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hram777 [196]
3 years ago
14

Temperature of 273K is the temperature at which water

Physics
1 answer:
hjlf3 years ago
5 0

Explanation:

273 K is 0 °C, the temperature at which water melts/freezes.

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A student practicing for a cross country meet runs 250 m in 30 s. What is the average speed
Kaylis [27]

Answer:8.3m/sec 30 sec,

Explanation:

A student practicing for a track meet, ran 250 m in 30 sec. a. What was her average speed? 250 m = 8.3 m/sec 30 sec.

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2 years ago
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Describe the importance of conservative forces to conservation of energy.
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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
3 years ago
If an object's mass increases, its------------------------- can be increased
otez555 [7]

Answer:

both kinetic energy and potential energy

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When you look at the light from distant stars you are really looking back in time explain what this means
Leona [35]

Answer: Stars are in space for very long time, much longer than that one night. You are looking back in time because those stars have been there for so long that it’s like looking back in time, to when those stars were there.

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