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3 years ago

What is free-fall acceleration on the surface of the moon?

1 answer:

7 0

R is the radius of the object. There is a negative sign in front of the equation because objects in free fall always fall downwards toward the center of the object. The acceleration due to gravity is <span>1.62 m/s2</span><span>. This is approximately 1/6 that of the acceleration due to gravity on Earth, </span><span>9.81 m/s2</span><span>.</span>

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Use the impulse-momentum theorem to find how long a stone falling straight down takes to increase its speed from 3.7 m/s to 9.90

11111nata11111 [884]

<h2>Time taken is 0.632 seconds</h2>

Explanation:

Impulse momentum theorem is change in momentum is impulse.

Change in momentum = Impulse

Final momentum - Initial momentum = Impulse

Mass x Final velocity - Mass x Initial Velocity = Force x Time

Mass x Final velocity - Mass x Initial Velocity =Mass x Acceleration x Time

Final velocity - Initial Velocity = Acceleration x Time

Final velocity = 9.9 m/s

Initial Velocity = 3.7 m/s

Acceleration = 9.81 m/s²

Substituting

9.9 - 3.7 = 9.81 x Time

Time = 0.632 seconds

Time taken is 0.632 seconds

8 0

2 years ago

An energy source forces a constant current of 2A to flow through a light bulbfilament for twenty seconds. If 4.6 kJ is given off

QveST [7]

Answer:

The voltage drop across the bulb is 115 V

Explanation:

The voltage drop equation is given by:

$V=\frac{\Delta W}{\Delta q}$

Where:

ΔW is the total work done (4.6kJ)

Δq is the total charge

We need to use the definition of electric current to find Δq

$I=\frac{\Delta q}{\Delta t}$

Where:

I is the current (2 A)

Δt is the time (20 s)

$2=\frac{\Delta q}{20}$

$q=40 C$

Then, we can put this value of charge in the voltage equation.

$V=\frac{4600}{40}=115 V$

Therefore, the voltage drop across the bulb is 115 V.

I hope it helps you!

6 0

2 years ago

A 720 g softball is traveling at 15.0 m/s when caught. If the force of the glove on the ball is 520 N, what is the time it takes

Sholpan [36]

Answer:

The time it takes the ball to stop is 0.021 s.

Explanation:

Given;

mass of the softball, m = 720 g = 0.72 kg

velocity of the ball, v = 15.0 m/s

applied force, F = 520 N

Apply Newton's second law of motion, to determine the time it takes the ball to stop;

$F = ma = \frac{mv}{t} \\\\t = \frac{mv}{F} \\\\t = \frac{0.72 \ \times \ 15}{520} \\\\t = 0.021 \ s \\$

Therefore, the time it takes the ball to stop is 0.021 s.

5 0

3 years ago

The Event Horizon Telescope needs a 22 micro-arcsecond resolution to view the event horizon regions around black holes. If the a

likoan [24]

Answer:

14869817.395 m

Explanation:

$\theta$ =22 microarcsecond

λ = Wavelength = 1.3 mm

Converting to radians we get

$22\times 10^{-6}\frac{\pi}{180\times 3600}\ radians$

From Rayleigh Criterion

$\theta=1.22\frac{\lambda}{D}\\\Rightarrow D=1.22\frac{\lambda}{\theta}\\\Rightarrow D=1.22\frac{1.3\times 10^{-3}}{22\times 10^{-6}\frac{\pi}{180\times 3600}}\\\Rightarrow D=14869817.395\ m$

Diameter of the effective primary objective is 14869817.395 m

It is not possible to build one telescope with a diameter of 14869817.395 m. But, we need this type of telescope. So, astronomers use an array of radio telescopes to achieve a virtual diameter in order to observe objects that are the size of supermassive black hole's event horizon.

7 0

2 years ago

e force acting between two charged particles A and B is 5.2 × 10-5 newtons. Charges A and B are 2.4 × 10-2 meters apart. If the

anastassius [24]

The force acting between the particles is

$F=k \frac{Q_{1}Q_{2}}{r^2}$
Then
$Q_{2}= \frac{5.2 \times 10^-^5 \times 0.024^2}{ 9.0 \times 10^9 7.2 \times 10^-^8} =4.622 \times 10^-^1^1C$

7 0

2 years ago

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