Answer:
opposite the sun. between the Earth and the sun. rising perpendicular to the sun.
Explanation:
Answer:
Explanation:
Recall the formula for acceleration:
, where is final velocity, is initial velocity, and is elapsed time (change in velocity over this amount of time).
Let's look at our time vs velocity graph. At t=0 seconds, V=25 m/s. So her initial velocity is 25 m/s.
We want to find the acceleration during the first 5 seconds of motion. Well, looking at our graph, at t=5 seconds, isn't our velocity still 25 m/s? Therefore, final velocity is 25 m/s (for this period of 5 seconds).
We are only looking from t=0 seconds to t=5 seconds which is a total period of 5 seconds. Therefore, elapsed time is 5 seconds.
Substituting values in our formula, we have:
Alternative:
Without even worrying about plugging in numbers, let's think about what acceleration actually is! Acceleration is the change in velocity over a certain period of time. If we are not changing our velocity at all, we aren't accelerating! In the graph, we can see that we have a straight line from t=0 seconds to t=5 seconds, the interval we are worried about. This indicates that our velocity is staying the same! At t=0 seconds, we have a velocity of 25 m/s and that velocity stays the same until t=5 seconds. Even though we are moving, we haven't changed velocity, which means our average acceleration is zero!
The amount of heat in the body in joule
<h2>Answer: True
</h2>
The <u>Doppler effect</u> refers to the change in a wave perceived frequency when the emitter of the waves, and the receiver (or observer in the case of light) move relative to each other.
In other words, it is the variation of the frequency of a wave due to the relative movement of the source of the wave with respect to its receiver.
It should be noted that this effect bears its name in honor of the Austrian physicist <u>Christian Andreas Doppler</u>, who in 1842 proposed the existence of this effect for the case of light in the stars. Another important aspect is that the effect occurs in all waves (including light and sound). However, it is more noticeable to humans with sound waves.
Answer:
mass of the neutron star =3.45185×10^26 Kg
Explanation:
When the neutron star rotates rapidly, a material on its surface to remain in place, the magnitude of the gravitational acceleration on the central material must be equal to magnitude of the centripetal acc. of the rotating star.
That is
M_ns = mass odf the netron star.
G= gravitational constant = 6.67×10^{-11}
R= radius of the star = 18×10^3 m
ω = 10 rev/sec = 20π rads/sec
therefore,
= 3.45185... E26 Kg
= 3.45185×10^26 Kg