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3 years ago

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At the surface of Mars the acceleration due to gravity is 3.8m/s . A book weighs 34 N at the surface of the earth. What is it’s

mass on the earth’s surface? What are it’s mass and weight on mar’s surface?

1 answer:

PilotLPTM [1.2K]3 years ago

8 0

Answer:

On the surface of Mars:

Mass, m = 3.47 kg

Weight, W' = 13.186 N

Solution:

As per the question:

Acceleration due to gravity at the surface of Mars, g' = $3.8\ m/s^{2}$

Weight of book on the Earth's surface, W = 34 N

Now,

We know that:

Weight, W = mg

where

m = mass of the body

a = acceleration due to gravity on the earth's surface = $9.8\ m/s^{2}$

Thus the mass of the body on the surface of the earth, m:

$m = \frac{W}{g} = \frac{34}{9.8} = 3.47\ kg$

<em>Also, we know that mass is constant and it is the weight of the body that varies with the acceleration due to gravity.</em>

Now,

Mass of the body on the surface of Mars will be same and is equal to m = 3.47 kg

Weight of the book on the surface of Mars, W' = mg'

$W' = 3.47\times 3.8 = 13.186\ N$

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A 3.50 cm tall object is held 24.8 cm from a lens of focal length 16.0 cm. What is the image height?

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Answer:

<h2>6.36 cm</h2>

Explanation:

Using the formula to first get the image distance

1/f = 1/u+1/v

f = focal length of the lens

u = object distance

v = image distance

Given f = 16.0 cm, u = 24.8 cm

1/v = 1/16 - 1/24.8

1/v = 0.0625-0.04032

1/v = 0.02218

v = 1/0.02218

v = 45.09 cm

To get the image height, we will us the magnification formula.

Mag = v/u = Hi/H

Hi = image height = ?

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A swimmer bounces straight up from a diving board and falls feet first into a pool. She starts with a velocity of 4.00 m/s, and

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Answer:

a = 1.152s

b = 0.817 m

c = 7.29m/s

Explanation: let the following

From the first equation of linear motion

V = u+at..........1

parameters be represented as :

t = Time taken

v = Final velocity

a = Acceleration due to gravity = 9.8m/s²

u = Initial velocity = 4 m/s

s = Displacement

V = 0

Substitute the values into equation 1

0 = 4-9.8(t)

-4 = -9.8t

t = 4/9.8

t = 0.408s

From : s = ut+1/2at^2.........2

S = 4×0.408+0.5(-9.8)×0.408^2

S= 1.632-4.9(0.166)

S = 1.632-0.815

S = 0.817m

Her highest height above the board is 0.817 m

Total height she would fall is 0.817+1.90 = 2.717 m

From equation 2

s = ut+1/2at^2

2.717 m = 0t+0.5(9.8)t^2

2.717 m = 0+4.9t^2

2.717 m = 4.9t^2

2.717/4.9 = t^2

0.554 =t^2

t =√0.554

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Hence, her feet were in the air for 0.744+0.408seconds

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V = 7.29m/s

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a = 1.152s

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Answer:

Part A:

The proton has a smaller wavelength than the electron.

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Part B:

The proton has a smaller wavelength than the electron.

$\lambda_{proton} = 1.29x10^{-13}m$ < $\lambda_{electron} = 5.525x10^{-12}m$

Explanation:

The wavelength of each particle can be determined by means of the De Broglie equation.

$\lambda = \frac{h}{p}$ (1)

Where h is the Planck's constant and p is the momentum.

$\lambda = \frac{h}{mv}$ (2)

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$\lambda = \frac{6.624x10^{-34} J.s}{(9.11x10^{-31}Kg)(6.55x10^{6}m/s)}$

But $J = Kg.m^{2}/s^{2}$

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(9.11x10^{-31}Kg)(6.55x10^{6}m/s)}$

$\lambda = 1.10x10^{-10}m$

Case for the proton:

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$\lambda = 6.05x10^{-14}m$

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<em>Part B </em>

For part b, the wavelength of the electron and proton for that energy will be determined.

First, it is necessary to find the velocity associated to that kinetic energy:

$KE = \frac{1}{2}mv^{2}$

$2KE = mv^{2}$

$v^{2} = \frac{2KE}{m}$

$v = \sqrt{\frac{2KE}{m}}$ (3)

Case for the electron:

$v = \sqrt{\frac{2(7.89x10^{-15}J)}{9.11x10^{-31}Kg}}$

but $1J = kg \cdot m^{2}/s^{2}$

$v = \sqrt{\frac{2(7.89x10^{-15}kg \cdot m^{2}/s^{2})}{9.11x10^{-31}Kg}}$

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Then, equation 2 can be used:

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(9.11x10^{-31}Kg)(1.316x10^{8}m/s)}$

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$v = \sqrt{\frac{2(7.89x10^{-15}J)}{1.67x10^{-27}Kg}}$

But $1J = kg \cdot m^{2}/s^{2}$

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$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(1.67x10^{-27}Kg)(3.07x10^{6}m/s)}$

$\lambda = 1.29x10^{-13}m$

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