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3 years ago

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Two laptop computers each have a mass of 2 kg. If they are 0.5 m apart on a desk, what is the gravitational force between them?

G = 6.67 × 10-11 N⋅(m/kg)2

1 answer:

Rudik [331]3 years ago

4 0

Answer:

$F = 1.067 \times 10^{-9} N$

Explanation:

As we know that the gravitational force between two masses is given as

$F = \frac{Gm_1m_2}{r^2}$

here we know that

$m_1 = m_2 = 2 kg$

also we know that

$r = 0.5 m$

$G = 6.67 \times 10^{-11} Nm^2/kg^2$

so the force between two masses is given as

$F = \frac{(6.67 \times 10^{-11})(2)(2)}{0.5^2}$

$F = 1.067 \times 10^{-9} N$

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Answer:

$x=\dfrac{r}{\sqrt2}$

Explanation:

Given that

Radius =r

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Q=Charge on the ring

The electric filed at distance x given as

$E=K\dfrac{Q}{(r^2+x^2)^{3/2}}$

For maximum condition

$\dfrac{dE}{dx}=0$

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For maximum condition

$\dfrac{dE}{dx}=0$

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At $x=\dfrac{r}{\sqrt2}$ the electric field will be maximum.

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3 years ago

1. The statement about Newton's reflecting telescope contains 1 error. Find the error, write it, and

Jlenok [28]

Answer:

i. The error is the rough convex mirror.

ii. This should be replaced with a smooth convex morror.

Explanation:

Reflection is dependent on the surface involved and has two types; diffuse and specular. When the surface is rough, diffused reflection is observed. The surface causes a distortion of the incident light (the rays would be reflected at different angles to one another) after reflection. This makes some rays to interfere with one another. While specular reflection is observed with a smooth surface.

In the statement, the rough convex mirror would produce a distorted reflection which would produce diffused reflection. The effect is that few or no rays (depending on the degree of how rough the surfce is) would be reflected to the other smooth, flat diagonal mirror.

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3 years ago

You push a desk with 245 N, but the desk doesn't move due to its friction with the ground. What is the magnitude of the friction

const2013 [10]

If the desk doesn't move, then it's not accelerating.

If it's not accelerating, then the net force on it is zero.

If the net force on it is zero, then any forces on it are balanced.

If there are only two forces on it and they're balanced, then they have equal strengths, and they point in opposite directions.

So the friction on the desk must be equal to your<em> 245N</em> .

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3 years ago

Write down short notes on:<br>c. Power

Fudgin [204]

Answer: In the International System of Units, the unit of power is the watt, equal to one joule per second. In older works, power is sometimes called activity. Power is a scalar quantity.

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3 years ago

A car's bumper is designed to withstand a 5.04 km/h (1.4-m/s) collision with an immovable object without damage to the body of t

emmasim [6.3K]

Answer:

the magnitude of the average force on the bumper is 3189.8 N

Explanation:

Given the data in the question;

In terms of force and displacement, work done is;

W = $F^>$ × $x^>$

W = $Fxcos\theta$ ------- let this be equation 1

where F is force applied, x is displacement and θ is angle between force and displacement.

Now, since the displacement of the bumper and force acting on it is in the same direction,

hence, θ = 0°

we substitute into equation 1

W = $Fxcos($ 0° $)$

W = $Fx$ ------- let this be equation 2

Now, using work energy theorem,

total work done on the system is equal to the change in kinetic energy of the system.

$W_{net$ = ΔKE

= $\frac{1}{2}$ mv² - $\frac{1}{2}$ mu² --------- let this be equation 3

where m is mass of object, v is final velocity, u is initial velocity.

from equation 2 and 3

$Fx$ = $\frac{1}{2}$ mv² - $\frac{1}{2}$ mu²

we make F, the subject of formula

F = $\frac{m}{2x}$ ( v² - u² )

given that mass of car m = 830 kg, x = 0.255 m, v = 0 m/s, and u = 1.4 m/s

so we substitute

F = $\frac{830}{(2)(0.255)}$ ( (0)² - (1.4)² )

F = 1627.45098 ( 0 - 1.96 )

F = 1627.45098 ( - 1.96 )

F = -3189.8 N

The negative sign indicates that the direction of the force was in opposite compare to the direction of the velocity of the car.

Therefore, the magnitude of the average force on the bumper is 3189.8 N

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3 years ago