Ask question

Ask question

All categories

3 years ago

5

More than 200 years later, Albert A. Michelson sent a beam of light from a revolving mirror to a stationary mirror 15 km away. S

how that the time interval between light leaving and returning to the revolving mirror was 0.0001 s.

2 answers:

MrRissso [65]3 years ago

4 0

Explanation:

distance in meter= 15000 m

velocity of light v=3×10^8 m/s

time interval t= d/v

= $\frac{15000}{ 3\times10^8}$

= 0.00005 sec

This is the time for the path in one direction. If you are looking for the time from the revolving mirror to the stationary mirror back to the revolving mirror, then it is twice the above value, which would be

2× 0.00005 = 0.0001 s

Aleks [24]3 years ago

3 0

Answer

given,

speed of light = 3 x 10⁸ m/s

distance of the mirror = 15 km

= 15000 m

time interval between light leaving and returning = ?

we know,

distance = speed x time

distance traveled by the light = 2 x 15000 = 30000 m

$t = \dfrac{distance}{speed}$

$t = \dfrac{30000}{3 \times 10^8}$

t = 0.0001 s

hence, it is shown that the time interval is t = 0.0001 s

You might be interested in

Which of the following is the most reasonable weight in units of newtons for an average adult?

Oliga [24]

Answer: The reason weight for average adult is 530N

Explanation: The clearly see this answer, let's convert this weight to kg using F=mg

530=m×9.8

g is a constant of 9.8m/s2

We divide both sides by 9.8 to get

54kg then this weight is reasonable for an average adult.

7 0

3 years ago

Read 2 more answers

A storage tank containing oil (SG=0.92) is 10.0 meters high and 16.0 meters in diameter. The tank is closed, but the amount of o

blagie [28]

Answer:

a-1 Graph is attached. The relation is linear.

a-2 The corresponding height for 68 kPa Pressure is 7.54 m

a-3 The corresponding weight for 68 kPa Pressure is 1394726kg

b The original height of the column is 5.98 m

Explanation:

Part a

a-1

The graph is attached with the solution. The relation is linear as indicated by the line.

a-2

By the equation

$P=\rho \times g \times h$

Here

P is the pressure which is given as 68 kPa.
ρ is the density of the oil whose SG is 0.92. It is calculated as

$\rho=S.G \times \rho_{water}\\\rho=0.92 \times 1000 kg/m^3\\\rho=920 kg/m^3\\$

g is the gravitational constant whose value is 9.8 m/s^2
h is the height which is to be calculated

$P=\rho \times g \times h\\h=\frac{P}{\rho \times g}\\h=\frac{68 \times 10^3}{920 \times 9.8}\\h=7.54m$

So the height of column is 7.54m

a-3

By the relation of volume and density

$M=\rho \times V$

Here

ρ is the density of the oil which is 920 kg/m^3
V is the volume of cylinder with diameter 16m calculated as follows

$V=\pi r^2h\\V=3.14\times (8)^2 \times 7.54\\V=1515.23 m^3$

Mass is given as

$M=\rho \times V\\M=920 \times 1515.23\\M=1394726kg$

So the mass of oil leading to 68kPa is 1394726kg

Part b

Pressure variation is given as

$\Delta P=P_{obs}-P_{atm}\\\Delta P=115-101 kPa\\\Delta P=14 kPa\\$

Now corrected pressure is as

$P_c=P_g-\Delta P\\P_c=68-14 kPa\\P_c=54 kPa$

Finding the value of height for this corrected pressure as

$P_c=\rho \times g \times h\\h=\frac{P_c}{\rho \times g}\\h=\frac{54 \times 10^3}{920 \times 9.8}\\h=5.98m$

The original height of column is 5.98m

4 0

3 years ago

You are standing in a building whose height (40m) you throw a ball downward at a angle of -30 at a speed of (10m/s) acceleration

jeyben [28]

Answer: 3.41 s

Explanation:

Assuming the question is to find the time $t$ the ball is in air, we can use the following equation:

$y=y_{o}+V_{o}sin \theta t-\frac{1}{2}gt^{2}$

Where:

$y=0m$ is the final height of the ball

$y_{o}=40 m$ is the initial height of the ball

$V_{o}=10 m/s$ is the initial velocity of the ball

$t$ is the time the ball is in air

$g=9.8 m/s^{2}$ is the acceleration due to gravity

$\theta=30\°$

Then:

$0 m=40 m+(10 m/s)(sin(30\°))t-\frac{1}{2}9.8 m/s^{2}t^{2}$

$0 m=40 m+5m/s t-4.9 m/s^{2}t^{2}$

Multiplying both sides of the equation by -1 and rearranging:

$4.9 m/s^{2}t^{2}-5m/s t-40 m=0$

At this point we have a quadratic equation of the form $at^{2}+bt+c=0$ , which can be solved with the following formula:

$t=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$

Where:

$a=4.9$

$b=-5$

$c=-40$

Substituting the known values:

$t=\frac{-(-5) \pm \sqrt{(-5)^{2}-4(4.9)(-40)}}{2(4.9)}$

Solving the equation and choosing the positive result we have:

$t=3.41 s$ This is the time the ball is in air

5 0

3 years ago

In the diagram, a force of 20 newtons is applied to a block. The block is in dynamic equilibrium. What is the magnitude and dire

sladkih [1.3K]

Answer:B 20 newtons opposite to the direction of the applied force

Explanation:

5 0

2 years ago

Read 2 more answers

1. A 3.5 kg object experiences an acceleration of 0.5 m/s2. What net force does the object experience

Leviafan [203]

Answer:

<h2>1.75 N</h2>

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 3.5 × 0.5 = 1.75

We have the final answer as

<h3>1.75 N</h3>

Hope this helps you

7 0

2 years ago