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Anika [276]
4 years ago
7

If you weigh 150 lbs. on the surface of Earth, how much would you weigh on Venus? On Mars?

Physics
1 answer:
Ghella [55]4 years ago
7 0

Answer:

weight on Venus =  604.15 N

weight on Venus = 252.01 N

Explanation:

Given:

Weight on the earth , W = 150 lbs

now,

1 lbs = 4.45 N

thus,

W = 150 × 4.45  = 667.5 N

now, mass on the earth, m = 667.5/9.8 = 68.11 kg

now,

⇒ weight on Venus,

the acceleration due to   gravity on venus = 8.87 m/s²

thus,

weight on Venus = mass × acceleration due to   gravity on venus

or

weight on Venus = 68.11 × 8.87 = 604.15 N

⇒ weight on mars,

the acceleration due to   gravity on mars = 3.70 m/s²

thus,

weight on Venus = mass × acceleration due to   gravity on venus

or

weight on Venus = 68.11 × 3.70 = 252.01 N

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Explanation:

It is given that,

A nerve signal travels 150 meters per second. It is the speed of the nerve signal. We need to convert the number of kilometers that the nerve signal will travel in the same time.

We know that,

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1 hour = 3600 seconds

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150\ m/s=540\ km/h

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3 years ago
A tire is filled with air at 10 ∘C to a gauge pressure of 250 kPa. Part A If the tire reaches a temperature of 45 ∘C, what fract
Alex_Xolod [135]

Answer:

The air fraction to be removed is 0.11

Given:

Initial temperature, T = 10^{\circ} = 283 K

Pressure, P = 250 kPa

Finally its temperature increases, T' = 45^{\circ} = 318 K

Solution:

Using the ideal gas equation:

PV = mRT

where

P = Pressure

V = Volume

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R = Rydberg's Constant

T = Temperature

Now,

Considering the eqn at constant volume and pressure, we get:

mT = m'T'

Thus

\frac{m}{m'} = \frac{T'}{T}                      (1)

Now, the fraction of the air to be removed for the maintenance of pressure at 250 kPa:

y = \frac{m - m'}{m} = 1 - \frac{m'}{m}

From eqn (1):

y = 1 - \frac{T}{T'}

y = 1 - \frac{283}{318} = 0.11

6 0
4 years ago
URGENT!! A 0.057 kg tennis ball and a tennis racket collide. The racket has an initial
babunello [35]

Answer:

a

Explanation:it is a

7 0
3 years ago
One of the most effective ways to evaluate data is to try to replicate it.
WINSTONCH [101]
It seems that you have missed the given options for the given statement above whether it is true or false. But anyway, the correct answer would be TRUE. It is true that one <span>of the most effective ways to evaluate data is to try to replicate it. Hope that this answer will help you. </span>
6 0
3 years ago
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An object sits at rest at some position to the left of the origin. Draw what you expect the position vs time, velocity vs time,
Andru [333]

Answer:

  see below

Explanation:

The position graph will be a horizontal line (of constant position) located at the negative value representing the position to the left of the origin.

The velocity graph will be a horizontal line at 0, since the object is at rest.

The acceleration graph will be a horizontal line at 0, since the object's velocity is not changing.

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