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Softa [21]
3 years ago
15

WOULUJUTUL RECIPECUIUS.

Physics
1 answer:
3241004551 [841]3 years ago
7 0

The force between the two objects is 19.73 nN.

<u>Explanation: </u>

Any force acting between two objects tends to be directly proportional to the product of their masses and inversely proportional to the square of the distance between the two objects. And this kind of attraction force between two objects is termed as gravitational force.

So if we consider M_{1} and M_{2} as the masses of both objects and let d be the distance of separation of two objects. Then the force between the two objects can be determined as below:

                      \text {Gravitational force}=\frac{G \times M_{1} \times M_{2}}{d^{2}}

As gravitational constant G=6.67 \times 10^{-11} \mathrm{m}^{3} \mathrm{kg}^{-1} \mathrm{s}^{-2}, M_{1} = 20 kg and  M_{2} = 100 kg, while d = 2.6 m, then

                    \text {Gravitational force}=\frac{6.67 \times 10^{-11} \times 20 \times 100}{(2.6)^{2}}=\frac{6.67 \times 20 \times 10^{-9}}{6.76}

Thus, we get finally,

                   \text {Gravitational force}=19.73 \times 10^{-9} \mathrm{N}

As we know, nano denoted by letter 'n' equals to 10^{-9}

So the force acting between two objects is 19.73 nN.

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I NEED HELP PLEASE, THANKS! :)
Yuliya22 [10]

Answer:

Radio waves have a wavelength between 10^{-9}m and 10^{-12}m

While,

X rays have a wavelength between 1m and 10km.

=> It is one of the condition of diffraction that the obstacle (coming in the way) must be comparable with the size of the wavelength.

=> This shows, that radio waves have a wavelength which is comparable with the size of buildings and can really easily diffract through it

=> While, X-rays are big enough to diffract through the wall.

So, if an X-ray technician stands behind a wall during the use of her machine, she will remain safe.

6 0
3 years ago
Physics question, help please!
vova2212 [387]

The change in potential energy when the block falls to ground is -480J.

The maximum change in kinetic energy of the ball is 480 J.

The initial kinetic energy of the ball is 0 J.

The final  kinetic energy of the ball is 0.148J.

The initial potential energy of the ball is 0.187 J.

The final  potential energy of the ball is 0 J.

The work done by the air resistance is 0.039 J.

<h3>Change in potential energy when the block falls to ground</h3>

ΔP.E = -mgh

ΔP.E = -Wh

ΔP.E = - 40 x 12

ΔP.E = -480 J

<h3>Maximum change in kinetic energy of the ball</h3>

ΔK.E = - ΔP.E

ΔK.E = - (-480 J)

ΔK.E = 480 J

<h3>Initial kinetic energy of the ball</h3>

K.Ei = 0.5mv²

where;

  • v is zero since it is initially at rest

K.Ei = 0.5m(0) = 0

<h3>Final kinetic energy</h3>

K.Ef =  0.5mv²

K.Ef = 0.5(0.0091)(5.7)²

K.Ef = 0.148 J

<h3>Initial potential energy of the ball</h3>

P.Ei = mghi

P.Ei = 0.0091 x 9.8 x 2.1

P.Ei = 0.187 J

<h3>Final potential energy</h3>

P.Ef = mghf

P.Ef = 0.0091 x 9.8 x 0

P.Ef = 0

<h3>Work done by the air resistance</h3>

W = ΔE

W = P.E - K.E

W = 0.187 J - 0.148 J

W = 0.039 J

Learn more about potential energy here: brainly.com/question/1242059

#SPJ1

<h3 />
7 0
1 year ago
A 50.0 kg child stands at the rim of a merry-go-round of radius 1.50 m, rotating with an angular speed of 3.00 rad/s. (a) what i
White raven [17]
Weight of the child m = 50 kg 
Radius of the merry -go-around r = 1.50 m
 Angular speed w = 3.00 rad/s
 (a)Child's centripetal acceleration will be a = w^2 x r = 3^2 x 1.50 => a = 9 x
1.5
 Centripetal Acceleration a = 13.5m/sec^2
 (b)The minimum force between her feet and the floor in circular path
 Circular Path length C = 2 x 3.14 x 1.50 => c = 3 x 3.14 => C = 9.424
 Time taken t = 2 x 3.14 / w => t = 6.28 / 3 => t = 2.09
 Calculating velocity v = distance / time = 9.424 / 2.09 m/s => v = 4.5 m/s
 Calculating force, from equation F x r = mv^2 => F = mv^2 / r => 50 x (4.5)^2

/ 1.5
 F = 50 x 3 x 4.5 => F = 150 x 4.5 => F = 675 N
 (c)Minimum coefficient of static friction u
 F = u x m x g => u = F / m x g => u = 675/ 50 x 9.81 => 1.376 
 u = 1.376
 Hence with the force and the friction coefficient she is likely to stay on merry-go-around.
8 0
3 years ago
An AM radio transmitter broadcasts 63.2 kW of power uniformly in all directions. (a) Assuming all of the radio waves that strike
egoroff_w [7]

Answer:

Intensity of the transmitted radio wave is 5.406 x 10⁻⁶ W/m²

Explanation:

Given;

power of radio transmitter, P = 63.2 kW = 63200 W

distance of transmission, r = 30.5 km

Intensity of the transmitted radio wave is calculated as follows;

I = \frac{P}{4\pi r^2}

where;

I is the intensity of the transmitted radio wave

Substitute the given values and calculate the intensity of the transmitted radio wave;

I = \frac{P}{4\pi r^2} = \frac{63200}{4\pi (30500)^2} = 5.406 *10^{-6} \ W/m^2

Therefore, Intensity of the transmitted radio wave is 5.406 x 10⁻⁶ W/m²

6 0
3 years ago
As you slide a heavy box across the floor, friction applies a force of -100 N
nirvana33 [79]

Answer:

A -500J

Explanation:

because W=Fs

100 × 5 = 500

4 0
3 years ago
Read 2 more answers
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