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3 years ago

A person weighing 785 newtons on the surface of Earth would weigh 298 newtons on the surface of Mars.

Physics

1 answer:

rodikova [14]3 years ago

3 0

Answer:

The gravitational field strength on the surface of Mars = 3.72 m/s²

Explanation:

Gravitational Field Strength: This can be defined as the force per unit mass which is exerted at that point. its direction is the force exerted on a mass in a gravitational field. The S.I unit of gravitational field strength is m/s²

Mathematically, Gravitational field is represented as,

g = F/m ..................... Equation 1.

m = F/g ..................... Equation 2.

Where g = gravitational Field Strength, F = force on the mass, m = mass of the body.

From the question,

Note: That The mass of the object is constant both on the surface of the earth and on the surface of Mars.

On the Surface of the earth,

Given: F = 785 N, g = 9.8 m/s²

Substituting this values into equation 2,

m = 785/9.8

m = 80.10 kg.

On the surface of Mars.

Given: m = 80.10 kg, F = 298.

Substituting into equation 2

g = 298/80.1

g = 3.72 m/s²

Thus the gravitational field strength on the surface of Mars = 3.72 m/s²

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A block is balanced on top of a frictionless sphere of radius R. When the block is given a slight nudge it starts to slide down

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Explanation:

Given

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When block is just about to leave then N=0

therefore

$mg\sin \theta =\frac{mv^2}{R}$

$v^2=gR\sin \theta$ -------------------1

Also by conserving Energy we get

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$mgh=\frac{mv^2}{2}$

here h=vertical distance traveled by block

From diagram

$h=R-R\sin \theta$

$h=R(1-\sin \theta )$

$mgR(1-\sin \theta )=\frac{mv^2}{2}$

$2gR(1-\sin \theta )=v^2$ -----------------2

From 1 and 2

$2(1-\sin \theta )=\sin \theta$

$3\sin \theta =2$

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Thus from this value of h is

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4 years ago

A speeding truck travels in the positive x direction. It hits a car traveling in the positive y direction. The vehicles are defo

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Answer:

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3 years ago

Read 2 more answers

PLEASE SOLVE QUICKLY!!!<br> Solve for A, B, and C from graph

hram777 [196]

A = 59.35cm

B = 196.56g

C = 74.65g

<u>Explanation:</u>

We know,

$x = \frac{L}{\frac{W}{F} +1}$

and L = x+y

Total length, L = 100cm

Weight of Beam, W = 71.8g

Center of mass, x = 49.2cm

Added weight, F = 240g

Position weight placed from fulcrum, y = ?

$L-y = \frac{L}{\frac{W}{F}+1 } \\100 - y = \frac{100}{\frac{71.8}{49.2}+1 } \\100 - y = \frac{100}{1.46+1}\\\\100 - y = \frac{100}{2.46} \\100-y = 40.65\\\\y = 59.35cm$

Therefore, position weight placed from fulcrum is 59.35cm

Total length, L = 100cm

Center of mass, x = 47.8 cm

Added weight, F = 180g

Position weight placed from fulcrum, y = 12.4cm

Weight of Beam, W = ?

$47.8 = \frac{100}{\frac{W}{180}+1 }\\\47.8 = \frac{100}{\frac{W+180}{180} } \\\\47.8 = \frac{100 X 180}{W+180}\\ \\47.8W + 47.8 X 180 = 18000\\47.8W = 18000 - 8604\\W = \frac{9396}{47.8}\\ W = 196.56g$

Therefore, weight of the beam is 196.56g

Total length, L = 100cm

Center of mass, x = 50.8 cm

Position weight placed from fulcrum, y = 9.8cm

Weight of Beam, W = 72.3g

Added weight, F = ?

$50.8 = \frac{100}{\frac{72.3}{F}+1 }\\\ 50.8 = \frac{100}{\frac{72.3+F}{F} } \\\\50.8 = \frac{100 X F}{72.3+F}\\ \\50.8 X 72.3 + 50.8 X F = 100F\\\\3672.84 = 100F-50.8F\\3672.84 = 49.2F\\F = 74.65g$

Therefore, Added weight F is 74.65g

A = 59.35cm

B = 196.56g

C = 74.65g

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