STATE 2 APPLICATIONS OF CIRCULAR MOTION

A solenoid of 2100 turns, area 10 cm2, and length 30 cm carries a current of 4.0 A. (a) Calculate the magnetic energy stored in

nikitadnepr [17]

Answer:

E = 0.1472 J

Explanation:

Given that,

The number of turns in the solenoid, N = 2100

Area of the solenoid, A = 10 cm² = 0.001 m²

The length of the solenoid, l = 30 cm = 0.3 m

Current in the solenoid, I = 4 A

We need to find the magnetic energy stored in the solenoid. The expression for the stored energy is :

$E=\dfrac{1}{2}LI^2$

Where

L is self inductance of the solenoid,

$L=\dfrac{\mu_oN^2A}{l}\\\\L=\dfrac{4\pi \times 10^{-7}\times 2100^2\times 0.001}{0.3}\\\\L=0.0184\ H$

So,

$E=\dfrac{1}{2}\times 0.0184\times 4^2\\\\E=0.1472\ J$

So, 0.1472 J of energy is stored in the solenoid.

3 0

3 years ago

A communication satellite is in a circular path orbit around Earth. If the speed of the satellite is constant, the net force act

allsm [11]

Answer:

is changing in direction, but constant in magnitude

Explanation:

This question is a bit tricky since the velocity of the satellite is changing, but the speed is constant.

Speed is simply a measure of how fast you are going. It doesn't matter where you're going, just how quickly.

Velocity, on the other hand, does care about which direction you're going. For example, it could be then when you travel right, your velocity is positive, and when you travel left, your velocity is negative. This is the similar for a 2D shape like a circular orbit

Since we know velocity is changing, there must be acceleration which changes that velocity (since acceleration <em>is</em><em> </em>the change in velocity: going from 0 to 60 mph, for example)

Thus, with a non-zero net acceleration, we know that there must be a force that is changing in direction, but constant in magnitude (since the orbit is a circle, and always attracted to the center of the Earth at equal distance).

4 0

3 years ago

A student stretches a spring, attaches a 1.20 kg mass to it, and releases the mass from rest on a frictionless surface. The resu

Sphinxa [80]

Answer:

the speed of the mass when it is halfway to the equilibrium position is 1.26 m/s

Explanation:

Given that;

mass of the object m = 1.20 kg

period of oscillation = 0.750 s

Amplitude ( A/x) = 15.0 cm = 0.15 m

now;

a) Determine the oscillation frequency;

oscillation frequency f = 1/T

we substitute

f = 1 / 0.750 s

f = 1.33 Hz

Therefore, the oscillation frequency is 1.33 Hz

b) Determine the spring constant;

we solve for spring constant from the following expression;

T = 2π√(m/k)

k = 4π²m / T²

so we substitute

k = (4π² × 1.20) / (0.750)²

k = 47.3741 / 0.5625

k = 84.22 N/m

Therefore, the spring constant is 84.22 N/m

c) determine the speed of the mass when it is halfway to the equilibrium position

So, at equilibrium, the energy is equal to K.E

such that;

1/2mv² = 1/2kx²

mv² = kx²

v² = kx² / m

v = √( kx²/m)

we substitute

v = √( 84.22×(0.15 m)²/ 1.2 )

v = √( 1.89495 / 1.2 )

v = √ 1.579125

v = 1.26 m/s

Therefore, the speed of the mass when it is halfway to the equilibrium position is 1.26 m/s

3 0

3 years ago

A small block is released from rest from the edge of a large hemispherical frictionless bowl with radius 3 m. As the mass passes

andreyandreev [35.5K]

Explanation:

It is given that,

Radius of bowl, r = 3 m

A small block is released from rest from the edge of a large hemispherical friction less bowl. Let v is the speed of the mass when it passes through the bottom of the bowl and u is the initial speed of the mass, u = 0. Using conservation of energy as :

$mgh=\dfrac{1}{2}mv^2$

h = r = 3 m

$v=\sqrt{2gh}$

$v=\sqrt{2\times 9.8\ m/s^2\times 3\ m}$

v = 7.66 m/s

or

v = 7.7 m/s

So, the speed of the mass at the bottom of the bowl is 7.7 m/s. Hence, this is the required solution.

4 0

3 years ago

Which graph showing constant acceleration correctly places the independent and dependent variables?

shtirl [24]

Answer:

graph 1

Explanation:

acceleration is defined as change in velocity/ change in time.

A constant acceleration should have a linear line.

Since velocity is in the numerator of the formula it is the dependent variable which will go on the y axis. thus making time the independent going on the x axis.

This makes sense because velocity should depend on time not the other way around. This therefore making the right answer graph 1

8 0

3 years ago

Read 2 more answers