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3 years ago

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Two charges are separated by a distance 'd' and exert a mutual attractive force of 'f' on each other. if the distance is decreas

ed to 1/3 of the original distance, what is the new force between the charges?

1 answer:

Zepler [3.9K]3 years ago

5 0

Peanut butter an jelly crackers are the best

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2 years ago

Read 2 more answers

A ball with a mass of 2000 g is floating on the surface of a pool of water. What is the minimum volume that the ball could have

Doss [256]

Answer:

$2000\; {\rm cm^{3}}$ .

Explanation:

When the ball is placed in this pool of water, part of the ball would be beneath the surface of the pool. The volume of the water that this ball displaced is equal to the volume of the ball that is beneath the water surface.

The buoyancy force on this ball would be equal in magnitude to the weight of water that this ball has displaced.

Let $m(\text{ball})$ denote the mass of this ball. Let $m(\text{water})$ denote the mass of water that this ball has displaced.

Let $g$ denote the gravitational field strength. The weight of this ball would be $m(\text{ball}) \, g$ . Likewise, the weight of water displaced would be $m(\text{water})\, g$ .

For this ball to stay afloat, the buoyancy force on this ball should be greater than or equal to the weight of this ball. In other words:

$\text{buoyancy} \ge m(\text{ball})\, g$ .

At the same time, buoyancy is equal in magnitude the the weight of water displaced. Thus:

$\text{buoyancy} = m(\text{water}) \, g$ .

Therefore:

$m(\text{water})\, g = \text{buoyancy} \ge m(\text{ball})\, g$ .

$m(\text{water}) \ge m(\text{ball})$ .

In other words, the mass of water that this ball displaced should be greater than or equal to the mass of of the ball. Let $\rho(\text{water})$ denote the density of water. The volume of water that this ball should displace would be:

$\begin{aligned}V(\text{water}) &= \frac{m(\text{water})}{\rho(\text{water})} \\ &\ge \frac{m(\text{ball}))}{\rho(\text{water})} \end{aligned}$ .

Given that $m(\text{ball}) = 2000\; {\rm g}$ while $\rho = 1.00\; {\rm g\cdot cm^{-3}}$ :

$\begin{aligned}V(\text{water}) &\ge \frac{m(\text{ball}))}{\rho(\text{water})} \\ &= \frac{2000\; {\rm g}}{1.00\; {\rm g\cdot cm^{-3}}} \\ &= 2000\; {\rm cm^{3}}\end{aligned}$ .

In other words, for this ball to stay afloat, at least $2000\; {\rm cm^{3}}$ of the volume of this ball should be under water. Therefore, the volume of this ball should be at least $2000\; {\rm cm^{3}}\!$ .

3 0

2 years ago

f the absolute pressure in a tank is 128 kPa , determine the pressure head in mm of mercury. The atmospheric pressure is 100 kPa

White raven [17]

Answer:

211 mmHg

Explanation:

Absolute Pressure = Gauge Pressure + Atmospheric pressure

128 = Gauge Pressure + 100

Gauge Pressure = 28 KPa = 28 × 10³ Pa

Also Gauge Pressure = ρgh

ρ = density = 13550 kg/m³

g = acceleration due to gravity = 9.8 m/s²

h = pressure head = ?

28 × 10³ = 13550 × 9.8 × h

h = 28000/(13550 × 9.8)

h = 0.211 m = 211 mm

5 0

3 years ago

A student carries a very heavy backpack. To lift the backpack off the ground, the student must apply 80 N of force to do so. The

Xelga [282]

The work done on the backpack by the student applies 80 N of force to lift the backpack 1.5 m is 120J.

<h3>How to calculate work done?</h3>

Work done is a measure of energy expended in moving an object; most commonly, force times distance.

It is said that no work is done if the object does not move, hence, the work done on an object can be calculated as follows:

Work done = Force × Distance

According to this question, a student carries a very heavy backpack and to lift the backpack off the ground, the student must apply 80 N of force to lift the backpack 1.5 m.

Work done = 80N × 1.5m

Work done = 120J

Therefore, the work done on the backpack by the student applies 80 N of force to lift the backpack 1.5 m is 120J.

Learn more about work done at: brainly.com/question/28172139

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2 years ago

The longest passenger liner ever built was the France, 66,348 tons and 315.5 long. Suppose its front end passes the edge of a pi

Kaylis [27]

Answer:

The back end of the vessel will pass the pier at 4.83 m/s

Explanation:

This is purely a kinetics question (assuming we're ignoring drag and other forces) so the weight of the boat doesn't matter. We're given:

Δd = 315.5 m

vi = 2.10 m/s

a = 0.03 m/s^2

vf = ?

The kinetics equation that incorporates all these variables is:

vf^2 = vi^2 + 2aΔd

vf = √(2.1^2 + 2(0.03)(315.5))

vf = 4.83 m/s

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3 years ago