What is the difference in the speed of sound on a warm day versus on a cold day?

What are some ways that light can be controlled and what does it do to the light

marusya05 [52]

-- pass the light through a lens
The path of the light is bent (refracted) to a new direction.

-- bounce the light off a shiny surface
The light is sent back (reflected) in the direction from which it arrived.

-- pass the light through a prism
The light is spread out according to the different wavelengths
that may be in it.

-- put something black in the light's path
The light is completely absorbed and is never seen again.

-- turn the light off
The source stops emitting light.

-- throw a towel over the lamp
The light is absorbed in the towel, and not seen outside of it.

7 0

3 years ago

Assuming 70% of Earth's surface is covered in water at an average depth of 2.5 mi, estimate the mass of the water on Earth in Ki

saw5 [17]

This is an excellent question that i do not have the answer to.

7 0

3 years ago

A solenoid coil with 22 turns of wire is wound tightly around another coil with 340 turns. The inner solenoid is 25.0 cm long an

LUCKY_DIMON [66]

Answer:

a) 1.34*10^-8 W

b) 1.18*10^-5 H

c) 20mV

Explanation:

a) To find the average magnetic flux trough the inner solenoid you the following formula:

$\Phi_B=BA=\mu_oNIA$

mu_o: magnetic permeability of vacuum = 4pi*10^-7 T/A

N: turns of the solenoid = 340

I: current of the inner solenoid = 0.100A

A: area of the inner solenoid = pi*r^2

r: radius of the inner solenoid = 2.00cm/2=1.00cm=10^-2m

You calculate the area and then replace the values of N, I, mu_o and A to find the magnetic flux:

$A=\pi(10^{-2}m)^2=3.141510^{-4}m^2\\\Phi_B=(4\pi*10^{-7}T/A)(340)(0.100A)(3.1415*10^{-4}m^2)=1.34*10^{-8}W\\$

the magnetic flux is 1.34*10^{-8}W

b) the mutual inductance is given by:

$M=\mu_o N_1 N_2 \frac{A_2}{l}$

N1: turns of the outer solenoid = 22

N2: turns of the inner solenoid

A_2: area of the inner solenoid

l: length of the solenoids = 25.0cm=0.25m

by replacing all these values you obtain:

$M=(4\pi*10^{-7}T/A)(340)(22)\frac{3.14*10^{-4}m^2}{0.25m}=1.18*10^{-5}H$

the mutual inductance is 1.18*10^{-5}H

c) the emf induced can be computed by using the mutual inductance and the change in the current of the inner solenoid:

$\epsilon_1=M\frac{dI_2}{dt}$

by replacing you obtain:

$\epsilon_1=(1.18*10^{-5}H)(1700A/s)=0.02V=20mV$

the emf is 20mV

7 0

3 years ago

Read 2 more answers

Describe the motion for each segment below. Include start position, relative speed, and direction. Then calculate the speed of e

ratelena [41]

Where is the picture?

3 0

3 years ago

Some of the highest tides in the world occur in the Bay of Fundy on the Atlantic Coast of Canada. At Hopewell Cape the water dep

Nady [450]

Answer:

(a) 1.939 m/h

(b) 0.926 m/h

(c) -0.315 m/h

(d) -1.21 m/h

Explanation:

Here, we have the water depth given by the function of time;

D(t) = 7 + 5·cos[0.503(t-6.75)]

Therefore, to find the velocity of the depth displacement with time, we differentiate the given expression with respect to time as follows;

D'(t) = $\frac{d(7 + 5\cdot cos[0.503(t-6.75)])}{dt}$

= 5×(-sin(0.503(t-6.75))×0.503

= -2.515×(-sin(0.503(t-6.75))

= -2.515×(-sin(0.503×t-3.395))

Therefore we have;

(a) at 5:00 AM = 5 - 0:00 = 5

D'(5) = -2.515×(-sin(0.503×5-3.395)) = 1.939 m/h

(b) at 6:00 AM = 6 - 0:00 = 6

D'(5) = -2.515×(-sin(0.503×6-3.395)) = 0.926 m/h

(c) at 7:00 AM = 7 - 0:00 = 7

D'(5) = -2.515×(-sin(0.503×7-3.395)) = -0.315 m/h

(d) at Noon 12:00 PM = 12 - 0:00 = 12

D'(5) = -2.515×(-sin(0.503×12-3.395)) = -1.21 m/h.

4 0

3 years ago