It is almost as if each outer planet is a solar system in its own right.<br><br> True or False

1) Beth rode the Flying Saucer ride at her favorite amusement park. The ride hung from a rod and swung back and forth while simu

quester [9]

#1

Since Beth rode the Flying saucer ride which is spinning on its axis while axis is moving to and fro as well, so here Beth will experience two motions in this case

(i) Spinning motion

(ii) Rotational motion

#2

Jill and Scott both traveled for 30 min

So time of motion for both is 0.5 h as we know that 1 h = 60 min

Now from the formula of speed we know that

$speed = \frac{distance}{time}$

Speed of Jill

$v_1 = \frac{5}{0.5} = 10 km/h$

speed of Scott

$v_2 = \frac{10}{0.5} = 20 km/h$

so correct answer is

B) Scott had the faster speed since he rode at 20 k/h while Jill only traveled 10 km/h.

#3

Since we know that <em>acceleration is rate of change in velocity</em>

So here we can say that every motion where velocity changes with time then it is an accelerated motion.

Now change in velocity occurs when either the magnitude will change or its direction will change because here velocity is a vector quantity and it will change with magnitude and direction both.

so here due to oval path the direction of velocity will changes at many points and hence it will be an accelerated motion.

so correct answer is

<em>C) James was right since the car changed direction during the race.</em>

6 0

3 years ago

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A small sphere with mass mcarries a positive chargeqand is attached to one end of a silk fiber of lengthL. The other end of the

Aleksandr-060686 [28]

Answer:

(a): The magnitude of the electric force on the small sphere = $\dfrac{q\sigma}{2\epsilon_o}.$

(b): Shown below.

Explanation:

<u>Given:</u>

m = mass of the small sphere.
q = charge on the small sphere.
L = length of the silk fiber.
$\sigma$ = surface charge density of the large vertical insulating sheet.

When the dimensions of the sheet is much larger than the distance between the charge and the sheet, then, according to Gauss' law of electrostatics, the electric field experienced by the particle due to the sheet is given as:

$\rm E = \dfrac{\sigma}{2\epsilon_o}.$

<em>where,</em>

$\epsilon_o$ is the electrical permittivity of the free space.

The electric field at a point is defined as the amount of electric force experienced by a unit positive test charge, placed at that point. The magnitude electric field at a point and the magnitude of the electric force on a charge q placed at that point are related as:

$\rm F_e=qE.$

Thus, the magnitude of the electric force on the small sphere is given by

$\rm F_e = q\times \dfrac{\sigma }{2\epsilon_o}=\dfrac{q\sigma}{2\epsilon_o}.$

The sheet and the small sphere both are positively charged, therefore, the electric force between these two is repulsive, which means, the direction of the electric force on the sphere is away from the sheet along the line which is perepndicular to the sheet and joining the sphere.

When the sphere is in equilibrium, the tension in the fiber is given by the resultant of the weight of the sphere and the electric force experienced by it as shown in the figure attached below.

According to the fig.,

$\rm \tan \theta = \dfrac{F_e}{W}.$

<em>where,</em>

$\rm F_e$ = electric force on the sphere, acting along left.
$\rm W$ = weight of the sphere, acting vertically downwards.

<em />

$\rm F_e = \dfrac{q\sigma}{2\epsilon_o}\\\\W=mg\\\\Therefore,\\\\\tan\theta = \dfrac{\dfrac{q\sigma}{2\epsilon_o}}{mg}=\dfrac{q\sigma}{2mg\epsilon_o}.\\\Rightarrow \theta=\tan^{-1}\left ( \dfrac{q\sigma}{2mg\epsilon_o}\right ) .$