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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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3 years ago

Read 2 more answers

At each corner of a square of side l there are point charges of magnitude Q, 2Q, 3Q, and 4Q.What is the magnitude and direction

lbvjy [14]

Answer:

$F_T=6k\frac{Q^2}{L}\hat{i}+10k\frac{Q^2}{L}\hat{j}=2k\frac{Q^2}{L}[3\hat{i}+5\hat{j}]$

$|F_T|=2\sqrt{34}k\frac{Q^2}{L}$

$\theta=tan^{-1}(\frac{5}{3})=59.03\°$

Explanation:

I attached an image below with the scheme of the system:

The total force on the charge 2Q is the sum of the contribution of the forces between 2Q and the other charges:

$F_T=F_Q+F_{3Q}+F_{4Q}\\\\F_T=k\frac{(Q)(2Q)}{R_1}\hat{i}+k\frac{(3Q)(2Q)}{R_2}\hat{j}+k\frac{(4Q)(2Q)}{R_3}[cos\theta \hat{i}+sin\theta \hat{j}]$

the distances R1, R2 and R3, for a square arrangement is:

R1 = L

R2 = L

R3 = (√2)L

θ = 45°

$F_T=k\frac{2Q^2}{L}\hat{i}+k\frac{6Q^2}{L}\hat{j}+k\frac{8Q^2}{\sqrt{2}L}[cos(45\°)\hat{i}+sin(45\°)\hat{j}]\\\\F_T=k\frac{2Q^2}{L}\hat{i}+k\frac{6Q^2}{L}\hat{j}+k\frac{8Q^2}{\sqrt{2}L}[\frac{\sqrt{2}}{2}\hat{i}+\frac{\sqrt{2}}{2}\hat{j}]\\\\F_T=6k\frac{Q^2}{L}\hat{i}+10k\frac{Q^2}{L}\hat{j}=2k\frac{Q^2}{L}[3\hat{i}+5\hat{j}]$

and the magnitude is:

$|F_T|=2k\frac{Q^2}{L}\sqrt{3^2+5^2}=2\sqrt{34}k\frac{Q^2}{L}$

the direction is:

$\theta=tan^{-1}(\frac{5}{3})=59.03\°$

4 0

3 years ago

What is the maximum value the string tension can have before the can slips? The coefficient of static friction between the can a

Naya [18.7K]

Answer:

T= 38.38 N

Explanation:

Here

mass of can = m = 3 kg

g= 9.8 m/sec2

angle θ = 40°

From figure we see the vertical and horizontal component of tension force T

If the can is to slip - then horizontal component of tension force should become equal to force of friction.

First we find force of friction

Fs= μ R

where

μ = 0.76

R = weight of can = mg = 3 × 9.8 = 29.4 N

Now horizontal component of tension

Tx= T cos 40 = T× 0.7660 N

==>T× 0.7660 = 29.4

==> T= 38.38 N

8 0

3 years ago