If speed = distance/time , then time = speed/distance.

So...

Speed of light = 3*10^8(m/s)
Average distance from Earth to Sun = 149.6*10^9(m)

Therefore, t=(3*10^8(m/s))/(149.6*10^9(m))

I hope this was a helpful explanation, please reply if you have further questions about the problem.

Good luck!

5 0

3 years ago

a person when asked to speak up,increases her sound level from 30dB to 60dB.The amount of power per unit area increased by? a)30

aivan3 [116]

Answer:

d) 1000 times

Explanation:

As we know that difference of sound level is given as

$L_2 - L_1 = 10 Log \frac{I_2}{I_1}$

so here we need to find the ratio of two intensity

it is given as

$Log\frac{I_2}{I_1} = \frac{(L_2- L_1)}{10}$

$Log\frac{I_2}{I_1} = \frac{60 - 30}{10}$

$Log\frac{I_2}{I_1} = 3$

now we have

$\frac{I_2}{I_1} = 10^3$

so it is

d) 1000 times

4 0

3 years ago

a stone is dropped from rest at an initial height h above the surface of the earth. Show that the speed with which it strikes th

Ksju [112]

Answer:

$v=\sqrt{2gh}$

Explanation:

We could use conversation of energy. Total distance the stone will cover will be

$h=\frac{1}{2} g t^{2}$

the Final velocity will be

$v=gt\\v=g\sqrt{\frac{2h}{g} }\\ v=\sqrt{2gh}$

5 0

3 years ago

A jet is travelling at a speed of 1200 km/h and drops cargo from a height of 2.5 km above the ground Calculate the time it takes

OLEGan [10]

a) Time of flight: 22.6 s

To calculate the time it takes for the cargo to reach the ground, we just consider the vertical motion of the cargo.

The vertical position at time t is given by

$y(t) = h +u_y t - \frac{1}{2}gt^2$

where

h = 2.5 km = 2500 m is the initial height

$u_y = 0$ is the initial vertical velocity of the cargo

g = 9.8 m/s^2 is the acceleration of gravity

The cargo reaches the ground when

$y(t) = 0$

So substituting it into the equation and solving for t, we find the time of flight of the cargo:

$0 = h - \frac{1}{2}gt^2\\t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2(2500)}{9.8}}=22.6 s$

b) 7.5 km

The range travelled by the cargo can be calculated by considering its horizontal motion only. In fact, the horizontal motion is a uniform motion, with constant velocity equal to the initial velocity of the jet:

$v_x = 1200 km/h \cdot \frac{1000 m/km}{3600 s/h}=333.3 m/s$

So the horizontal distance travelled is

$d=v_x t$

And if we substitute the time of flight,

t = 22.6 s

We find the range of the cargo:

$d=(333.3)(22.6)=7533 m = 7.5 km$

7 0

3 years ago