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3 years ago

Gravity____? Help please

Physics

2 answers:

Anika [276]3 years ago

8 0

Gravity is described as a fundamental force and therefore does not depend on other forces

HACTEHA [7]3 years ago

3 0

Gravity is a fundamental force. I understand why you asked the question, as the answers are challenging. Gravity <span>does not</span><span> depend upon velocity, nor does friction make gravity work.
The force of gravity affects systems of mass due to the </span>acceleration it creates in a system.

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Determine the change in electric potential energy of a system of two charged objects when a -2.1-C charged object and a -5.0-C c

elena55 [62]

Answer:

Change in electric potential energy ∆E = 365.72 kJ

Explanation:

Electric potential energy can be defined mathematically as:

E = kq1q2/r ....1

k = coulomb's constant = 9.0×10^9 N m^2/C^2

q1 = charge 1 = -2.1C

q2 = charge 2 = -5.0C

∆r = change in distance between the charges

r1 = 420km = 420000m

r2 = 160km = 160000m

From equation 1

∆E = kq1q2 (1/r2 -1/r1) ......2

Substituting the given values

∆E = 9.0×10^9 × -2.1 ×-5.0(1/160000 - 1/420000)

∆E = 94.5 × 10^9 (3.87 × 10^-6) J

∆E = 365.72 × 10^3 J

∆E = 365.72 kJ

6 0

3 years ago

A girl throws a pebble into a deep well at 4.0 m/s (downward). It hits the water in 2.0 s. How far below the ground is the water

alexandr402 [8]

Answer:

27.6 m
13.8 m/s

Explanation:

(b) The initial velocity is added to that due to acceleration by gravity. The velocity is increased linearly by gravity at the rate of 9.8 m/s². The average velocity of the pebble will be its velocity halfway through the 2-second time period.* That is, it will be ...

4 m/s + (9.8 m/s²)(2 s)/2 = 13.8 m/s . . . . average velocity

(a) The distance covered in 2 seconds at an average velocity of 13.8 m/s is ...

d = vt

d = (13.8 m/s)(2 s) = 27.6 m

The water is about 27.6 m below ground.

_____

* We have chosen to make use of the fact that the velocity curve is linear, so the average velocity is half the sum of initial and final velocities:

vAvg = (vInit + vFinal)/2 = (vInit + (vInit +at))/2 = vInit +at/2

If you work this in a straightforward way, you would find distance as the integral of velocity, then find average velocity from the distance and time.

$\displaystyle d=\int_0^t{(v_0+at)}\,dt=v_0t+\dfrac{1}{2}at^2=t\left(v_0+a\dfrac{t}{2}\right)\\\\v_{avg}=\dfrac{d}{t}=v_0+a\dfrac{t}{2}\qquad\text{the formula we started with}$

8 0

2 years ago

A solid conducting sphere of radius 2.00 cm has a charge of 6.88 μC. A conducting spherical shell of inner radius 4.00 cm and ou

zepelin [54]

Explanation:

Given that,

Radius R= 2.00

Charge = 6.88 μC

Inner radius = 4.00 cm

Outer radius = 5.00 cm

Charge = -2.96 μC

We need to calculate the electric field

Using formula of electric field

$E=\dfrac{kq}{r^2}$

(a). For, r = 1.00 cm

Here, r<R

So, E = 0

The electric field does not exist inside the sphere.

(b). For, r = 3.00 cm

Here, r >R

The electric field is

$E=\dfrac{kq}{r^2}$

Put the value into the formula

$E=\dfrac{9\times10^{9}\times6.88\times10^{-6}}{(3.00\times10^{-2})^2}$

$E=6.88\times10^{7}\ N/C$

The electric field outside the solid conducting sphere and the direction is towards sphere.

(c). For, r = 4.50 cm

Here, r lies between R₁ and R₂.

So, E = 0

The electric field does not exist inside the conducting material

(d). For, r = 7.00 cm

The electric field is

$E=\dfrac{kq}{r^2}$

Put the value into the formula

$E=\dfrac{9\times10^{9}\times(-2.96\times10^{-6})}{(7.00\times10^{-2})^2}$

$E=5.43\times10^{6}\ N/C$

The electric field outside the solid conducting sphere and direction is away of solid sphere.

Hence, This is the required solution.

6 0

3 years ago

Let n represent the number of similar resistors in a circuit. Each resistor has a resistance of r. When the resistors are connec

jekas [21]

Resistance=(r1+r2+r3...)n
=n×r

8 0

3 years ago

A satellite of mass 5460 kg orbits the Earth and has a period of 6520 s

Nikolay [14]

Answer:

what if I do and b then someone else c I don't have enough time pls

8 0

2 years ago