How much power was used if it took them 60s

If the distance between us and a star is doubled, with everything else remaining the same, the luminosity If the distance betwee

natima [27]

Answer:

The luminosity remains the same, but the apparent brightness is decreased by a factor of four.

Explanation:

The apparent brightness, F = L/4πr² where L = luminosity and r = distance between us and the star.

Since L is independent of the distance between us and the star, it is constant, then

F ∝ 1/r²

So, F₁/F₂ = r₂²/r₁² where F₁ = apparent brightness at r₁ and F₂ = apparent brightness at r₂

If the distance is doubled, that is r₂ = 2r₁, then

F₁/F₂ = r₂²/r₁²

F₁/F₂ = (2r₁)²/r₁²

F₁/F₂ = 4r₁²/r₁²

F₁/F₂ = 4

F₂ = F₁/4

So, since the luminosity is constant, <u>the luminosity remains the same, but the apparent brightness is decreased by a factor of four.</u>

5 0

3 years ago

What is the Asch experiment

Leokris [45]

Answer:

the experiment that Asch was studying was how individuals yielded to or defied a majority of groups and the effect of such influences on beliefs and options

4 0

4 years ago

A pendulum is observed to complete 23 full cycles in 58 seconds. Determine the period and

Fynjy0 [20]

Answer:

Frequency=0.39Hz

Explanation:

A pendulum completes 23 full cycles in 58 seconds

A pendulum completes 1 full cycle in 58/23=2.52 seconds

Time taken(t)=2.52 seconds

Now,

Frequency=1/time taken

Frequency=1/2.25

Frequency=0.39Hz

6 0

2 years ago

A ball rolls off a level table with a speed of 2.2 m/s. The table is 1.2 meters high. How far away from the base of the table do

r-ruslan [8.4K]

The base of table is "1.08856 m" far away from the ball land.

Given:

Distance travelled by ball,

S = 1.2 m

Initial velocity,

u = 0 m/s

Acceleration,

a = 9.8 m/s²

Constant speed,

2.2 m/s

As we know the formula,

→ $S = ut+\frac{1}{2}at^2$

By substituting the values, we get

→ $1.2=\frac{1}{2}\times 9.8t^2$

→ $1.2 = 4.9 t^2$

→ $t = \sqrt{\frac{1.2}{4.9} }$

→ $= 0.4948 \ seconds$

Let,

Horizontal distance will be "x".
Time = t

→ $x = 2.2\times t$

$= 2.2\times 0.4948$

$= 1.08856 \ m$

Thus the solution above is right.

Learn more:

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8 0

3 years ago

Two cars start a race with initial velocity 7ms-1 and 4ms-1 respectively.Their acceleration are 0.4ms-2 and 0.5ms-2 respectively

Ronch [10]

The distance of the track is 600 m

Explanation:

The two cars move by uniformly accelerated motion, so we the distance they cover after a time t is described by the following suvat equation

$d=ut+\frac{1}{2}at^2$

where

u is the initial velocity

a is the acceleration

For car 1, we have

$d_1 = u_1 t + \frac{1}{2}a_1 t^2$

where

$u_1 = 7 m/s$ is the initial velocity of car 1

$a_1 = 0.4 m/s^2$ is the acceleration of car 1

So the equation can be rewritten as

$d_1 = 7t + 0.2t^2$

For car 2, we have

$d_2 = u_2 t + \frac{1}{2}a_2 t^2$

where

$u_2 = 4 m/s$ is the initial velocity of car 2

$a_2 = 0.5 m/s^2$ is the acceleration of car 2

So the equation can be rewritten as

$d_2= 5t + 0.25t^2$

The two cars finish the race at the same time: this means that they cover the same distance in the same time t, so we can write

$d_1 = d_2\\7t + 0.2t^2=5t + 0.25t^2$

And solving for t, we find

$2t - 0.05t^2= 0\\t(2-0.05t)=0$

Which gives two solutions:

t = 0 (initial instant)

t = 40 s

Therefore, the distance of the track is

$d_1 = 7t +0.2t^2 = 7\cdot 40+0.2\cdot 40^2=600 m$

Learn more about accelerated motion:

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7 0

4 years ago