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Andre45 [30]
3 years ago
15

The distance between slits on a diffraction grating is 0.60 mm, and one of the angles of diffraction is 0.30°. The light forms a

second-order bright band. How large is the path difference?
Physics
2 answers:
Anastaziya [24]3 years ago
7 0

Answer:  1.57 × 10⁻⁶ m

Explanation:

For constructive interference i.e. formation of bright fringe through a diffraction grating is given by:

d sin θ = n λ

here, d is the distance between slits.

θ is the angle of diffraction

n is the order of fringe

and λ is the path difference.

λ = (d sin θ)/ n = (0.60 ×10 ⁻³m)(sin 0.30°)÷2 = 1.57 × 10⁻⁶ m

bixtya [17]3 years ago
3 0
Just took the test and they are right it is C welcome
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A proton travels with a speed of 4.2×106 m/s at an angle of 30◦ west of north. A magnetic field of 2.5 T points to the north. Fi
Arada [10]

Answer:

8.4\cdot 10^{-13} N

Explanation:

The magnitude of the magnetic force on the proton is given by:

F=qvB sin \theta

where:

q=1.6\cdot 10^{-19} C is the proton charge

v=4.2\cdot 10^6 m/s is the proton velocity

B=2.5 T is the magnetic field

\theta=30^{\circ} is the angle between the direction of v and B

Substituting into the formula, we find

F=(1.6\cdot 10^{-19}C)(4.2\cdot 10^6 m/s)(2.5 T) sin 30^{\circ}=8.4\cdot 10^{-13} N

6 0
3 years ago
A motorcycle is capable of accelerating at 5.1 m/s starting feom rest how far can it travel im 1.5 seconds
saul85 [17]

5.1 m/ s


~Multiply 1.5 on both sides since you’re trying to finding out how far it can travel in 1.5 seconds


5.1 x 1.5= 7.65

s x 1.5= 1.5


So a motorcycle can travel 7.65 m in 1.5 seconds.



6 0
3 years ago
Seasons are caused by differences in daylight, temperature, and weather patterns due to :
Neko [114]

... the angular tilt of the Earth's position on its axis relative to the sun


4 0
4 years ago
An astronaut goes out for a space walk. Her mass (including space suit, oxygen tank, etc.) is 100 kg. Suddenly, disaster strikes
Marina CMI [18]

Answer:

<u>Part A:</u>

Unknown variables:

velocity of the astronaut after throwing the tank.

maximum distance the astronaut can be away from the spacecraft to make it back before she runs out of oxygen.

Known variables:

velocity and mass of the tank.

mass of the astronaut after and before throwing the tank.

maximum time it can take the astronaut to return to the spacecraft.

<u>Part B: </u>

To obtain the velocity of the astronaut we use this equation:

-(momentum of the oxygen tank) = momentum of the astronaut

-mt · vt = ma · vt

Where:

mt = mass of the tank

vt = velocity of the tank

ma = mass of the astronaut

va = velocity of the astronaut

To obtain the maximum distance the astronaut can be away from the spacecraft we use this equation:

x = x0 + v · t

Where:

x = position of the astronaut at time t.

x0 = initial position.

v = velocity.

t = time.

<u>Part C:</u>

The maximum distance the astronaut can be away from the spacecraft is 162 m.

Explanation:

Hi there!

Due to conservation of momentum, the momentum of the oxygen tank when it is thrown away must be equal to the momentum of the astronaut but in opposite direction. In other words, the momentum of the system astronaut-oxygen tank is the same before and after throwing the tank.

The momentum of the system before throwing the tank is zero because the astronaut is at rest:

Initial momentum = m · v

Where m is the mass of the astronaut plus the equipment (100 kg) and v is its velocity (0 m/s).

Then:

initial momentum = 0

After throwing the tank, the momentum of the system is the sum of the momentums of the astronaut plus the momentum of the tank.

final momentum = mt · vt + ma · va

Where:

mt = mass of the tank

vt = velocity of the tank

ma = mass of the astronaut

va = velocity of the astronaut

Since the initial momentum is equal to final momentum:

initial momentum = final momentum

0 = mt · vt + ma · va

- mt · vt = ma · va

Now, we have proved that the momentum of the tank must be equal to the momentum of the astronaut but in opposite direction.

Solving that equation for the velocity of the astronaut (va):

- (mt · vt)/ma = va

mt = 15 kg

vt = 10 m/s

ma = 100 kg - 15 kg = 85 kg

-(15 kg · 10 m/s)/ 85 kg = -1.8 m/s

The velocity of the astronaut is 1.8 m/s in direction to the spacecraft.

Let´s place the origin of the frame of reference at the spacecraft. The equation of position for an object moving in a straight line at constant velocity is the following:

x = x0 + v · t

where:

x = position of the object at time t.

x0 = initial position.

v = velocity.

t = time.

Initially, the astronaut is at a distance x away from the spacecraft so that

the initial position of the astronaut, x0, is equal to x.

Since the origin of the frame of reference is located at the spacecraft, the position of the spacecraft will be 0 m.

The velocity of the astronaut is directed towards the spacecraft (the origin of the frame of reference), then, v = -1.8 m/s

The maximum time it can take the astronaut to reach the position of the spacecraft is 1.5 min = 90 s.

Then:

x = x0 + v · t

0 m = x - 1.8 m/s · 90 s

Solving for x:

1.8 m/s · 90 s = x

x = 162 m

The maximum distance the astronaut can be away from the spacecraft is 162 m.

6 0
3 years ago
A metal ring 4.60 cm in diameter is placed between the north and south poles of large magnets with the plane of its area perpend
telo118 [61]

Answer:

A. Ein = 8.05*10^-4 V/m

B. Clockwise sense

Explanation:

A. the magnitude of the electric field induced in the ring is obtaind by using the following formula:

\int E_{in} \cdot ds=-\frac{d\Phi_B}{dt}            (1)

Ein: induced electric field

ds: differential of a path of the ring

ФB: magnetic flux in the ring

The Ein vector is parallel to ds in the complete ring. Furthermore, the area of the ring is constant, hence, you have in the equation (1):

\int E_{in}ds=E_{in}(2\pi r)=-A\frac{dB}{dt}\\\\E_{in}=-\frac{A}{2\pi r}\frac{dB}{dt}   (2)

dB/dt = -0.280T/s     (it is decreasing)

A: area of the ring = π(r/2)^2= (π/4) r^2

r: radius of the ring = 4.60/2 = 2.30 cm

Then, you replace the values of all variables in the equation (2):

E_{in}=-\frac{(\pi/4)r^2}{2\pi r}\frac{dB}{dt}=\frac{r}{8}\frac{dB}{dt}\\\\E_{in}=-\frac{0.0230m}{8}(-0.280T)=8.05*10^{-4}\frac{V}{m}

hence, the induced electric field is 8.05*10^-4 V/m

B. The induced current in the ring produced a magnetic field that is opposite to the magnetic field of the magnet. The, in this case you have that the induced current is in a clockwise sense.

6 0
4 years ago
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