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3 years ago

Why is the absolute magnitude of some stars greater than their apparent magnitude

Physics

1 answer:

SSSSS [86.1K]3 years ago

6 0

<span>answer under the link: http: //briskrange.com/7gAl</span>

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Which of the following is a unit of acceleration?

olasank [31]

${\tt{\red{\underline{\underline{\huge{Answer:}}}}}}$

$\longrightarrow$ The rate of change of velocity per unit time is called acceleration.

$\longrightarrow$ Its SI unit is m/s².

$\huge\boxed{\fcolorbox{blue}{red}{Thank you}}$

7 0

3 years ago

A 12 kg<br> mass is lifted to a height of 2 m. What is its potential energy<br> at this position?

Romashka-Z-Leto [24]

Answer:

Explanation:

Potential energy is the energy stored within an object, due to the object's position, arrangement or state

4 0

3 years ago

Read 2 more answers

A 65-kg woman in an elevator is accelerating upward at a rate of 0.6 m/s2. The gravitational force is ___ N?

Nastasia [14]

Answer:

The gravitational force is 130.

Explanation:

During this problem you have to multiply the 65 and the 0.6.

5 0

3 years ago

Which example has the least kinetic energy

Stells [14]

Answer:

Explanation:

Let’s calculate the kinetic energy for all of the choices.

a. (1/2)(100)(100)^2 = 50(10000)=500,000

b. (1/2)(100)(1)^2 = 50

c. (1/2)(10)(100)^2 = 5(10000) = 50,000

d. (1/2)(1)(1)^2 = 0.5

We can see that (d) has the least kinetic energy.

5 0

2 years ago

A 5,000 kg satellite is orbiting the Earth in a circular path. The height of the satellite above the surface of the Earth is 800

Arada [10]

Explanation:

The given data is as follows.

m = 5000 kg, h = 800 km = $0.8 \times 10^{6} m$

$R_{e} = 6.37 \times 10^{6} m$ , r = R + h = $7.17 \times 10^{6} m$

$M_{e} = 5.98 \times 10^{24}$ kg, G = $6.67 \times 10^{-11} Nm^{2}/kg^{2}$

As we know that,

$\frac{mv^{2}}{r} = \frac{GmM_{e}}{r^{2}}$

v = $\sqrt{\frac{GM_{e}}{r^{2}}}$

And, it is known that formula to calculate angular velocity is as follows.

$\omega = \frac{v}{r} = \sqrt{\frac{GM_{e}}{r^{3}}}$

v = $\sqrt{\frac{GM_{e}}{r^{3}}}$

= $\sqrt{\frac{6.67 \times 10^{-11} Nm^{2}/kg^{2} \times 5.98 \times 10^{-24} kg^{-2}}{(7.17 \times 10^{6} m)^{3}}}$

= $1.0402 \times 10^{-3} rad/s$

Thus, we can conclude that speed of the satellite is $1.0402 \times 10^{-3} rad/s$ .

6 0

3 years ago