Ask question

Ask question

All categories

2 years ago

13

When a single source of light shines through an extremely thin rectangular slit and projects on a far away viewing screen, a sin

gle rectangular region of the viewing screen is illuminated (matching the shape of the thin rectangular slit).
O True O False

1 answer:

saveliy_v [14]2 years ago

7 0

True
If it helps you plz brainlest me

You might be interested in

An object with a mass of 2.0 kg accelerates 2.0 m/s 2when an unknown force is applied to it. What is the amount of force?

Musya8 [376]

Answer:

4 N

Explanation:

mass = 2 kg

acceleration = 2 m/s^2

Force = mass * acceleration

= 2 *2

= 4 N

5 0

2 years ago

2. A solid plastic cube of side 0.2 m is submerged in a liquid of density 0.8 hgm calculate the

kotegsom [21]

Answer:

vpg = 0.064 N

Explanation:

Upthrust = Volume of fluid displaced

upthrust liquid on the cube g=10ms−2

vpg =0.2 x 0.2 x 0.2 x0.8 x 10= 0.064N

vpg = 0.064 N

hope it helps.

3 0

3 years ago

What is the resistance of a 3.5 m copper wire (Rho= 1.7x10-8 Ohm·m) that 1 point

VikaD [51]

Answer:

(D)

Explanation:

Given :

l=3.5 m

$A=5.26*10^{-6}$ $m^{2}$

$p=1.7*10^{-8} ohm.m$

Resistance can be calculated as :

$R=p\frac{l}{A} \\R=1.7*10^{-8} \frac{3.5}{5.26*10^{-6} }$

$R=\frac{5.95*10^{-2} }{5.26} \\R=1.13*10^{-2}$

Resistance of the wire will be 1.1× $10^{-2}$ ohms

Option D is correct

4 0

2 years ago

Your friend decides to generate electrical power by rotating a 100,000 turn coil of wire around an axis in the plane of the coil

MakcuM [25]

Answer:

a) I=35mA

b) P=1.73W

Explanation:

a) The max emf obtained in a rotating coil of N turns is given by:

$emf_{max}=NBA\omega$

where N is the number of turns in the coil, B is the magnitude of the magnetic field, A is the area and w is the angular velocity of the coil.

By calculating A and replacing in the formula (1G=10^{-4}T) we get:

$A=\pi r^2 =\pi(0.23m)^2=0.16m^2$

$emf_{max}=(100000)(0.3*10^{-4}T)(0.166m^2)(140\frac{rev}{s})=69.72V$

Finally, the peak current is given by:

$I=\frac{emf}{R}=\frac{69.72V}{1400\Omega}=49.8mA$

b)

we have that

$I_{rms}=\frac{I}{\sqrt{2}}=\frac{0.0498A}{\sqrt{2}}=0.035A$

$P_{rms}=I^2{rms}R=(0.035A)^2(1400\Omega)=1.73W$

hope this helps!!

6 0

2 years ago

Energy that does not involve the large-scale motion or the position of objects in a system is called a. potential energy. b. mec

docker41 [41]

Answer:

C

Explanation:

4 0

2 years ago

Read 2 more answers