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3 years ago

A. A man of 100 kg mass is standing on the surface of the earth of mass 6 × 1024 kg and

Physics

1 answer:

Alexxx [7]3 years ago

7 0

Answer:

m1=100kg

m2=6×10^24kg

r=6400km

G=6.6×10^-11

F=?

as we know that

F=G m1.m2/r²

F=6.6×10^-11×100×6×10^24/(6400)²

F=6.6×10^-11×6×10^26/4.096×10^7

F=6.6×10^-11×1.465×10^19

F=9.67×10^8N

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If the magnitude of a positive charge is tripled, what is the ratio of the original value of the electric field at a point to th

ipn [44]

Answer:

b)1 :3

Explanation:

Lets that

The value of a positive charge = q

As we know that electric filed on a point charge given as

$E=\dfrac{Kq}{r^2}$

Where ,K=Constant

q=Charge ,r=Distance

If the value of the charge gets tripled ,q'= 3 q

Then electric filed E'

$E'=\dfrac{Kq'}{r^2}$

$E'=\dfrac{3Kq}{r^2}$

E' = 3 E

Therefore we can say that

$\dfrac{E}{E'}==\dfrac{1}{3}$

therefore the answer will be --

b)1 :3

3 0

2 years ago

An 18.5-cm-diameter loop of wire is initially oriented perpendicular to a 1.5-T magnetic field. The loop is rotated so that its

nikdorinn [45]

Answer:

The average induced emf in the loop is 0.20 V

Explanation:

Given:

Radius of loop $r = \frac{d}{2} = 9.25 \times 10^{-2}$ m

Magnetic field $B = 1.5$ T

Change in time $\Delta t = 0.20$ sec

According to the faraday's law,

Induced emf is given by

$\epsilon = -\frac{\Delta \phi}{\Delta t}$

Where $\phi =$ magnetic flux

$\phi = BA\cos0$ ( here $\theta = 0$ )

Where $A = \pi r^{2}$

We neglect minus sign because it's shows lenz law

$\epsilon = \frac{B \pi r^{2} }{\Delta t}$

$\epsilon = \frac{1.5 \times 3.14 \times (9.25 \times 10^{-2} )^{2} }{0.20}$

$\epsilon = 0.20$ V

Therefore, the average induced emf in the loop is 0.20 V

4 0

3 years ago

True or false Sound waves do not require a medium in which to travel

Delicious77 [7]

False. That's why sound cannot be heard in space.

7 0

3 years ago

Read 2 more answers

If two children, with masses of 16 kg and 24 kg , sit in seats opposite one another, what is the moment of inertia about the rot

Elena-2011 [213]

Answer:

The moment of inertia about the rotation axis is 117.45 kg-m²

Explanation:

Given that,

Mass of one child = 16 kg

Mass of second child = 24 kg

Suppose a playground toy has two seats, each 6.1 kg, attached to very light rods of length r = 1.5 m.

We need to calculate the moment of inertia

Using formula of moment of inertia

$I=I_{1}+I_{2}$

$I=(m+m_{1})\times r^2+(m+m_{2})\times r^2$

m = mass of seat

m₁ =mass of one child

m₂ = mass of second child

r = radius of rod

Put the value into the formula

$I=(16+6.1)\times(1.5)^2+(24+6.1)\times(1.5)^2$

$I=117.45\ kg-m^2$

Hence, The moment of inertia about the rotation axis is 117.45 kg-m²

8 0

3 years ago

I WILL GIVE BRAINLIEST!!!

astra-53 [7]

Answer:

is that high school work??? cause I don't know it and I'm about to go to high school

7 0

3 years ago