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2 years ago

If an astronaut has a mass of 80kg on earth, what is their mass on the moon?

Physics

1 answer:

stealth61 [152]2 years ago

6 0

Answer:

130 N basically

Explanation:

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During a rainy day, as a result of colliding clouds an observer saw lighting and a heard thunder sound. The time between seeing

hodyreva [135]

Answer:

d = 1700 meters

Explanation:

During a rainy day, as a result of colliding clouds an observer saw lighting and a heard thunder sound. The time between seeing the lighting and hearing the sound was 5 second, t = 5 seconds

Speed of sound, v = 340 m/s (say)

Let d is the distance of the colliding cloud from the observer. The distance covered by the object. It is given by :

$d=v\times t$

$d=340\ m/s\times 5\ s$

d = 1700 meters

So, the distance of the colliding cloud from the observer is 1700 meters. Hence, this is the required solution.

8 0

2 years ago

Is it possible for a nazi style government to exist in the modern world? Explain why or why not.

nevsk [136]

I think only if they were too overpowered maybe, but the modern world doesn't except this kind of dictatorship. Most armies are much more powerful than in the past.

7 0

3 years ago

Differentiate between s-block and p-block

alexandr402 [8]

Explanation:

hope this helps you dear friend.

5 0

2 years ago

Read 2 more answers

An electron and a proton are held on an x axis, with the electron at x = + 1.000 m and the proton at x = - 1.000 m. Part A How m

r-ruslan [8.4K]

PART A)

Electrostatic potential at the position of origin is given by

$V = \frac{kq_1}{r_1} + \frac{kq_2}{r_2}$

here we have

$q_1 = 1.6 \times 10^{-19} C$

$q_2 = -1.6 \times 10^{-19} C$

$r_1 = r_2 = 1 m$

now we have

$V = \frac{Ke}{r} - \frac{Ke}{r}$

$V = 0$

Now work done to move another charge from infinite to origin is given by

$W = q(V_f - V_i)$

here we will have

$W = e(0 - 0) = 0$

so there is no work required to move an electron from infinite to origin

PART B)

Initial potential energy of electron

$U = \frac{Kq_1e}{r_1} + \frac{kq_2e}{r_2}$

$U = \frac{9\times 10^9(-1.6\times 10^{-19}(-1.6 \times 10^{-19})}{19} + \frac{9\times 10^9(1.6\times 10^{-19}(-1.6 \times 10^{-19})}{21}$

$U = (2.3\times 10^{-28})(\frac{1}{19} - \frac{1}{21})$

$U = 1.15\times 10^{-30}$

Now we know

$KE = \frac{1}{2}mv^2$

$KE = \frac{1}{2}(9.1\times 10^{-31}(100)^2$

$KE = 4.55 \times 10^{-27} kg$

now by energy conservation we will have

So here initial total energy is sufficient high to reach the origin

PART C)

It will reach the origin

4 0

2 years ago

How high was a brick dropped from if if falls in 2.5 seconds?

jenyasd209 [6]

Using the kinematic equation d = V_0 * t + 1/2 * a * t^2, where d is height you can rewrite this to be d = 1/2*g*t^2 or 4.9t^2
g = a because this is a free fall
d = 1/2 * 9.81m/s^2 * 2.5^2
d = 30.65625m
d = 30.7m

7 0

2 years ago