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3 years ago

You have given a power supply, copper wire, and an iron nail. What should you do to decrease the strength of the electro magnet?

1 answer:

8 0

If you wrap some of the wire around the nail in one direction and some of the wire in the other direction, the magnetic fields from the different sections fight each other and cancel out, reducing the strength of your magnet.

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If you know that the period of a pendulum is 1.87 seconds, what is the length of that pendulum? (assume that we are on earth and

laiz [17]

The length of the pendulum is 0.087 m. Option d is correct.

<h3>What is Simple harmonic motion?</h3>

Simple harmonic motion is periodic motion caused by a restoring force that is proportionate to the deviation from equilibrium.

Simple harmonic motion is periodic motion but many other conditions are dependent.

The time period of the pendulum is found as;

$\rm T= 2 \pi \sqrt{\frac{L}{g} } \\\\ \rm 1.87 \ sec= 2 \times 3.14 \sqrt{\frac{L}{9.81 m/s^2} } \\\\ L=0.087 \ m$

The length of the pendulum is 0.087 m

Hence, option d is correct.

To learn more about the simple harmonic motion refer to the link;

brainly.com/question/17315536

#SPJ1

4 0

2 years ago

It has been suggested that rotating cylinders several miles in length and several miles in diameter be placed in space and used

stepladder [879]

Answer:

the required revolution per hour is 28.6849

Explanation:

Given the data in the question;

we know that the expression for the linear acceleration in terms of angular velocity is;

$a_{c}$ = rω²

ω² = $a_{c}$ / r

ω = √( $a_{c}$ / r )

where r is the radius of the cylinder

ω is the angular velocity

given that; the centripetal acceleration equal to the acceleration of gravity a $a_{c}$ = g = 9.8 m/s²

so, given that, diameter = 4.86 miles = 4.86 × 1609 = 7819.74 m

Radius r = Diameter / 2 = 7819.74 m / 2 = 3909.87 m

so we substitute

ω = √( 9.8 m/s² / 3909.87 m )

ω = √0.002506477 s²

ω = 0.0500647 ≈ 0.05 rad/s

we know that; 1 rad/s = 9.5493 revolution per minute

ω = 0.05 × 9.5493 RPM

ω = 0.478082 RPM

1 rpm = 60 rph

ω = 0.478082 × 60

ω = 28.6849 revolutions per hour

Therefore, the required revolution per hour is 28.6849

7 0

3 years ago

A mass of 5kg starts from rest and pulls down vertically on a string wound around a disk-shaped, massive pulley. The mass of the

bekas [8.4K]

Answer:

1.98 m/s

Explanation:

To solve this, we would be using the law of conservation of energy, i.e total initial energy is equal to total final energy.

E(i) = E(f)

mgh = ½Iw² + ½mv²

Recall, v = wr, thus, w = v/r

Also, I = ½mr²

I = 0.5 * 5 * 2²

I = 10 kgm²

Remember,

mgh = ½Iw² + ½mv²

Substituting w for v/r, we have

mgh = ½I(v/r)² + ½mv²

Now, putting the values in the equation, we have

5 * 9.8 * 0.3 = ½ * 10 * (v/2)² + ½ * 5 * v²

14.7 = 1.25 v² + 2.5 v²

14.7 = 3.75 v²

v² = 14.7/3.75

v² = 3.92

v = √3.92

v = 1.98 m/s

Thus, the speed is 1.98 m/s

5 0

3 years ago

How does deformation by drawing of a semicrystalline polymer affect its tensile strength?

Elenna [48]

This effect is explained by increased chain entanglements at higher molecular weights. Increasing the degree of crystallinity of a semicrystalline polymer leads to an enhancement of the tensile strength. Deformation by drawing increases the tensile strength of a semicrystalline polymer.

4 0

3 years ago

A flywheel with a very low friction bearing takes 1.6 h to stop after the motor power is turned off. The flywheel was originally

nadezda [96]

Answer: 5.76 rads/s

Explanation:

The initial rotation is 55 rpm

1 rev = 2π radians

55 revs = 55 × 2π/1 = 345.58 radians/min

345.58 rads/min = 345.58rads/60s = 5.76 rads/s

3 0

4 years ago