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While an object is moving at a constant 20 m/s, a 5 N forcepushes the object to the left. At the same time, a 5 N force ispushin

Nostrana [21]

Answer: 20m/s.

Explanation:

Remember the second Newton's law:

F = a*m

This is:

The net force acting on an object is equal to the mass of the object times the acceleration of the object.

In this case, we have a force of 5N pushing the object to the right.

We also have a force of 5N pushing the object to the left.

These forces act on opposite directions.

Then the net force will be equal to the difference of these forces, this is:

F = 5N - 5N = 0N

Then the net force is 0N, then we have:

0N = m*a

0N/m = 0m/s^2 = a

This means that the acceleration of the object is 0, then the velocity of the object does not change.

This means that if the object was moving at a constant velocity of 20m/s, the velocity of the object will still be equal to 20m/s. (because the net force acting on the object is zero)

5 0

3 years ago

The same fluid flows through four different branching pipes. It enters each pipe from the left with the same speed, v0, and flow

Anvisha [2.4K]

Answer:

The correct option is A.

Explanation:

Following the equation of continuum, AV remains constant.

Case a

(3A)(V0) = AV1 + AV1 + AV1

3AV0 = 3AV1

V1 = V0

Case b

(A)(V0) = (A/3)V2 + (A/3)V2 + (A/3)V2 + (A/3)V2

AV0 = 4V2/3

V2 = 3/4V0

Case c

(A/2)(V0) = AV3 + AV3 + AV3

AV0/2 = 3AV3

V3 = V0/6

Case d

(3A)(V0) = 2AV4 + 2AV4

3AV0 = 4AV4

V4 = 3V0/4

Comparing all the flow speeds, V1 is the largest.

Thus, the correct option is A.

6 0

3 years ago

The suspension system of a 1700 kg automobile "sags" 7.7 cm when the chassis is placed on it. Also, the oscillation amplitude de

spin [16.1K]

Answer:

the spring constant k = $5.409*10^4 \ N/m$

the value for the damping constant $\\ \\b = 1.518 *10^3 \ kg/s$

Explanation:

From Hooke's Law

$F = kx\\\\k =\frac{F}{x}\\\\where \ F = mg\\\\k = \frac{mg}{x}\\\\given \ that:\\\\mass \ of \ each \ wheel = 425 \ kg\\\\x = 7.7cm = 0.077 m\\\\g = 9.8 \ m/s^2\\\\Then;\\\\k = \frac{425 \ kg * 9.8 \ m/s^2}{0.077 \ m}\\\\k = 5.409*10^4 \ N/m$

Thus; the spring constant k = $5.409*10^4 \ N/m$

The amplitude is decreasing 37% during one period of the motion

$e^{\frac{-bT}{2m}}= \frac{37}{100}\\\\e^{\frac{-bT}{2m}}= 0.37\\\\\frac{-bT}{2m} = In(0.37)\\\\\frac{-bT}{2m} = -0.9943\\\\b = \frac{2m(0.9943)}{T}\\\\b = \frac{2m(0.9943)}{\frac{2 \pi}{\omega}}\\\\b = \frac{m(0.9943) \ ( \omega) )}{ \pi}$

$b = \frac{m(0.9943)(\sqrt{\frac{k}{m})}}{\pi}\\\\b = \frac{425*(0.9943)(\sqrt{\frac{5.409*10^4}{425}) } }{3.14}\\\\b = 1518.24 \ kg/s\\\\b = 1.518 *10^3 \ kg/s$

Therefore; the value for the damping constant $\\ \\b = 1.518 *10^3 \ kg/s$

5 0

3 years ago

Describe liquids and GASSES IN TERMS OF THERE VOLUME AND SHAPES

antiseptic1488 [7]

Answer:

Explanation:

liquids have definite volume

liquids do not have definite shape. The take the shape of the container in which they are kept.

gases do not have definite volume.

gases do not have definite shape. They take the shape of the container in which they are kept.

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3 0

3 years ago

Read 2 more answers

a pebble is dropped down a well and hits the water 1.5 seconds later. using the equations for motion with constant acceleration,

Effectus [21]

Let h = the distance from the edge of the wall to the water surface (m).

Use g = 9.8 m/s² and neglect air resistance.

The initial vertical velocity of the pebble is zero.
Because the pebble hits the surface of the water after 1.5 s, therefore
h = (1/2)*(9.8 m/s²)*(1.5 s)² = 11.025 m

Answer: 11.025 m

7 0

4 years ago

Read 2 more answers