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BaLLatris [955]
2 years ago
10

Consider a variety of colors of visible light (say 400 nm to 700 nm) falling onto a pair of slits.

Physics
1 answer:
babymother [125]2 years ago
6 0

Answer:

Explanation:

The relationship between angle and wavelength for maxima and minima in Young's double slit experiment is given by

For constructive interference

d\sin \theta =m\lambda

For Destructive interference

d\sin \theta =(m+\frac{1}{2})\lambda

where \lambda =wavelength

d=slit\ width

m=order of maxima and minima

for second order maxima i.e. m=2

For smallest separation taking \lambda =400 nm, \theta =90^{\circ}

d\sin 90=2\times 400\times 10^{-9}

d=0.8\times 10^{-6}

d=0.8\mu m

   

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Some gliders are launched from the ground by means of a winch, which rapidly reels in a towing cable attached to the glider. Wha
aliina [53]

Answer:

P=627.47W

Explanation:

To solve this problem we have to take into account, that the work done by the winch is

W=Fh

the force, at least must equal the gravitational force

F=Mg=(156kg)(9.8\frac{m}{s^2})=1258.8N

with force the tension in the cable makes the winch go up.

The work done is

W=(1258.8N)(58.0m)=73010.4J

To calculate the power we need to know what is the time t. But first we have to compute the acceleration

The acceleration will be

v_f^2=v_0+2ah\\a=\frac{v_f^2}{2h}=\frac{(24.9\frac{m}{s})}{2(58.0m)}=0.214\frac{m}{s^2}

and the time t

v_f=v_0+at\\t=\frac{v_f}{a}=116.35s

The power will be

P=\frac{W}{t}=\frac{73010.4J}{116.35s}=627.47W

HOPE THIS HELPS!!

6 0
2 years ago
A large box of mass M is pulled across a horizontal, frictionless surface by a horizontal rope with tension T. A small box of ma
Nesterboy [21]

Answer:

T = g μ_s ( M+m )

78.4 N

Explanation:

When both of them move with the same acceleration , small box will not slip over the bigger one. When we apply force on the lower box, it starts moving with respect to lower box. So a frictional force arises on the lower box which helps it too to go ahead . The maximum value that this force can attain is mg μ_s . As a reaction of this force, another force acts on the lower box in opposite direction .

Net force on the lower box

= T - mg μ_s = M a    ( a is the acceleration created by net force in M )

Considering force on the upper box

mg μ_s = ma

a = g μ_s

Put this value of a in the equation above

T - m gμ_s = M g μ_s

T = mg μ_s + M g μ_s

=  g μ_s ( M+m )

2 )

Largest tension required

T = 9.8 x  .50 x ( 10+6 )

= 78.4 N

5 0
2 years ago
A quantity y is to be determined from the equation y=(px)/q^2
ki77a [65]

Answer:

heya answer option b

Explanation:

please mark me brainliest

4 0
1 year ago
If a sound with frequency fs is produced by a source traveling along a line with speed vs. If an observer is traveling with spee
Alexus [3.1K]

Answer:

457.81 Hz

Explanation:

From the question, it is stated that it is a question under Doppler effect.

As a result, we use this form

fo = (c + vo) / (c - vs) × fs

fo = observed frequency by observer =?

c = speed of sound = 332 m/s

vo = velocity of observer relative to source = 45 m/s

vs = velocity of source relative to observer = - 46 m/s ( it is taking a negative sign because the velocity of the source is in opposite direction to the observer).

fs = frequency of sound wave by source = 459 Hz

By substituting the the values to the equation, we have

fo = (332 + 45) / (332 - (-46)) × 459

fo = (377/ 332 + 46) × 459

fo = (377/ 378) × 459

fo = 0.9974 × 459

fo = 457.81 Hz

7 0
2 years ago
A thin, horizontal, 18-cm-diameter copper plate is charged to -3.8 nC. Assume that the electrons are uniformly distributed on th
son4ous [18]

Answer:

Part a)

E = 8436.7 N/C

Part b)

E = 8436.7 N/C

Explanation:

Part a)

Electric field due to large sheet is given as

E = \frac{\sigma}{2\epsilon_0}

\sigma = \frac{Q}{A}

Q = -3.8 nC

A = \pi(0.09)^2

A = 0.025 m^2

\sigma = \frac{-3.8\times 10^{-9}}{0.025}

\sigma = -1.5 \times 10^{-7} C/m^2

now the electric field is given as

E = \frac{-1.5 \times 10^{-7}}{2(8.85 \times 10^{-12})}

E = 8436.7 N/C

Part b)

Now since the electric field is required at same distance on other side

so the field will remain same on other side of the plate

E = 8436.7 N/C

5 0
2 years ago
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