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Lubov Fominskaja [6]
3 years ago
11

PLEASE HELP!!! i’ll mark brainliest if you’re correct

Physics
1 answer:
dedylja [7]3 years ago
7 0

\qquad \qquad\huge \underline{\boxed{\sf Answer}}

The net force acting on the block is ~

\qquad \sf  \dashrightarrow \: 10 + 5

\qquad \sf  \dashrightarrow \: 15 \:N

So, the Answer in the boxes will be ~

\boxed{ \sf15} \:  \boxed{ \sf N}

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What is a prediction
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An educated guess about something. (What might happen in the future)
7 0
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Heredity appears to have a large effect on the intelligence of adults than environment. True False
Annette [7]
I JUST WANT THE ANSWER
5 0
3 years ago
A skater extends her arms, holding a 2 kg mass in each hand. She is rotating about a vertical axis at a given rate. She brings h
Usimov [2.4K]

Explanation:

It is known that relation between torque and angular acceleration is as follows.

                    \tau = I \times \alpha

and,       I = \sum mr^{2}

So,      I_{1} = 2 kg \times (1 m)^{2} + 2 kg \times (1 m)^{2}

                       = 4 kg m^{2}

      \tau_{1} = 4 kg m^{2} \times \alpha_{1}

     \tau_{2} = I_{2} \alpha_{2}

So,      I_{2} = 2 kg \times (0.5 m)^{2} + 2 kg \times (0.5 m)^{2}

                     = 1 kg m^{2}

 as \tau_{2} = I_{2} \alpha_{2}

                   = 1 kg m^{2} \times \alpha_{2}        

Hence,     \tau_{1} = \tau_{2}

                  4 \alpha_{1} = \alpha_{2}

            \alpha_{1} = \frac{1}{4} \alpha_{2}

Thus, we can conclude that the new rotation is \frac{1}{4} times that of the first rotation rate.

8 0
3 years ago
A car traveling at 27.4 m/s hits a bridge abutment. A passenger in the car, who has a mass of 65.0 kg, moves forward a distance
Minchanka [31]

Answer:

F=43570.9N

Explanation:

We can calculate the acceleration experimented by the passenger using the formula v_f^2=v_i^2+2ad, taking the initial direction of movement as the positive direction and considering it comes to a rest:

a=\frac{v_f^2-v_i^2}{2d}=\frac{-v_i^2}{2d}

Then we use Newton's 2nd Law to calculate the force the passenger of mass m experimented to have this acceleration:

F=ma=\frac{-mv_i^2}{2d}

Which for our values is:

F=\frac{-(65kg)(27.4m/s)^2}{2(0.56m)}=43570.9N

6 0
3 years ago
Explain why more stars are circumpolar for observers at higher latitudes.
PIT_PIT [208]

Explanation:

When you observe the night sky you will notice that the stars are moving. They rise from eastern horizon and set in the western horizon. It happens due to rotation of Earth. When observed closely you will notice that the all the stars seem to go around the pole star. Out of all the stars there are some stars which neither set not rise, such stars are called as Circumpolar stars. This means that they are always above the horizon. If we trace the path of such stars they will appear to make complete circle around the pole star.

Also, you will notice that the altitude of pole star (separation of pole star from the horizon in degrees) will depend on the location of observe on the Earth. This happens due to Earth being spherical. So if you are on equator the pole star will be on the horizon i.e. 0° altitude. If you are at Poles, altitude of the pole star will be 90°. Technically the altitude of pole star at any place on Earth is equal to the latitude of the place.

If the altitude of pole star varies and increases as you move towards higher latitude on Earth, the distance between horizon and pole star will also increase. This will result in more stars being circumpolar.

If you are at Poles, all the stars will be circumpolar and if you are at equator no star will be circumpolar.

8 0
3 years ago
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