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mezya [45]
3 years ago
12

Help help help help help

Physics
1 answer:
nalin [4]3 years ago
3 0
I am not sure on the rest but the answer to question 4 is volts. sorry i couldn’t be of more help!
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What’s the answer because I have no clue
Snezhnost [94]

Answer:

C. The number of protons.

Explanation:

All atoms of mercury will remain the same while different isotopes may have different number of neutrons.

For example, mercury has 80 protons, correlating with its atomic number. However, the number of protons, neutrons, and electrons could vary between each atom. If there are 80 protons, then there has to be 80 electrons. This means that the protons have to be the same in order to apply to all atoms of mercury

8 0
4 years ago
If you change the mass of a moving object, you change its _______.
Paul [167]
Momentum and inertia, Momentum=P. P=MV. M=mass and V=velocity. Mass is related to inertia so inertia=F. F=MA. A=acceleration.
6 0
3 years ago
Read 2 more answers
The net force acting on an object equals the applied force plus the force of friction.
Georgia [21]

Answer:

False

Explanation:

The net force is equal to the applied force minus the force of friction. It is possible for friction to act in the same direction as an applied force, but that would mean there would have to be more than two forces acting on the object.

3 0
3 years ago
The Moon requires about 1 month (0.08 year) to orbit Earth. Its distance from us is about 400,000 km (0.0027 AU). Use Kepler’s t
dem82 [27]

Answer:

\frac{M_e}{M_s} = 3.07 \times 10^{-6}

Explanation:

As per Kepler's III law we know that time period of revolution of satellite or planet is given by the formula

T = 2\pi \sqrt{\frac{r^3}{GM}}

now for the time period of moon around the earth we can say

T_1 = 2\pi\sqrt{\frac{r_1^3}{GM_e}}

here we know that

T_1 = 0.08 year

r_1 = 0.0027 AU

M_e = mass of earth

Now if the same formula is used for revolution of Earth around the sun

T_2 = 2\pi\sqrt{\frac{r_2^3}{GM_s}}

here we know that

r_2 = 1 AU

T_2 = 1 year

M_s = mass of Sun

now we have

\frac{T_2}{T_1} = \sqrt{\frac{r_2^3 M_e}{r_1^3 M_s}}

\frac{1}{0.08} = \sqrt{\frac{1 M_e}{(0.0027)^3M_s}}

12.5 = \sqrt{(5.08 \times 10^7)\frac{M_e}{M_s}}

\frac{M_e}{M_s} = 3.07 \times 10^{-6}

4 0
3 years ago
The pressure of a fluid at a specific depth depends only on the type of fluid?
timurjin [86]
I think yes the density
5 0
3 years ago
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