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Gnesinka [82]
3 years ago
5

What happens to the water after it goes down the drain?

Physics
1 answer:
valentina_108 [34]3 years ago
6 0

Answer:

, it flows through your community's sanitary sewer system to a wastewater treatment facility.

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Which property of light remains unchanged when it enters a different medium?
Amiraneli [1.4K]
The FREQUENCY of light remains unchanged once it leaves the source.
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Hii guys follow me please please.​
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Heck yeah i'll follow!!!!!!
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3 years ago
Compute the size of the charge necessary for two spheres separated by 1m to be attached with the force of 1N. How many electrons
yarga [219]

Answer:

q\approx 6.6\cdot 10^{13}~electrons

Explanation:

<u>Coulomb's Law</u>

The force between two charged particles of charges q1 and q2 separated by a distance d is given by the Coulomb's Law formula:

\displaystyle F=k\frac{q_1q_2}{d^2}

Where:

k=9\cdot 10^9\ N.m^2/c^2

q1, q2 = the objects' charge

d= The distance between the objects

We know both charges are identical, i.e. q1=q2=q. This reduces the formula to:

\displaystyle F=k\frac{q^2}{d^2}

Since we know the force F=1 N and the distance d=1 m, let's find the common charge of the spheres solving for q:

\displaystyle q=\sqrt{\frac{F}{k}}\cdot d

Substituting values:

\displaystyle q=\sqrt{\frac{1}{9\cdot 10^9}}\cdot 1

q = 1.05\cdot 10^{-5}~c

This charge corresponds to a number of electrons given by the elementary charge of the electron:

q_e=1.6 \cdot 10^{-19}~c

Thus, the charge of any of the spheres is:

\displaystyle q = \frac{1.05\cdot 10^{-5}~c}{1.6 \cdot 10^{-19}~c}

\mathbf{q\approx 6.6\cdot 10^{13}~electrons}

5 0
3 years ago
Ahsoka pulls a wounded soldier (mass 64 kg) across the ground with a force 18 Newtons. If they have an acceleration of 1.5 m/s2
Kryger [21]

Answer:

0.124

Explanation:

18-(9.81*64)*k=64*1.5

4 0
3 years ago
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Review. As an astronaut, you observe a small planet to be spherical. After landing on the planet, you set off, walking always st
anyanavicka [17]

To find the mass of the planet we will apply the relationship of the given circumference of the planet with the given data and thus find the radius of the planet. From the kinematic equations of motion we will find the gravitational acceleration of the planet, and under the description of this value by Newton's laws the mass of the planet, that is,

The circumference of the planet is,

\phi = 25.1m

Under the mathematical value the radius would be

\phi = 2\pi r

r = \frac{25}{2\pi}

r = 3.9788km

Using second equation of motion

x = \frac{1}{2} at^2

Replacing the values given,

1.4 = \frac{1}{2} a (29.2)^2

Rearranging and solving for 'a' we have,

a = 0.003283m/s^2

Using the value of acceleration due to gravity from Newton's law we have that

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

Here,

r = Radius of the planet

G = Gravitational Universal constant

M = Mass of the Planet

\frac{(6.67*10^{-11})*M}{(3.9788*10^3)^2} = 0.003283

M = 7.79201*10^{14}kg

Therefore the mass of this planet is 7.79201*10^{14}kg

5 0
4 years ago
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