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omeli [17]
3 years ago
15

the distance between any two bodies is 10 M and the gravitational force between them is 3.2×10-⁹m. if the mass of one object is

40 kg, calculate the mass of another body. (And:119.9 kg)​
Physics
1 answer:
cluponka [151]3 years ago
3 0

Answer:

0.8 x 10^-9 kg

Explanation:

Given,

Distance ( R ) = 10 m

Force ( F ) = 3.2 x 10^-9 N

Mass ( m1 ) = 40 kg

To find : Mass ( m2 ) = ?

Formula : -

F = m1.m2 / R^2

m2 = FR^2 / m1

= 3.2 x 10^-9 x 10 / 40

= 3.2 x 10^-9 / 4

= ( 3.2 / 4 ) x 10^-9

m2 = 0.8 x 10^-9 kg

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Two identical satellites orbit the earth in stable orbits. One satellite orbits with a speed v at a distance r from the center o
jolli1 [7]

Answer:

c)At a distance greater than r

Explanation:

For a satellite in orbit around the Earth, the gravitational force provides the centripetal force that keeps the satellite in motion:

G\frac{Mm}{r^2}=m\frac{v^2}{r}

where

G is the gravitational constant

M is the Earth's mass

m is the satellite's mass

r is the distance between the satellite and the Earth's centre

v is the speed of the satellite

Re-arranging the equation, we write

r = \frac{GM}{v^2}

so we see from the equation that when the speed is higher, the distance from the Earth's centre is smaller, and when the speed is lower, the distance from the Earth's centre is larger.

Here, the second satellite orbit the Earth at a speed less than v: this means that its orbit will have a larger radius than the first satellite, so the correct answer is

c)At a distance greater than r

7 0
3 years ago
A spherical, conducting shell of inner radius r1= 10 cm and outer radius r2 = 15 cm carries a total charge Q = 15 μC . What is t
lutik1710 [3]

a) E = 0

b) 3.38\cdot 10^6 N/C

Explanation:

a)

We can solve this problem using Gauss theorem: the electric flux through a Gaussian surface of radius r must be equal to the charge contained by the sphere divided by the vacuum permittivity:

\int EdS=\frac{q}{\epsilon_0}

where

E is the electric field

q is the charge contained by the Gaussian surface

\epsilon_0 is the vacuum permittivity

Here we want to find the electric field at a distance of

r = 12 cm = 0.12 m

Here we are between the inner radius and the outer radius of the shell:

r_1 = 10 cm\\r_2 = 15 cm

However, we notice that the shell is conducting: this means that the charge inside the conductor will distribute over its outer surface.

This means that a Gaussian surface of radius r = 12 cm, which is smaller than the outer radius of the shell, will contain zero net charge:

q = 0

Therefore, the magnitude of the electric field is also zero:

E = 0

b)

Here we want to find the magnitude of the electric field at a distance of

r = 20 cm = 0.20 m

from the centre of the shell.

Outside the outer surface of the shell, the electric field is equivalent to that produced by a single-point charge of same magnitude Q concentrated at the centre of the shell.

Therefore, it is given by:

E=\frac{Q}{4\pi \epsilon_0 r^2}

where in this problem:

Q=15 \mu C = 15\cdot 10^{-6} C is the charge on the shell

r=20 cm = 0.20 m is the distance from the centre of the shell

Substituting, we find:

E=\frac{15\cdot 10^{-6}}{4\pi (8.85\cdot 10^{-12})(0.20)^2}=3.38\cdot 10^6 N/C

4 0
3 years ago
If the period of a spring is 5 seconds what is the frequency
Scilla [17]

Answer:

C. 0.2 Hertz

Explanation:

The frequency of a spring is equal to the reciprocal of the period:

f=\frac{1}{T}

where

f is the frequency

T is the period

For the spring in this problem,

T = 5 s

therefore, the frequency is

f=\frac{1}{5 s}=0.2 Hz

7 0
3 years ago
What is the current when the voltage is 18 volts and the resistance is 6 ohms?
juin [17]
Voltage = current(I) * resistance (R)
V = 18
R = 6

18 = I * 6
I = 18/6 = 3 Amps or D
6 0
3 years ago
Read 2 more answers
Hi!
dangina [55]

Answer:

the moe weight you have in the marble, the higher the speed on the way down

Explanation:

4 0
3 years ago
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