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Anon25 [30]
2 years ago
12

What force is necessary to keep a mass of 0.8 kg revolving in a horizontal circle of radius 0.7 m with a period of 0.5 s? What i

s the direction of this force? ​
Physics
1 answer:
inn [45]2 years ago
8 0

Answer:

88.34 N directed towards the center of the circle

Explanation:

Applying,

F = mv²/r................... Equation 1

F = Force needed to keep the mass in a circle, m = mass of the mass, v = velocity of the mass, r = radius of the circle.

But,

v = 2πr/t................... Equation 2

Where t = time, π = pie

Substitute equation 2 into equation 1

F = m(2πr/t)²/r

F = 4π²r²m/t²r

F = 4π²rm/t²............. Equation 3

From the question,

Given: m = 0.8 kg, r = 0.7 m, t = 0.5 s

Constant: π = 3.14

Substitute these values into equation 3

F = 4(3.14²)(0.7)(0.8)/0.5²

F = 88.34 N directed towards the center of the circle

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Find the force required to do 25 joule work when the force causes a displacement of 0.5 m​
Eduardwww [97]

Answer:

<h2>50 N</h2>

Explanation:

The force required can be found by using the formula

f =  \frac{w}{d}  \\

w is the workdone

d is the distance

From the question we have

f =  \frac{25}{0.5}  \\

We have the final answer as

<h3>50 N</h3>

Hope this helps you

6 0
2 years ago
5 description of a motion​
yuradex [85]

Answer:

<h2>Displacement</h2><h2>Distance</h2><h2>Velocity</h2><h2> Acceleration</h2><h2>Speed</h2><h2> Time</h2>

Explanation:

HOPE IT HELPS

8 0
3 years ago
What colors form the spectrum of visible light?
USPshnik [31]

Answer:

Explanation:

ROYGBIV

Red

Orange

Yellow

Green

Blue

Indigo

Violet

3 0
3 years ago
Read 2 more answers
Calculate the cenrtripetal acceleration of the Earth in
miv72 [106K]

Answer:

0.00594\ m/s^2\ \text{towards the Sun}

3.54737\times 10^{22}\ N

Explanation:

r = Distance between Earth and Sun = 1.5\times 10^{11}\ m

Time taken to complete one rotation around Sun is given by

T=365.25\times 24\times 3600

Centripetal acceleration is given by

a=\dfrac{v^2}{r}\\\Rightarrow \\\Rightarrow a=\dfrac{(\dfrac{2\pi r}{T})^2}{r}\\\Rightarrow a=\dfrac{4\pi^2r}{T^2}\\\Rightarrow a=\dfrac{4\pi^2\times 1.5\times 10^{11}}{(365.25\times 24\times 3600)^2}\\\Rightarrow a=0.00594\ m/s^2\ \text{towards the Sun}

The centripetal acceleration of the Earth in its orbit is 0.00594\ m/s^2\ \text{towards the Sun}

Force is given by

F=ma\\\Rightarrow F=5.972\times 10^{24}\times 0.00594\\\Rightarrow F=3.54737\times 10^{22}\ N

The force on the Earth is 3.54737\times 10^{22}\ N

4 0
2 years ago
The pilot of an airplane traveling 45 m/s wants to drop supplies to flood victims isolated on a patch of land 160 m below. The s
Sidana [21]

Answer:

<h2>254.56 m</h2>

Explanation:

       A object dropped from a plane from a certain height will follow a parabolic trajectory because it has a horizontal velocity equal to plane's velocity.

       So, if supplies are to be dropped from a plane from a height of 160 m, let us calculate the time it takes to reach the ground.

       H=\frac{1}{2}gt^{2}\\160=\frac{1}{2}\times10\times t^{2}\\t=\sqrt{32}=4\sqrt{2}\text{ }sec

       So, in this time, the supply moves a horizontal distance of (4\sqrt{2}\text{ }sec)\times(45\text{ }\frac{m}{sec})=180\sqrt{2}\text{ }m=254.56\text{ }m.

∴ The supply must be dropped when the plane is 255 m away.

5 0
2 years ago
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