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Marina CMI [18]

1 year ago

9

Newton’s cradle is a contraption where metal balls hang from a frame. When one ball is pulled and released, the collision causes

one ball on the opposite end to be released.
Which type of force can decrease the total momentum of this system?

(1 point)
Responses

A friction force
B strong nuclear force
C electrostatic force
D magnetic force

1 answer:

Marysya12 [62]1 year ago

6 0

Answer: A

Explanation: the principle of friction with swinging and colliding balls.

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A small object of mass 3.82 g and charge -16.5 µC is suspended motionless above the ground when immersed in a uniform electric f

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2271.16N/C upward

Explanation:

The diagram well illustrate all the forces acting on the mass. The weight is acting downward and the force is acting upward in other to balance the weight.since the question says it is motionless, then indeed the forces are balanced.

First we determine the downward weight using

$W=mg\\g=9.81m/s^{2}$

Hence for a mass of 3.82g 0r 0.00382kg we have the weight to be

$W=0.00382kg*9.81m/s^{2}$

$W=0.0375N$

To calculate the electric field,

$E=f/q\\E=0.0375/16.5*10^{-6} \\E=2271.16N/C$

Since the charge on the mass is negative, in order to generate upward force, there must be a like charge below it that is repelling it, Hebce we can conclude that the electric field lines are upward.

Hence the magnitude of the electric force is 2271.16N/C and the direction is upward

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3 years ago

Technician A says the manual lever position (MLP) switch is sometimes called a transmission range switch or a neutral safety swi

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3 years ago

A horizontal pipe contains water at a pressure of 110 kPa flowing with a speed of 1.4 m/s. When the pipe narrows to one half its

Pavel [41]

Answer:

a

$v_2 = 5.6 \ m/s$

b

$P_2 = 80600 \ Pa$

Explanation:

From the question we are told that

The pressure of the water in the pipe is $P_1= 110 \ kPa = 110 *10^{3 } \ Pa$

The speed of the water is $v_1 = 1.4 \ m/s$

The original area of the pipe is $A_1 = \pi \frac{d^2 }{4}$

The new area of the pipe is $A_2 = \pi * \frac{[\frac{d}{2} ]^2}{4} = \pi * \frac{\frac{d^2}{4} }{4} = \pi \frac{d^2}{16}$

Generally the continuity equation is mathematically represented as

$A_1 * v_1 = A_2 * v_2$

Here $v_2$ is the new velocity

So

$\pi * \frac{d^2}{4} * 1.4 = \pi * \frac{d^2}{16} * v_2$

=> $\frac{d^2}{4} * 1.4 = \frac{d^2}{16} * v_2$

=> $d^2 * 1.4 = \frac{d^2}{4} * v_2$

=> $1.4 = 0.25 * v_2$

=> $v_2 = 5.6 \ m/s$

Generally given that the height of the original pipe and the narrower pipe are the same , then we will b making use of the Bernoulli's equation for constant height to calculate the pressure

This is mathematically represented as

$P_1 + \frac{1}{2} * \rho * v_1 ^2 = P_2 + \frac{1}{2} * \rho * v_2 ^2$

Here $\rho$ is the density of water with value $\rho = 1000 \ kg /m^3$

$P_2 = P_1 + \frac{1}{2} * \rho [ v_1^2 - v_2^2 ]$

=> $P_2 = 110 *10^{3} + \frac{1}{2} * 1000 * [ 1.4 ^2 - 5.6 ^2 ]$

=> $P_2 = 80600 \ Pa$

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2 years ago