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3 years ago

Weight measures:

Physics

1 answer:

GrogVix [38]3 years ago

3 0

C. Gravity's effect on an object. Weight = the mass of your object X the acceleration of gravity (9.8).

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What role did gravity played in the formation of the solar system

hodyreva [135]

Gravity is the cause of . . .

-- star formation,

-- planet formation,

-- orbits.

So without gravity, there pretty much wouldn't be any solar or any system.

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3 years ago

You want to build a snowman, so you accelerate a 2kg snowball across your yard at a rate of 0.5m/s2. Calculate the amount of for

tamaranim1 [39]

Answer:

Units:

Newtons

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2 years ago

I attach a 4.1 kg block to a spring that obeys Hooke's law and supply 3.8 J of energy to stretch the spring. I release the block

borishaifa [10]

Answer:

The amplitude of the oscillation is 2.82 cm

Explanation:

Given;

mass of attached block, m = 4.1 kg

energy of the stretched spring, E = 3.8 J

period of oscillation, T = 0.13 s

First, determine the spring constant, k;

$T = 2\pi \sqrt{\frac{m}{k} }$

where;

T is the period oscillation

m is mass of the spring

k is the spring constant

$T = 2\pi \sqrt{\frac{m}{k} } \\\\k = \frac{m*4\pi ^2}{T^2} \\\\k = \frac{4.1*4*(3.142^2)}{(0.13^2)} \\\\k = 9580.088 \ N/m\\\\$

Now, determine the amplitude of oscillation, A;

$E = \frac{1}{2} kA^2$

where;

E is the energy of the spring

k is the spring constant

A is the amplitude of the oscillation

$E = \frac{1}{2} kA^2\\\\2E = kA^2\\\\A^2 = \frac{2E}{k} \\\\A = \sqrt{\frac{2E}{k} } \\\\A = \sqrt{\frac{2*3.8}{9580.088} }\\\\A = 0.0282 \ m\\\\A = 2.82 \ cm$

Therefore, the amplitude of the oscillation is 2.82 cm

8 0

3 years ago

The acceleration of a block attached to a spring is given by a=−(0.324m/s2)cos([2.50rad/s]t) a = − ( 0.324 m / s 2 ) c o s ( [ 2

allsm [11]

Answer:

Looks like you have:

a = -.324 cos 2.5 t

In this case ω^2 A = .324

ω = 2.5

f = ω / (2 * pi) = 2.5 / 6.28 = .40 / sec

5 0

2 years ago

A small circular coil of 5 turns of wire lies in a uniform magnetic field of 0.8 T, so that the normal to the plane of the coil

Travka [436]

Complete question:

A small circular coil of 5 turns of wire lies in a uniform magnetic field of 0.8 T, so that the normal to the plane of the coil makes an angle of 100◦ with the direction of B~ . The radius of the coil is 4 cm, and it carries a current of 1 A.

What is magnitude of the magnetic moment of the coil? Answer in units of A · m2.

Answer:

The magnetic moment of the coil is 0.0252 A.m²

Explanation:

Given;

radius of the coil, r = 4 cm = 0.04 m

number of turns of the coil, N = 5 turns

magnetic field strength B = 0.8 T

current in the coil, I = 1 A

Area of the coil, A = πr² = π(0.04)² = 0.00503 m²

magnetic moment of the coil, μ = NIA

where;

N is the number of turns

I is the current in the coil

A is the area of the coil

magnetic moment of the coil, μ = 5 x 1 x 0.00503 = 0.0252 A.m²

Therefore, the magnetic moment of the coil is 0.0252 A.m²

8 0

3 years ago