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2 years ago

The speed of light can be measured by;

Physics

1 answer:

Bezzdna [24]2 years ago

4 0

Answer:

it's the distance between objects in space

Explanation: Light travels super fast; but it still takes a long time to travel between objects in space. This is because distances between objects in space are enormous.

And can i please receive a brainliest and have a good day

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A pendulum consists of a large balanced mass hanging on the end of a long wire. At the point where a 28-kg pendulum has the grea

Ray Of Light [21]

Answer:

The length of the wire is approximately 67.1 m

Explanation:

The parameters of the pendulum are;

The mass of the pendulum, m = 28 kg

The angle between the pendulum weight and the wire, θ = 89°

The magnitude of the torque exerted by the pendulum's weight, τ = 1.84 × 10⁴ N·m

We have;

Torque, τ = F·L·sinθ = m·g·l·sinθ

Where;

F = The applies force = The weight of the pendulum = m·g

g = The acceleration due to gravity ≈ 9.8 m/s²

l = The length of the wire

Plugging in the values of the variables gives;

1.84 × 10⁴ N·m = 28 kg × 9.8 m/s² × l × sin(89°)

Therefore;

l = 1.84 × 10⁴ N·m/(28 kg × 9.8 m/s² × sin(89°)) = 67.0656080029 m ≈ 67.1 m

The length of the wire, l ≈ 67.1 m

6 0

2 years ago

Explain why machine can"t be <br>100% efficient practically?

Zigmanuir [339]

Answer:

Heat

Explanation:

Heat is the main component in why everything is not 100% efficient. Heat is a byproduct of everything and cannot be avoided.

5 0

3 years ago

Read 2 more answers

A girls throws a rock horizontaly with velocity=10m/s from a bridge .It falls 45 m

kherson [118]

Answer:

the answer is 45 m plus 10

4 0

2 years ago

magine an astronaut on an extrasolar planet, standing on a sheer cliff 50.0 m high. She is so happy to be on a different planet,

Mama L [17]

Answer:

$\Delta t=(\frac{20}{g'}+\sqrt{\frac{400}{g'^2}+\frac{100}{g'} } )-(\frac{20}{g}+\sqrt{\frac{400}{g^2}+\frac{100}{g} } )$

Explanation:

Given:

height above which the rock is thrown up, $\Delta h=50\ m$

initial velocity of projection, $u=20\ m.s^{-1}$

let the gravity on the other planet be g'

The time taken by the rock to reach the top height on the exoplanet:

$v=u+g'.t'$

where:

$v=$ final velocity at the top height = 0 $m.s^{-1}$

$0=20-g'.t'$ (-ve sign to indicate that acceleration acts opposite to the velocity)

$t'=\frac{20}{g'}\ s$

The time taken by the rock to reach the top height on the earth:

$v=u+g.t$

$0=20-g.t$

$t=\frac{20}{g} \ s$

Height reached by the rock above the point of throwing on the exoplanet:

$v^2=u^2+2g'.h'$

where:

$v=$ final velocity at the top height = 0 $m.s^{-1}$

$0^2=20^2-2\times g'.h'$

$h'=\frac{200}{g'}\ m$

Height reached by the rock above the point of throwing on the earth:

$v^2=u^2+2g.h$

$0^2=20^2-2g.h$

$h=\frac{200}{g}\ m$

The time taken by the rock to fall from the highest point to the ground on the exoplanet:

$(50+h')=u.t_f'+\frac{1}{2} g'.t_f'^2$ (during falling it falls below the cliff)

here:

$u=$ initial velocity= 0 $m.s^{-1}$

$\frac{200}{g'}+50 =0+\frac{1}{2} g'.t_f'^2$

$t_f'^2=\frac{400}{g'^2}+\frac{100}{g'}$

$t_f'=\sqrt{\frac{400}{g'^2}+\frac{100}{g'} }$

Similarly on earth:

$t_f=\sqrt{\frac{400}{g^2}+\frac{100}{g} }$

Now the required time difference:

$\Delta t=(t'+t_f')-(t+t_f)$

$\Delta t=(\frac{20}{g'}+\sqrt{\frac{400}{g'^2}+\frac{100}{g'} } )-(\frac{20}{g}+\sqrt{\frac{400}{g^2}+\frac{100}{g} } )$

3 0

2 years ago

A solid cylinder with a radius of 10 cm and a mass of 3.0 kg is rotating about its center with anangular speed of 3.5 rad/s. Wha

Lynna [10]

Answer:

kE=0.0735 J

Explanation:

Given that

Radius ,R=10 cm = 0.1 m

Mass ,m= 3 kg

Angular speed ,ω = 3.5 rad/s

We know that moment of inertia for solid sphere given as

$I=\dfrac{2}{5}mR^2$

Kinetic energy

$KE=\dfrac{1}{2}I\omega^2$

Now by putting the values

$KE=\dfrac{1}{2}\times \dfrac{2}{5}mR^2\times \omega^2$

$KE=\dfrac{1}{2}\times \dfrac{2}{5}\times 3\times 0.1^2\times 3.5^2\ J$

kE=0.0735 J

Therefore the kinetic energy will be 0.0735 J

3 0

2 years ago