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Zolol [24]
4 years ago
10

An elephant and a mouse would both have zero weight in gravity-free space. If they were moving toward you with the same speed, w

ould they bump into you with the same effect? Explain.
Physics
1 answer:
Lunna [17]4 years ago
8 0

The elephant and the mouse having zero weight in a gravity free space will not bump into you at the same effect.

<u>Explanation: </u>

When both are in a gravity free space, the weights are zero, as we know that the\text {weight of the body}=\text {mass of the body} \times \text {acceleration due to gravity}

\text {here, the weight of elephant}=\text {mass of elephant } \times \text {zero gravti} y=zero

\text {similarly,weight of mouse}=\text {mass of mouse } \times \text {zero gravity}=zero

But when they will acquire the speed of same magnitude, say v, their different masses will acquire different momentum, which will make the difference in effect while bumping.  

\text { momentum of elephant }=\text { mass of elephant } \times v  \text { momentum of mouse = mass of mouse } \times v

And as we know \text { mass of elephant }>\text { mass of mouse }  Therefore, effect of impact by elephant will be more than that of mouse . An elephant breaking into you will take you back faster than a mouse in space hits you.

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Answer:

4

Units:

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3 0
3 years ago
A. How long does it take light to travel through a 3.0-mm-thick piece of window glass?
hodyreva [135]

Answer:

a) 1.517\times10^{-11} s

b) 3.41 mm

Explanation:

a)

We take the speed of light, c = 3.0\times10^8 m/s and the refractive index of glass as 1.517.

Speed = distance/time

Time = distance/speed

Refractive index, n = speed of light in vacuum / speed of light in medium

n=\dfrac{c}{s}

s=\dfrac{c}{n}

t=\dfrac{d}{c/n}

t=\dfrac{dn}{c}

t=\dfrac{3\times10^{-3}\times1.517}{3.0\times10^8}

t=1.517\times10^{-11}

b)

We take the refractive index of water as 1.333.

Speed in water = speed in vacuum / refractive index of water

Distance = speed * time

d=s\times t

d=\dfrac{c}{n_w}\times \dfrac{3\times10^{-3}\times1.517}{c}

d=\dfrac{3\times10^{-3}\times 1.517}{1.333}

d = 3.41 mm

6 0
4 years ago
The cornea behaves as a thin lens of focal lengthapproximately 1.80 {\rm cm}, although this varies a bit. The material of whichi
Keith_Richards [23]

Answer:

Explanation:

  a )

from lens makers formula

\frac{1}{f} =(\mu-1)(\frac{1}{r_1} -\frac{1}{r_2})

f is focal length , r₁ is radius of curvature of one face and r₂ is radius of curvature of second face

putting the values

\frac{1}{1.8} =(1.38-1)(\frac{1}{.5} -\frac{1}{r_2})

1.462 = 2 - 1 / r₂

1 / r₂ = .538

r₂ = 1.86 cm .

= 18.6 mm .

b )

object distance u = 25 cm

focal length of convex lens  f  = 1.8 cm

image distance  v   = ?

lens formula

\frac{1}{v} - \frac{1}{u} = \frac{1}{f}

\frac{1}{v} - \frac{1}{-25} = \frac{1}{1.8}

\frac{1}{v} = \frac{1}{1.8} -\frac{1}{25}

.5555 - .04

= .515

v = 1.94 cm

c )

magnification = v / u

= 1.94 / 25

= .0776

size of image = .0776 x size of object

= .0776 x 10 mm

= .776 mm

It will be a real image and it will be inverted.

 

5 0
3 years ago
What is the angle of reflection?
hoa [83]
I can guarantee you that it is not
C.<span>the angle that the incident ray makes with a line drawn perpendicular to the reflecting surface  I hope this somewhat helps</span>
4 0
3 years ago
A car starts moving from the position of rest with uniform acceleration of 8m/s^calculate the distance travelled by it during 10
ad-work [718]

We calculate the coordinates at t₁ = 9 min and t₂ = 10 min, since the 10th minute is between t₁ and t₂.

As it leaves from rest, it means that the initial speed is zero


t₁=9 min=540 s

t₂=10 min=600 s

x₁=at₁²/2=8*540²/2=4*291600=1166400 m

x₂=at₂²/2=8*600²/2=4*360000=1440000 m

Δx=x₂-x₁=1440000-1166400=273600 m represents the distance traveled by the car in the 10th minute of travel




8 0
3 years ago
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