What is a force that opposes motion between two surfaces that are in contact?

The plates on a capacitor have a radius of 1.5 mm and are separated by a distance of 0.43 mm. The space between the plates is fi

zhenek [66]

So, C = kE°A/d

putting the values,

C
= 3.8 × 8.85×10^(-12) × 3.14×1.5×1.5 × 10^(-6)/0.43 × 10^(-3)

so, 1.02 × 10^(-13)

so the most appropriate answer is 2 ...that is
1.4 × 10^(-13) ....answer !!

3 0

3 years ago

In a standard tensile test a steel rod of 22-mm diameter is subjected to a tension force of 75kN. Knowing that v = 0.30 and E =

VladimirAG [237]

Answer:

(a) Elongation of rod = 0.19732 mm

(b) Change in diameter = 0.00651 mm

Explanation:

Circular area at end of steel rod = pi * diameter^2 / 4

$Area = pi * (22 * 10^-^3)/4$

Area = 3.801 * 10^(-4) meter squared

Stress = force / area

Stress = 75000 / (3.801 * 10^-4)

Stress = 197316495 Pa OR 0.197 GPa

Modulus of elasticity = stress / strain

200 = 0.19732 / Strain

Strain = 0.0009866 (longitudinal)

(a) Strain = change in length / initial length

0.0009866 = Elongation of rod / 200

Elongation of rod = 0.19732 mm

(b) Poisson ratio = lateral strain / longitudinal strain

0.3 = lateral strain / 0.0009866

Lateral strain = 0.000296

Lateral strain = Change in diameter / original diameter

0.000296 = Change in diameter / 22

Change in diameter = 0.00651 mm

4 0

3 years ago

Read 2 more answers

The quantity of charge Q in coulombs (C) that has passed through a point in a wire up to time t (measured in seconds) is given b

Mnenie [13.5K]

Explanation:

The quantity of charge Q in coulombs (C) that has passed through a point in a wire up to time t (measured in seconds) is given by :

$Q=t^3-2t^2+4t+4$

We need to find the current flowing. We know that the rate of change of electric charge is called electric current. It is given by :

$I=\dfrac{dQ}{dt}\\\\I=\dfrac{d(t^3-2t^2+4t+4)}{dt}\\\\I=3t^2-4t+4$

At t = 1 s,

Current,

$I=3(1)^2-4(1)+4\\\\I=3\ A$

So, the current at t = 1 s is 3 A.

For lowest current,

$\dfrac{dI}{dt}=0\\\\\dfrac{d(3t^2-4t+4)}{dt}=0\\\\6t-4=0\\\\t=0.67\ s$

Hence, this is the required solution.

7 0

3 years ago

EARTH SCIENCE PLEASE ANSWER

Rzqust [24]

The Earth Science answers are shown below.

Explanation:

1. The movement of the sun will change the angle it has on the sky in 30 minutes, it is always moving from the east to the west, so in 30 minutes it would move more west, no matter at what time you make the experiment. From Earth, the Sun looks like it moves across the sky in the daytime and appears to disappear at night. This is because the Earth is spinning towards the east. The Earth spins about its axis, an imaginary line that runs through the middle of the Earth between the North and South poles

2. No, both marks are the same distance from the ground. the amount of stick above the mark will not affect the distance that the shadow of the mark moves at all. The Sun's clockwise motion is an apparent motion caused by the rotation of the Earth. The counterclockwise rotation of the Earth in the Sun's light causes the shadow of the gnomon to move clockwise. As the Sun appears to move higher above the horizon before solar noon, the shadow grows shorter and shorter.

3. In the summer the shadows are shorter, and in the winter the shadows are longer. In the morning your shadow will point west and in the afternoon it will point east. If your shadow is long, it is near sunrise or sunset. Your shadow is shortest around noon.

4. If the sun rises in the east and sets in the west, then the Earth should rotate in the opposite direction from west to east (anti-clockwise). Earth's spin (or rotation) on its axis. Earth rotates or spins toward the east, and that's why the Sun, Moon, planets, and stars all rise in the east and make their way westward across the sky.

4 0

3 years ago

An automobile of mass 2000 kg moving at 30 m/s is braked suddenly with a constant braking force of 10000 N. How far does the car

saveliy_v [14]

Answer:

The car traveled the distance before stopping is 90 m.

Explanation:

Given that,

Mass of automobile = 2000 kg

speed = 30 m/s

Braking force = 10000 N

For, The acceleration is