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kondaur [170]
3 years ago
9

((GIVING 40 POINTS AND BRAINLIEST TO CORRECT ANSWER!)) What is shown in the diagram?

Physics
2 answers:
lara31 [8.8K]3 years ago
8 0
ITs an electromagnet that creates a motor?
NNADVOKAT [17]3 years ago
5 0
Hello!!
The Answer is A! Motor.

Hope this helps ~Nayiah~
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Pressure = Force/ Area = 3000/2 = 1500 pascal.
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How does radioactive decay work
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Radioactive decay is the spontaneous breakdown of an atomic nucleus resulting in the release of energy and matter from the nucleus. Remember that a radioisotope has unstable nuclei that does not have enough binding energy to hold the nucleus together.

Explanation:

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When you observe a distant galaxy whose photons have traveled for 10 billion years before reaching earth we are seeing that gal?
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A roller coaster car rapidly picks up speed as it
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3 years ago
A train starts at rest in a station and accelerates at a constant 0.987 m/s2 for 182 seconds. Then the train decelerates at a co
algol [13]

Answer:

\displaystyle X_T=66.6\ km

Explanation:

<u>Accelerated Motion </u>

When a body changes its speed at a constant rate, i.e. same changes take same times, then it has a constant acceleration. The acceleration can be positive or negative. In the first case, the speed increases, and in the second time, the speed lowers until it eventually stops. The equation for the speed vf at any time t is given by

\displaystyle V_f=V_o+a\ t

where a is the acceleration, and vo is the initial speed .

The train has two different types of motion. It first starts from rest and has a constant acceleration of 0.987 m/s^2 for 182 seconds. Then it brakes with a constant acceleration of -0.321 m/s^2 until it comes to a stop. We need to find the total distance traveled.

The equation for the distance is

\displaystyle X=V_o\ t+\frac{a\ t^2}{2}

Our data is

\displaystyle V_o=0,a=0.987m/s^2,\ t=182\ sec

Let's compute the first distance X1

\displaystyle X_1=0+\frac{0.987\times 182^2}{2}

\displaystyle X_1=16,346.7\ m

Now, we find the speed at the end of the first period of time

\displaystyle V_{f1}=0+0.987\times 182

\displaystyle V_{f1}=179.6\ m/s

That is the speed the train is at the moment it starts to brake. We need to compute the time needed to stop the train, that is, to make vf=0

\displaystyle V_o=179.6,a=-0.321\ m/s^2\ ,V_f=0

\displaystyle t=\frac{v_f-v_o}{a}=\frac{0-179.6}{-0.321}

\displaystyle t=559.5\ sec

Computing the second distance

\displaystyle X_2=179.6\times559.5\ \frac{-0.321\times 559.5^2}{2}

\displaystyle X_2=50,243.2\ m

The total distance is

\displaystyle X_t=x_1+x_2=16,346.7+50,243.2

\displaystyle X_t=66,589.9\ m

\displaystyle \boxed{X_T=66.6\ km}

8 0
4 years ago
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