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jenyasd209 [6]
3 years ago
14

The universal law of gravitation states that the force of attraction between two objects depends on which quantities?

Physics
1 answer:
Ronch [10]3 years ago
4 0

Answer:

D. the masses of the objects and the distance between them

Explanation:

Gravitation is a force, a force doesn't care about the shape or density of objects, only about their masses... and distances.

And you can get it using the following equation:

f = \frac{Gm_{1}m_{2} }{d^{2} }

Where :

G is the universal gravitational constant : G = 6.6726 x 10-11N-m2/kg2

m represent the mass of each of the two objects

d is the distance between the centers of the objects.

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A particle with charge − 2.74 × 10 − 6 C −2.74×10−6 C is released at rest in a region of constant, uniform electric field. Assum
s2008m [1.1K]

Answer:

241.7 s

Explanation:

We are given that

Charge of particle=q=-2.74\times 10^{-6} C

Kinetic energy of particle=K_E=6.65\times 10^{-10} J

Initial time=t_1=6.36 s

Final potential difference=V_2=0.351 V

We have to find the time t after that the particle is released and traveled through a potential difference 0.351 V.

We know that

qV=K.E

Using the formula

2.74\times 10^{-6}V_1=6.65\times 10^{-10} J

V_1=\frac{6.65\times 10^{-10}}{2.74\times 10^{-6}}=2.43\times 10^{-4} V

Initial voltage=V_1=2.43\times 10^{-4} V

\frac{\initial\;voltage}{final\;voltage}=(\frac{initial\;time}{final\;time})^2

Using the formula

\frac{V_1}{V_2}=(\frac{6.36}{t})^2

\frac{2.43\times 10^{-4}}{0.351}=\frac{(6.36)^2}{t^2}

t^2=\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}

t=\sqrt{\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}}

t=241.7 s

Hence, after 241.7 s the particle is released has it traveled through a potential difference of 0.351 V.

6 0
3 years ago
If an atomic nucleus were the size of a dime how far away might one of its electrons be?
mihalych1998 [28]
The radius of a nucleus of hydrogen is approximately r_{n1}=1\cdot 10^{-15}m, while we can use the Borh radius as the distance of an electron from the nucleus in a hydrogen atom: r_{e1}=5.3 \cdot 10^{-11}m

The radius of a dime is approximately r_{n2} = 9\cdot 10^{-3}m: if we assume that the radius of the nucleus is exactly this value, then we  can find how far is the electron by using the proportion
r_{n1}:r_{e1}=r_{n2}:r_{e2}
from which we find
r_{e2}= \frac{r_{e1} r_{n2}}{r_{n1}}= \frac{(5.3 \cdot 10^{-11}m)(9\cdot 10^{-3}m)}{1 \cdot 10^{-15}m}=477 m

So, if the nucleus had the size of a dime, we would find the electron approximately 500 meters away.
6 0
3 years ago
A car’s engine provides a forward force of 500 N. It is opposed by a resistive force of 200 N. The car accelerates forwards for
wariber [46]

Answer:

Explanation:

W = Fd

Engine 500(20) = 10000 J

friction 200(-20) = -4000 J

Energy increase is 10000 - 4000 = 6000 J

5 0
3 years ago
When a constant force acts upon an object, the acceleration of the object varies inversely with its mass 2kg. When a certain con
BARSIC [14]

Answer:

The acceleration of the second object is 4 m/s².

Explanation:

Given that,

when the a constant force acts upon an object, the acceleration varies inversely with its mass .

mass\propto \frac{1}{acceleration}

We can rewrite the equation as

\frac{m_1}{m_2}=\frac{a_2}{a_1}

When an a certain constant force acts upon an object with mass 2 kg , the acceleration of the object is 26 m/s².

Here, m_1 = 2 kg, a_1 = 26 m/s²,m_2 = 13 kg and a_2=?

\frac{m_1}{m_2}=\frac{a_2}{a_1}

\frac{2}{13}=\frac{a_2}{26}

\Rightarrow \frac{a_2}{26}= \frac{2}{13}

\Rightarrow {a_2}= \frac{2}{13}\times 26

\Rightarrow a_2=4 m/s²

The acceleration of the second object is 4 m/s².

3 0
3 years ago
An airplane traveling at half the speed of sound emits a sound of frequency 5.84 kHz. (a) At what frequency does a stationary li
sergejj [24]

Answer:

Stationary listener frequency = 77.68 kHz

Explanation:

Given:

Speed of airplane = 344/2 = 172 m/s

Sound of frequency = 5.84 kHz

Computation:

f0 = fs[(v+v0)/(v-vs)]

f0 = 5.84[(344+0)/(344-172)]

f0 = 5.84[(344)/172]

Stationary listener frequency = 77.68 kHz

7 0
2 years ago
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