Ask question

Ask question

All categories

2 years ago

9

An experiment to measure the speed of light uses an apparatus similar to Fizeau's. The distance between the light source and the

mirror is 10 m, and the wheel has 800 notches. If the wheel rotates at 9000 rev/s when the light from the source is extinguished, what is the experimental value for c (in m/s)?

1 answer:

Marina86 [1]2 years ago

4 0

Answer:

$2.88*10^{8} m/s$

Explanation:

The speed of light is given by

$c=\frac {2d}{t}$ and $t=\frac {\theta}{\omega}$ hence

$c=\omega\frac {2d}{\theta}$

Speed of light is given by

$c=900\times 2\pi(\frac {2\times 10}{\frac {2\pi}{2*800}}})=2.88*10^{8} m/s$

You might be interested in

An individual white LED (light-emitting diode) has an efficiency of 20% and uses 1.0 W of electric power. a. How many LEDs must

Neporo4naja [7]

Answer:

8, 8 W

Explanation:

The useful power of 1 Light Emitting Diode is

$0.2\times 1=0.2\ W$

Total power required is 1.6 W

Number of Light Emitting Diodes would be

$n=\dfrac{1.6}{0.2}\\\Rightarrow n=8$

The number of Light Emitting Diodes is 8.

Power would be

$P=8\times 1=8\ W$

The power that is required to run the Light Emitting Diodes is 8 W

7 0

2 years ago

Kevin Tan's Balance Sheet. Total assets are 13,200 dollars. Total liabilities are 9,150 dollars. What is Kevin’s net worth on Ma

Alexus [3.1K]

Answer:

4,050

Explanation:

I did the math.

5 0

2 years ago

A target in a shooting gallery consists of a vertical square wooden board, 0.250 m on a side and with mass 0.750 kg, that pivots

Alenkasestr [34]

Here in this case since there is no torque about the hinge axis for the system of bullet and block then we can say that angular momentum of this system will remain conserved

$L_i = L_f$

$mv \frac{L}{2} = (I_1 + I_2)\omega$

here we will have

L = 0.250 m

v = 385 m/s

m = 1.90 gram

now moment of inertia of the plate will be

$I_1 = \frac{ML^2}{3}$

$I_1 = \frac{0.750 (0.250)^2}{3} = 0.0156 kg m^2$

$I_2 = m(\frac{L}{2})^2 = 0.0019(0.125)^2 = 2.97 \times 10^{-5} kg m^2$

now from above equation

$0.0019 (385)(0.125) = (0.0156 + 2.97 \times 10^{-5})\omega$

$\omega = 5.85 rad/s$

8 0

2 years ago

A man pushes his child in a grocery cart. The total mass of the cart and child is 30.0 kg. If the force resisting the carts moti

-BARSIC- [3]

The force applied by the man is 60 N

Explanation:

We can solve this problem by applying Newton's second law, which states that:

$\sum F = ma$ (1)

where

$\sum F$ is the net force acting on the child+cart

m is the mass of the child+cart system

a is their acceleration

In this problem, we have:

m = 30.0 kg is the mass

$a=1.50 m/s^2$

And there are two forces acting on the child+cart system:

The forward force of pushing, F
The force resisting the cart motion, R = 15.0 N

Therefore we can write the net force as

$\sum F = F -R$

where R is negative since its direction is opposite to the motion

So eq.(1) can be rewritten as

$F-R=ma$

And solving for F,

$F=ma+R=(30.0)(1.50)+15.0=60 N$

Learn more about Newton's second law:

brainly.com/question/3820012

#LearnwithBrainly

4 0

2 years ago

Charge q1 is distance r from a positive point charge q. charge q2=q1/3 is distance 2r from q. what is the ratio u1/u2 of their p

makvit [3.9K]

We need the power law for the change in potential energy (due to the Coulomb force) in bringing a charge q from infinity to distance r from charge Q. We are only interested in the ratio U₁/U₂, so I'm not going to bother with constants (like the permittivity of space).

<span>The potential energy of charge q is proportional to </span>
<span>∫[s=r to ∞] qQs⁻²ds = -qQs⁻¹|[s=r to ∞] = qQr⁻¹, </span>

<span>so if r₂ = 3r₁ and q₂ = q₁/4, then </span>
<span>U₁/U₂ = q₁Qr₂/(r₁q₂Q) = (q₁/q₂)(r₂/r₁) </span>
<span>= 4•3 = 12.</span>

5 0

2 years ago