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weeeeeb [17]
3 years ago
11

In a laboratory experiment, a diffraction grating produces an interference pattern on a screen. If the number of slits in the gr

ating is increased, with everything else (including the slit spacing) the same, thenA.The fringes stay the same brightness and get closer together.B.The fringes stay the same brightness and get farther apart.C.The fringes stay in the same positions but get brighter and narrower.D.The fringes stay in the same positions but get dimmer and wider.E.The fringes get brighter, narrower, and closer together.
Physics
1 answer:
8_murik_8 [283]3 years ago
4 0

Answer:

C

Explanation:

Diffraction grating is an optics tool used when there is a need to separate light of different wavelengths using a high resolution. This tool is used in:

  • Optical pulse compressing devices
  • Spectrometers
  • Lasers
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A car is traveling at 120 km/h (75 mph). When applied the braking system can stop the car with a deceleration rate of 9.0 m/s2.
Bumek [7]

Answer:

the number of additional car lengths approximately it takes the sleepy driver to stop compared to the alert driver is 15

Explanation:

Given that;

speed of car V  = 120 km/h = 33.3333 m/s

Reaction time of an alert driver = 0.8 sec

Reaction time of an alert driver = 3 sec

extra time taken by sleepy driver over an alert driver = 3 - 0.8 = 2.2 sec

now, extra distance that car will travel in case of sleepy driver  will be'

S_d = V × 2.2 sec

S_d = 33.3333 m/s × 2.2 sec

S_d = 73.3333 m

hence, number of car of additional car length  n will be;

n = S_n / car length

n = 73.3333 m / 5m

n = 14.666 ≈ 15

Therefore, the number of additional car lengths approximately it takes the sleepy driver to stop compared to the alert driver is 15

8 0
3 years ago
Two rams run toward each other. One ram has a mass of 49 kg and runs west
olga nikolaevna [1]

Answer: (d)

Explanation:

Given

Mass of the first ram m_1=49\ kg

The velocity of this ram is v_1=-7\ m/s

Mass of the second ram m_2=52\ kg

The velocity of this ram v_2=9\ m/s

They combined after the collision

Conserving the momentum

\Rightarrow m_1v_1+m_2v_2=(m_1+m_2)v\\\Rightarrow 49\times (-7)+52\times (9)=(52+49)v\\\Rightarrow v=\dfrac{125}{101}\ m/s \quad[\text{east}]

Momentum after the collision will be

\Rightarrow 101\times \dfrac{125}{101}=125\ kg-m/s\ \text{East}

Therefore, option (d) is correct

4 0
3 years ago
What is the value of work done on an object when a
tino4ka555 [31]
W = F * s
Here, F = 50 N
s = 15 m

Substitute their values, 
W = 50 * 15
W = 750 J

In short, Your Answer would be 750 Joules

Hope this helps!
7 0
3 years ago
Why is it that the two possible faults with inductors are short circuit and open circuit?
lukranit [14]
These are the most common type of faults not just inductors but also with other elements too like resistors,transformers, generators etc.
open circuit fault means the flow of current is disrupted some how in the circuit and the circuit stops operating. and for short circuit fault the current in the system will be pretty high and this short circuit current or fault current will always run back to the fault location, if the inductor got short circuited somehow then the fault current will only run through it because it will then provide a very low impedence path
5 0
3 years ago
A 0.877 mol sample of N2(g) initially at 298 K and 1.00 atm is held at constant volume while enough heat is applied to raise the
MissTica

Answer : The value of q\text{ and }\Delta U is 286.2 J and 286.2 J respectively.

Explanation : Given,

Moles of sample = 0.877 mol

Change in temperature = 15.7 K

First we have to calculate the heat absorbed by the system.

Formula used :

q=n\times c_v\times \Delta T

where,

q = heat absorbed by the system = ?

n = moles of sample = 0.877 mol

\Delta T = Change in temperature = 15.7 K

c_v = heat capacity at constant volume of N_2 (diatomic molecule) = \frac{5}{2}R

R = gas constant = 8.314 J/mol.K

Now put all the given value in the above formula, we get:

q=0.877mol\times \frac{5}{2}\times 8.314J/mol.K\times 15.7K

q=286.2J

Now we have to calculate the change in internal energy of the system.

\Delta U=q+w

As we know that, work done is zero at constant volume. So,

\Delta U=q=286.2J

Therefore, the value of q\text{ and }\Delta U is 286.2 J and 286.2 J respectively.

8 0
4 years ago
Read 2 more answers
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