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3 years ago

Differences differences between gravity and acceleration due to gravity

Physics

1 answer:

Nina [5.8K]3 years ago

4 0

Look at this table. Accleration of gravity means gravitational constant.

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(15 points) (Asap!!) In what two ways can you increase the elastic potential energy of a spring?

adelina 88 [10]

Hello There

Answers: The elastic potential energy can be increased by: 

1) Getting a spring with a higher spring constant

2) Increasing the length at which the spring is compressed.

Reasons: Getting a stronger spring makes it stronger which equals more energy. While increasing the compression on the spring, increases the stored energy which makes it more powerful when its released

I hope this helps
-Chris

5 0

3 years ago

Read 2 more answers

Two spherical shells have a common center. A -1.50 × 10-6 C charge is spread uniformly over the inner shell, which has a radius

Murrr4er [49]

Answer:

a) At 0.20 m, the magnitude of the field is 675.0 kV

The direction of the field is acting outwards from the center of the charged spheres

b) At 0.10 m, the magnitude of the field is 135 kV

The direction is acting outwards from the center of the charged spheres

c) At 0.025 m

The magnitude of the field, V = -270.0 kV

The direction of the field is inwards, towards the center of the charged spheres

Explanation:

The charged spherical shell parameters are;

The charge on the inner sphere, q₁ = -1.50 × 10⁻⁶ C

The radius of the inner shell, R₁ = 0.050 m

The charge on the outer sphere, q₂ = +4.50 × 10⁻⁶ C

The radius of the outer shell, R₂ = 0.15 m

Let 'r', represent the distance at which the electric field is measured, the following relationships can be obtained;

When r < R₁ < R₂,

$V = k \cdot \left( \dfrac{q_1}{R_1} + \dfrac{q_2}{R_2} \right )$

When R₁ < r < R₂,

$V = k \cdot \left( \dfrac{q_1}{r} + \dfrac{q_2}{R_2} \right )$

When R₁ < R₂ < r,

$V = k \cdot \left( \dfrac{q_1 + q_2}{r^2} \right )$

a) When r = 0.20 m, we have;

R₁ < R₂ < r, therefore

$V = k \cdot \left( \dfrac{q_1 + q_2}{r^2} \right )$

By plugging in the values, we get;

$V = 9 \times 10^9 \times \left( \dfrac{-1.50 \times 10^{-6} + 4.50\times 10^{-6} }{0.20^2} \right ) = 675.0 \ kV$

Therefore, the magnitude of the field, V = 675.0 kV

The direction of the field is outwards

b) When r = 0.10 m, we have;

When R₁ < r < R₂, therefore;

$V = k \cdot \left( \dfrac{q_1}{r} + \dfrac{q_2}{R_2} \right )$

By plugging in the values, we get;

$V = 9 \times 10^9 \times \left( \dfrac{-1.50 \times 10^{-6} }{0.10} + \dfrac{4.50\times 10^{-6}}{0.15} \right ) = 135 \ kV$

Therefore, the magnitude of the field, V = 135 kV

The direction of the field is outwards from the center

c) When r = 0.025 m, we have;

When r < R₁ < R₂, therefore;

$V = k \cdot \left( \dfrac{q_1}{R_1} + \dfrac{q_2}{R_2} \right )$

By plugging in the values, we get;

$V = 9 \times 10^9 \times \left( \dfrac{-1.50 \times 10^{-6} }{0.05} + \dfrac{4.50\times 10^{-6}}{0.15} \right ) = -270 \ kV$

Therefore, the magnitude of the field, V = -270.0 kV

The direction of the field is inwards, towards the center of the charged spheres.

4 0

3 years ago

A student is experimenting with some insulated copper wire and a power supply. She winds a single layer of the wire on a tube wi

OverLord2011 [107]

Answer:

$P=214.7187\,W$

Explanation:

Given that:

Diameter of the solenoid, $D=10\,cm=0.1\,m$

length of the solenoid, $L=90\,cm=0.9\,m$

diameter of the wire, $d=0.1\,cm=10^{-3}\,m$

magnetic field at the center of the solenoid, $B=7.4\times 10^{-3}\,T$

Now we need the no. of turns incorporated in the length of 90 cm:

$N=\frac{Length\,\,of\,\,solenoid}{diameter\,\,of\,\, wire}$

$N=\frac{L}{d}$

$N=\frac{0.9}{10^{-3}}$

$N=900\,\,turns$

For solenoids we have:

$B=\mu.n.I$ ...............................(1)

where:

$\mu=$ permeability of the medium

n = no. of turns per unit length

I = current in the coil

So,

$n=\frac{900}{0.9}$

$n=1000\,turns\,.\,m^{-1}$

Now putting the respective values in the eq. (1)

$7.4\times 10^{-3}=4\pi\times10^{-7}\times 1000\times I$

$I=5.8887\,A$

For copper we have resistivity:

$\rho=1.72\times 10^{-8}\, \Omega.m$

We know that resistance is given by:

$R=\rho.\frac{l}{a}$ .....................................(2)

where:

l = length of the conducting wire

a = cross sectional area of the conducting wire

Now we need the length (l) of the wire:

Circumference of the solenoid,

$C=\pi.D$

$C=0.1\pi\,m$

$\therefore l=C\times N$

$l=90\pi\,m$

Cross-sectional area of wire:

$a=\pi.\frac{d^2}{4}$

$a=\pi. \frac{(10^{-3})^2}{4}\,m^2$

Resistance from eq. (2):

$R=1.72\times 10^{-8}\times \frac{90\pi}{\pi. \frac{(10^{-3})^2}{4}}$

$R=6.192 \,\Omega$

For power we have:

$P=I^2.R$

$P=5.8887^2 \times 6.192$

$P=214.7187\,W$

6 0

3 years ago

I need help with question 8 .

Vesnalui [34]

The bike is maintaining "constant velocity". He's moving at 15 m/s when we see him for the first time, 15 m/s later that day, and 15 m/s next week.

The car starts from zero, and goes 4.0 m/s FASTER each second. After one second, it's going 4.0 m/s. After 2 seconds, it's going 8 m/s. And after 3 seconds, it's going 12 m/s.

This is the point at which the question wants us to compare them ... 3 seconds. The bike is moving at 15 m/s and the car has sped up to 12 m/s. The bike is moving faster than the car.

If we hung around and kept watching for another second, the car would then be moving at 16 m/s, and would be moving faster than the bike. But we lost interest after answering the question, and we left at 3 seconds.

5 0

3 years ago

A ball thrown with 50N of force accelerates at 25 m/s2. What is the mass of the ball?

MrRissso [65]

F=ma
Force is 50N. Acceleration is 25 m/s^2.
50N=m*25 m/s^2
Divide both sides by 25.
mass=2 kg

7 0

4 years ago

Differences differences between gravity and acceleration due to gravity​

Differences differences between gravity and acceleration due to gravity