat conduction.

In the process of heat conduction, thermal energy is usually transferred from fast moving particles to slow moving particles during the collision of these particles. Also, thermal energy is typically transferred between objects that has different degrees of temperature and materials (particles) that are directly in contact with each other but differ in their ability to accept or give up electrons.

Any material or object that allow the conduction (transfer) of electric charge or thermal energy is generally referred to as a conductor. Conductors include metal, steel, aluminum, copper, frying pan, pot, spoon etc.

In conclusion, conduction typically involves the transfer of heat energy by direct contact between two or more conductors such as a pot and electric cooker.

6 0

2 years ago

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Using Newton's third law of motion, explain what happens when you let an untied balloon go.

mash [69]

Answer:

balloon pushes you back

Explanation:

3rd Law: Every action has an equal and opposite reaction

So, when you let go of the balloon it's pushed forward so the balloon pushes you back

7 0

3 years ago

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Consider a cyclotron in which a beam of particles of positive charge q and mass m is moving along a circular path restricted by

Ulleksa [173]

A) $v=\sqrt{\frac{2qV}{m}}$

B) $r=\frac{mv}{qB}$

C) $T=\frac{2\pi m}{qB}$

D) $\omega=\frac{qB}{m}$

E) $r=\frac{\sqrt{2mK}}{qB}$

Explanation:

A)

When the particle is accelerated by a potential difference V, the change (decrease) in electric potential energy of the particle is given by:

$\Delta U = qV$

where

q is the charge of the particle (positive)

On the other hand, the change (increase) in the kinetic energy of the particle is (assuming it starts from rest):

$\Delta K=\frac{1}{2}mv^2$

where

m is the mass of the particle

v is its final speed

According to the law of conservation of energy, the change (decrease) in electric potential energy is equal to the increase in kinetic energy, so:

$qV=\frac{1}{2}mv^2$

And solving for v, we find the speed v at which the particle enters the cyclotron:

$v=\sqrt{\frac{2qV}{m}}$

B)

When the particle enters the region of magnetic field in the cyclotron, the magnetic force acting on the particle (acting perpendicular to the motion of the particle) is

$F=qvB$

where B is the strength of the magnetic field.

This force acts as centripetal force, so we can write:

$F=m\frac{v^2}{r}$

where r is the radius of the orbit.

Since the two forces are equal, we can equate them:

$qvB=m\frac{v^2}{r}$

And solving for r, we find the radius of the orbit:

$r=\frac{mv}{qB}$ (1)

C)

The period of revolution of a particle in circular motion is the time taken by the particle to complete one revolution.

It can be calculated as the ratio between the length of the circumference ( $2\pi r$ ) and the velocity of the particle (v):

$T=\frac{2\pi r}{v}$ (2)

From eq.(1), we can rewrite the velocity of the particle as

$v=\frac{qBr}{m}$

Substituting into(2), we can rewrite the period of revolution of the particle as:

$T=\frac{2\pi r}{(\frac{qBr}{m})}=\frac{2\pi m}{qB}$

And we see that this period is indepedent on the velocity.

D)

The angular frequency of a particle in circular motion is related to the period by the formula

$\omega=\frac{2\pi}{T}$ (3)

where T is the period.

The period has been found in part C:

$T=\frac{2\pi m}{qB}$

Therefore, substituting into (3), we find an expression for the angular frequency of motion:

$\omega=\frac{2\pi}{(\frac{2\pi m}{qB})}=\frac{qB}{m}$

And we see that also the angular frequency does not depend on the velocity.

E)

For this part, we use again the relationship found in part B:

$v=\frac{qBr}{m}$

which can be rewritten as

$r=\frac{mv}{qB}$ (4)

The kinetic energy of the particle is written as

$K=\frac{1}{2}mv^2$

So, from this we can find another expression for the velocity:

$v=\sqrt{\frac{2K}{m}}$

And substitutin into (4), we find:

$r=\frac{\sqrt{2mK}}{qB}$

So, this is the radius of the cyclotron that we must have in order to accelerate the particles at a kinetic energy of K.

Note that for a cyclotron, the acceleration of the particles is achevied in the gap between the dees, where an electric field is applied (in fact, the magnetic field does zero work on the particle, so it does not provide acceleration).

6 0

3 years ago

6. A satellite is orbiting Earth just above its surface. The centripetal force making the satellite follow a circular trajectory

Svetllana [295]

Answer:

Engular velocity: $w=1,24*10^{-3}/s$

Linear velocity: $V=7905 m/s$

The time it takes:

$t=5060s=84min$

Explanation:

The magnitude of the centripetal acceleration can be related to the angular velocity and radius as:

(1) $a=r*w^{2}$

Solving for w:

(2) $w=\sqrt{\frac{a}{r} }$

Replacing a=9,8m/s2 and r=6,375,000m:

(3) $w=\sqrt{\frac{9,8m/s^{2}}{6375000m} }=1,24*10^{-3}/s$

And the angular velocity relates to the linear velocity:

$V=w*r=1,24*10^{-3}/s*6375000m=7905 m/s$

The perimeter of the orbit is:

$P=\pi *2*r=\pi *2*6375000m=40.05*10^{6}m$

The time it takes:

$t=\frac{P}{V} =\frac{40.05*10^{6}m}{7905 m/s}=5060s=84min$

5 0

3 years ago

The two cyclists travel at the same speed on level ground. They approach a low hill and decide to coast up instead of hard pedal

Sergeeva-Olga [200]

Answer:

<u>Option-(E): </u>Bike B; its wheels have a smaller moment of inertia.

Explanation:

The two cyclists travel at the same speed on level ground, as they approach a low hill and decide to coast up instead of hard pedaling. At the top of the hill, as bike "B" will have a larger speed, as compared to the bike "A".Due, to the effect of having smaller moment of inertia, I=m×r².

Having no friction nor air resistance, and all the wheels roll on the ground without slipping. While, the B have larger speed,v due to the fact that bike "B" will have a larger speed,"v" as compared to the bike "A".

3 0

3 years ago