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Anastaziya [24]
3 years ago
14

On Pluto, a baby would weigh 2.7 newtons. How much does this baby weigh on Earth? Give your answer in

Physics
1 answer:
Klio2033 [76]3 years ago
3 0
The acceleration of gravity on Pluto is 0.62 m/s² .
The acceleration of gravity on Earth is 9.81 m/s² .

Earth's gravity is  (9.81/0.62) = 15.8 times as strong as Pluto's gravity.

So any object that weighs 2.7N on Pluto would weigh

                                (2.7N) x (15.8) = 42.7 N on Earth .

42.7 Newtons  =  about  9.6 pounds .
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The length of second hand of clock is 14cm, an ant sits on the top of second hand. find the following
Anastaziya [24]

Answer:

The answer is below

Explanation:

i) Since the length of the second clock (radius) is 14 cm = 0.14 m, the distance covered by the second hand in one revelution is:

Distance covered = 2πr = 2π(0.14) = 0.88 m

The time taking to complete one revolution = 60 seconds, hence;

Speed = distance covered in one revolution / time take o complete a revolution

Speed = 0.88 m / 60 s = 0.0147 m/s

ii) Distance covered in 150 s = speed * 150 s = 0.0147 * 150 = 2.2 m

iii) Displacement in 150 seconds = distance from initial position to final position

At 150 s, the hand has covered 2 revolutions and moved 30 s. Hence:

Displacement in 150 seconds = speed * 30 s = 0.0147 * 30 = 0.44 m

4 0
3 years ago
To form a hydrogen atom, a proton is fixed at a point and an electron is brought from far away to a distance of 0.529 ✕ 10−10 m,
Elodia [21]
1) First of all, let's calculate the potential difference between the initial point (infinite) and the final point (d=0.529x10-10 m) of the electron.
This is given by:
\Delta V =- \int\limits^{d}_{\infty} {E} \, dr
Where E is the electric field generated by the proton, which is
E=k_e  \frac{q}{r^2} 
where k_e=8.99\cdot10^9~Nm^2C^{-2} is the Coulomb constant and q=1.6\cdot10^{-19}~C is the proton charge.
Replacing the electric field formula inside the integral, we obtain
\Delta V =- \int\limits^{d}_{\infty} {k_e  \frac{q}{r^2} } \, dr = k_e  \frac{q}{d}= 27~V

2) Then, we can calculate the work done by the electric field to move the electron (charge q_e=-1.6\cdot10^{-19}C) through this \Delta V. The work is given by
W=-q_e \Delta V = - (-1.6\cdot10^{-19}C) (27V)=4.35\cdot10^{-18}~J

5 0
3 years ago
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A cheetah can run at 30 m/s but only for about 12s. how far will it run in that time
Mashutka [201]
About 360 meters I hope this helps 
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3 years ago
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Based on the free-body diagram, the net force acting<br> on this firework is
Strike441 [17]

0N. The net force acting on this firework is 0.

The key to solve this problem is using the net force formula based on the diagram shown in the image. Fnet = F1 + F2.....Fn.

Based on the free-body diagram, we have:

The force of gases is Fgases = 9,452N

The force of the rocket Frocket = -9452

Then, the net force acting is:

Fnet = Fgases + Frocket

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3 years ago
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If two objects have the same volume but one has a greater mass, the one with greater mass
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Explain more. I can't understand

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