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antoniya [11.8K]

3 years ago

6

Thomson observed that cathode beams can pass undeflected through crossed electric and magnetic fields. Which of the following qu

antities must then be common to the particles making up these beams? A. Mass B. Size C. Magnitude of charge D. Sign of charge E. Sign and magnitude of charge F. Velocity

1 answer:

user100 [1]3 years ago

3 0

Answer:

e must have the charge of the particle

Explanation:

In this experiment, the particles that the cathode rays are found where from an electric field and another magnetic field, for the particle to pass without deviating the magnetic force must have the same modulus of the entricar force, and be in the opposite direction.

the expression for the magnetic force is

F_m = q v xB

the expression for the electric force is

F_e = q E

when warning the check two formulas we must have the charge of the particle

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You pull down on a rope to raise a flag on a flag pole which describes why the rope and the flag pole act as a machine

ikadub [295]

Answer: A-they change the direction of the forces

Explanation:

Any device which makes our work simpler is known as a machine. A machine may increase the amount of output force which is determined by its mechanical advantage or a machine changes the direction of the force.

In the given situation the input force acts in the downward direction to pull the rope. The output force acts on the flag being pulled upwards at the other end. Thus, the direction of the forces varies. Correct option is A.

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4 years ago

A person in bare feet is standing under a tree during a thunderstorm, seeking shelter from the rain. A lightning strike hits the

Anni [7]

Answer:

$I=38.181\ A$ is the current through the body of the man.

$E=34.5\ J$ energy dissipated.

Explanation:

Given:

time for which the current lasted, $t=43\times 10^{-6}\ s$

potential difference between the feet, $V=21000\ V$

resistance between the feet, $R=550\ \Omega$

<u>Now, from the Ohm's law we have:</u>

$I=\frac{V}{R}$

$I=\frac{21000}{550}$

$I=38.181\ A$ is the current through the body of the man.

<u>Energy dissipated in the body:</u>

$E=I^2.R.t$

$E=38.181^2\times 550\times 43\times 10^{-6}$

$E=34.5\ J$

5 0

4 years ago

Upon its return voyage from a space mission, the spacecraft has a velocity of 25000 km/h at point A, which is 7130 km from the c

Nadya [2.5K]

Answer:

the velocity at point B is 27120.08 km/hr

Explanation:

check the picture attached for explanation

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4 years ago

1. A heat engine operates between two reservoirs at T2 = 600 K and T1 = 350 K. It takes in 1.00 x 103 J of energy from the highe

Alex73 [517]

Answer:

$\Delta S_u=2.1429\ J.K^{-1}$

$W_c=416.67\ J$

Explanation:

Given:

temperature of source reservoir, $T_H=600\ K$

temperature of sink reservoir, $T_L=350\ K$

energy absorbed from the source, $Q_{in}=1000\ J$

work done, $W=250\ J$

a.

<u>Now change in entropy of the surrounding:</u>

$\Delta S_u=\frac{dQ_L}{T_L}$

<em>Since heat engine is a device that absorbs heat from a high temperature reservoir and does some work giving out heat in the universe as the byproduct.</em>

$\Delta S_u=\frac{Q_H-W}{T_L}$

$\Delta S_u=\frac{1000-250}{350}$

$\Delta S_u=2.1429\ J.K^{-1}$

b.

<u>We know Carnot efficiency is given as:</u>

$\eta_c=1-\frac{T_L}{T_H}$

$\eta_c=1-\frac{350}{600}$

$\eta_c=0.4167$

<u>Now the Carnot work done:</u>

$W_c=Q_H\times \eta_c$

$W_c=1000\times 0.4167$

$W_c=416.67\ J$ .......................(1)

c.

From eq. (1) we have the Carnot work, so the difference:

$\Delta W=W_c-W$

$\Delta W=416.67-250$

$\Delta W=166.67\ J$

Now, we find:

$T_L.\Delta S_u=350\times 2.1429$

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3 years ago

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ELEN [110]

Answer:

Explanation:

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3 years ago