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BabaBlast [244]
3 years ago
5

A future space explorer has been kidnapped and is held prisoner on a planet in our solar system. With nothing else to do, our pr

isoner amuses herself by dropping her watch from eye level (170 cm) to the floor. She observes that the watch takes 0.36 s to fall. Ignore air resistance.On which of the following planets is she most likely being held? a. Saturn b. Pluto c. Venus d. Jupiter
Physics
1 answer:
DIA [1.3K]3 years ago
4 0

To develop this problem we will apply the linear motion kinematic equations. Specifically, the second law that describes the position of a body as a function of its initial velocity, time and acceleration.

y = ut+\frac{1}{2}gt^2

Here,

u = Initial velocity

t = Time

g = Acceleration due to gravitation

If we replace the values to find the gravitational acceleration we have then,

1.7m = 0+\frac{1}{2} g*0.36^2

g= 26.2346m/s^2

Recall that the force of gravity on the planet Jupiter is 24.79 m / s² so the measure is closer to this planet. It is likely that you are in Jupiter.

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A 7.5 cm object is 14 cm from a concave lens, which has a focal length of –7 cm. Its image is 5 cm in front of the lens. What is
Rama09 [41]
Given conditions:
height of object = 7.5cmdistance of object from mirror  = 14 cmfocus length = -7 cmimage distance =  ?
Using mirror formula: 
1/(focus length) = 1/(object distance) + 1/(image distance)
or, -1/7 = 1/14 + 1/(image distance)
or, image distance = -4.66cm (the image formed is a virtual image)


Also, magnification of image is:
image height /height of object = - image distance /object distance
or, image height = - image distance / object distance * height of object
or, image height = -(-4.66) / 14 * 7.5 = 2.49 = 3(nearest whole number)
3 0
3 years ago
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What is the weight of a 7.0 kilogram bowling ball on the surface of the moon
Artist 52 [7]
The acceleration on surface of moon =1.67m/s^2

Weight =mass ×acceleration
=7×1.67
=11.69N
8 0
3 years ago
A gas is compressed from an initial volume of 5.55 L to a final volume of 1.22 L by an external pressure of 1.00 atm. During the
algol13

Answer:

Explanation:

change in the volume of the gas = 5.55 - 1.22

= 4.33 X 10⁻³ m³

external pressure ( constant ) P = 1 x 10⁵ Pa

work done on the gas

=external pressure x change in volume

= 10⁵ x  4.33 X 10⁻³

=4.33 x 10²

433 J

Using the formula

Q = ΔE + W , Q is heat added , ΔE is change in internal energy , W is work done by the gas

Given

Q = - 124 J ( heat is released so negative )

W = - 433 J . ( work done by gas is negative, because it is done on gas  )

- 124  = ΔE - 433

ΔE = 433  - 124

= 309 J

There is increase of 309 J in the internal energy of the gas.

3 0
3 years ago
A satellite moves in a circular orbit at a constant speed v0 around Earth at a distance R from its center. The force exerted on
postnew [5]

Answer:

vT = v0/3

Explanation:

The gravitational force on the satellite with speed v0 at distance R is F = GMm/R². This is also equal to the centripetal force on the satellite F' = m(v0)²/R

Since F = F0 = F'

GMm/R² = m(v0)²/R

GM = (v0)²R  (1)

Also, he gravitational force on the satellite with speed vT at distance 3R is F1 = GMm/(3R)² = GMm/27R². This is also equal to the centripetal force on the satellite at 3R is F" = m(vT)²/3R

Since F1 = F'

GMm/27R² = m(vT)²/3R

GM = 27(vT)²R/3

GM = 9(vT)²R (2)

Equating (1) and (2),

(v0)²R = 9(vT)²R

dividing through by R, we have

9(vT)² = (v0)²

dividing through by 9, we have

(vT)² = (v0)²/9

taking square-root of both sides,

vT = v0/3

6 0
3 years ago
Numbers in exponential form
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Do the same thing you were doing
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