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2 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]2 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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A motorboat traveling 4 m/s, East encounters a current traveling 3.0 m/s, North. a) What is the resultant velocity of the motorb

Nostrana [21]

Explanation:

It is given that,

Velocity in East, $v_1=4\ m/s$

Velocity in North, $v_2=3\ m/s$

(a) The resultant velocity is given by :

$v=\sqrt{4^2+3^2}=5\ m/s$

(b) The width of the river is, d = 80 m

Let t is the time taken by the boat to travel shore to shore. So,

$t=\dfrac{d}{v}$

$t=\dfrac{80\ m}{5\ m/s}$

t = 16 seconds

(c) Let x is the distance covered by the boat to reach the opposite shore. So,

$x=v_2\times t$

$x=3\ m/s\times 16\ s$

x = 48 meters

Hence, this is the required solution.

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2 years ago

At a depth of 10.9 km, the Challenger Deep in the Marianas Trench of the Pacific Ocean is the deepest site in any ocean. Yet, in

bearhunter [10]

Answer:

$P = 1.09 \times 10^8 Pa$

Explanation:

As we know that the pressure inside the liquid level is given as

$P = \rho g h + P_o$

here we have

$\rho = 1024 kg/m^3$

h = 10.9 km

also we know that

$P_o = 1.01 \times 10^5 Pa$

now we have

$P = (1.01 \times 10^5) + (1024)(9.81)(10.9 \times 10^3)$

$P = 1.09 \times 10^8 Pa$

8 0

2 years ago

The masses of blocks A and B are 20kg and 10kg, respectively. The blocks are initially at rest on the floor and are connecte by

Goryan [66]

Answer

mass of the block A = 20 Kg

mass of block B = 10 Kg

acceleration of Block A = ?

acceleration of Block B = ?

Assuming the magnitude of force = 124 N

Applying newton's second law

F = 2 T

T = F/2

now, Tension in the string =

T = 124/2 = 62 N

Weight of the block A

W = m₁ g

W = 20 x 9.8 = 196 N

Weight of the block B

W = m₂ g

W = 10 x 9.8

W = 98 N

Weight of blocks is greater than force applied by the pulley so, the blocks will not move.

Hence, acceleration of the block A and B = 0 m/s²

If the Force magnitude is increased to 294 N

T = F/2 = 294/2 = 147 N

Since this Force is less than Weight A acceleration of the block A = 0 m/s².

For Block B

Tension is more than Weight hence block will move

Net Force = 147 - 98 = 49 N

acceleration = $\dfrac{F}{m}$

$a = \dfrac{49}{10}$

a = 4.9 m/s²

8 0

2 years ago

A window washer is standing on a scaffold supported by vertical ropes at each end. The scaffold weighs 200. N and is 3.00 m long

seropon [69]

<h2>Tension in ropes are 333.33 N and 566.67 N.</h2>

Explanation:

Refer the given figure showing the arrangement.

For having stability we have torque at A and B are zero.

Sum of Torques at A = 0

700 x 1 + 200 x 1.5 - Q x 3 = 0

3Q = 1000

Q = 333.33 N

Sum of Torques at B = 0

200 x 1.5 + 700 x 2 - P x 3 = 0

3P = 1700

P = 566.67 N

Tension in ropes are 333.33 N and 566.67 N.

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2 years ago

Does the sun give plants the nutrients to grow yes or no!!!

Vlad1618 [11]

Your is going to be yes.

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2 years ago