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KengaRu [80]
3 years ago
10

An astronaut on another planet drops a 1-kg rock from rest and finds that it falls a vertical distance of 2.5 meters in one seco

nd. on this planet, the rock has a weight of: (hint: use d= 1/2 g t^2 to find g; then use w = mg to find weight.)
Physics
2 answers:
amid [387]3 years ago
8 0

Answer:

Weight = 5 N

Explanation:

As we know that rock falls a distance of 2.5 m in one second so now we can use kinematics to find the acceleration of the rock

d = \frac{1}{2}gt^2

here we know

d = 2.5 m

t = 1 s

now from above equation we know

2.5 = \frac{1}{2}g(1)^2

g = 5 m/s^2

So here the weight of the rock is given as

W = mg

W = 1\times 5

W = 5 N

kow [346]3 years ago
4 0
X=.5(a)t^2 can be used: 2.5m=.5(g)(1), g=5m/s^2. 
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Answer:

at the beginning: 2.3\cdot 10^{-10} F

when the plates are pulled apart: 1.1\cdot 10^{-10} F

Explanation:

The capacitance of a parallel-plate capacitor is given by

C=k \epsilon_0 \frac{A}{d}

where

k is the relative permittivity of the medium (for air, k=1, so we can omit it)

\epsilon_0 = 8.85\cdot 10^{-12} F/m is the permittivity of free space

A is the area of the plates of the capacitor

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In this problem, we have:

A=0.210 m^2 is the area of the plates

d=0.815 cm=8.15\cdot 10^{-3} m is the separation between the plates at the beginning

Substituting into the formula, we find

C=(1)(8.85\cdot 10^{-12}F/m)\frac{0.210 m^2}{8.15\cdot 10^{-3} m}=2.3\cdot 10^{-10} F

Later, the plates are pulled apart to d=1.63 cm=0.0163 m, so the capacitance becomes

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4 0
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Nat2105 [25]
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jasenka [17]

Answer:

d. Not enough information is given to answer this question.

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If work done on the gas:

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