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3 years ago

What is the low end of the range of surface temperature of white stars

1 answer:

8 0

The way astronomers determine the surface temperature of a star which is determined by the spectral lines present in a stars spectrum

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The fast train known as the TGV (Train à Grande Vitesse) that runs south from Paris, France, has a scheduled average speed of 21

andrezito [222]

Answer:

Explanation:

Given

average speed of train $(v_{avg})=216 kmph\approx 60 m/s$

Maximum acceleration=0.05g

Now centripetal acceleration is

$a_c=\frac{v^2}{r}$

$0.05\times 9.8=\frac{60^2}{r}$

r=7346.93 m

(b)Radius of curvature=900 m

therefore $a_c=\frac{v^2}{r}$

$v=\sqrt{a_cr}$

$v=\sqrt{0.05\times 9.8\times 900}$

$v=\sqrt{441}=21 m/s$

8 0

3 years ago

PLEASE PLEASE HELP! this is on newton’s 2nd law! i will award 20 points and make you brainliest! ASAP PLEASE! ;)

horsena [70]

Answer: The correct answer is B

Explanation: The string is pulling right and the string is unraveling causing it to accelerate left

5 0

3 years ago

Read 2 more answers

The change in momentum for an object is equal to

koban [17]

Answer:

We conclude that the change in momentum of a body is equal to the impulse experienced by a body.

Explanation:

Considering the equation

F • t = m • Δ v

Here,

m • Δ v is basically a change in momentum of a body which is equal to the mass of the object multiplied by the change in its velocity.

Also,

F • t is called the impulse of the object.

In the formula, it is clear that the impulse experienced by a body during the collision is basically a change in the momentum of the body.

In other words, the change in momentum of a body is equal to the impulse experienced by a body.

Therefore, we conclude that the change in momentum of a body is equal to the impulse experienced by a body.

4 0

3 years ago

This procedure has been used to "weigh" astronauts in space: A 42.5 kg chair is attached to a spring and allowed to oscillate. W

Bingel [31]

Answer:

The mass of the astronaut is approximately 119.74 kg

Explanation:

Assuming this problem as a Simple Harmonic Motion of a mass-spring system, the period (T) of the oscillations for a mass (m) and spring constant (k) is:

$T=2\pi\sqrt{\frac{m}{k}}$ (1)

First, we have to calculate the spring constant using equation (1) and the data provided for the oscillations without the astronaut:

(it’s important to note that one complete vibration is the period of the movement)

$1.3=2\pi\sqrt{\frac{42.5}{k}}\Longrightarrow k=42.5(\frac{2\pi}{1.3})^{2}$

$k\approx992.8\,\frac{N}{m}$

Now with the value of k, we can use again (1) to find the mass of the astronaut (Ma) that makes the period to be 2.54 seconds

$2.54=2\pi\sqrt{\frac{42.5+M_{a}}{992.8}}\Longrightarrow M_{a}=992.8(\frac{2.54}{2\pi})^{2}-42.5$

$\mathbf{M_{a}\approx119.74\,kg}$

3 0

3 years ago

The photo shows falling water droplets.

levacccp [35]

I Think The answer is d I hope it helps My friend Message Me if I’m wrong and I’ll change My answer and fix it for you

4 0

3 years ago