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Orlov [11]
2 years ago
15

By examining a wave's pattern after it interacts with a barrier or gap, measurements can be made to better understand certain wa

ve phenomena. What is a broad question you can answer by doing this experiment?
Physics
2 answers:
rjkz [21]2 years ago
8 0

Sample Response: How does diffraction occur?

Tanya [424]2 years ago
7 0

Explanation :

When a ray of light passes through the narrow opening then the light diffracts. The process is known as the diffraction of light.

By examining a wave's pattern after it interacts with a barrier or gap, measurements can be made to better understand diffraction wave phenomena.

After diffracting the light forms a pattern consisting of dark and bright bands.

Hence, the question could be " How diffraction of wave occurs? Explain ".

You might be interested in
What units are given to the right of the equals sign
zhuklara [117]
The answer

2y + 14 = 17

The 17 is to the right of the = sign
It is also the answer
7 0
2 years ago
A child is running his 46.1 g toy car down a ramp. The ramp is 1.73 m long and forms a 40.5° angle with the flat ground. How lon
frosja888 [35]

Answer:

t=0.704s

Explanation:

A child is running his 46.1 g toy car down a ramp. The ramp is 1.73 m long and forms a 40.5° angle with the flat ground. How long will it take the car to reach the bottom of the ramp if there is no friction?

from newton equation of motion , we look for the y component of the speed and look for the x component of the speed. we can then find the resultant of the speed

V^{2} =u^{2} +2as

Vy^2=0+2*9.8*1.73sin40.5

Vy^2=22.021

Vy=4.69m/s

Vx^2=u^2+2*9.81*cos40.5

Vy^2=25.81

Vy=5.08m/s

V=(Vy^2+Vx^2)^0.5

V=47.71^0.5

V=6.9m/s

from newtons equation of motion we know that force applied is directly proportional to the rate of change in momentum on a body.

f=force applied

v=velocity final

u=initial velocity

m=mass of the toy, 0.046

f=ma

f=m(v-u)/t

v=u+at

6.9=0+9.8t

t=6.9/9.81

t=0.704s

4 0
2 years ago
Question
Jobisdone [24]

Answer:

A wavelength is a measure of distance between two consecutive crests or trough. So , the unit of wavelength is same as unit of distance. I.e metre

7 0
2 years ago
An object of mass 24kg is accelerated up a frictionless place incline at an angle of 37° with horizontal by a constant force, st
RoseWind [281]

a) Average power: 1425 W

b) Instantaneous power at 3.0 sec: 2850 W

Explanation:

a)

The motion of the object along the ramp is a uniformly accelerated motion (because the force applied is constant), so we can use the suvat equation

s=ut+\frac{1}{2}at^2

where

s = 18 m is the displacement along the ramp

u = 0 is the initial velocity

t = 3.0 s is the time taken

a is the acceleration of the object along the ramp

Solving for a,

a=\frac{2s}{t^2}=\frac{2(18)}{(3.0)^2}=4 m/s^2

Now we can apply Newton's second law to find the net force on the object:

F=ma=(24 kg)(4 m/s^2)=96 N

This net force is the resultant of the applied force forward (F_a) and the component of the weight acting backward (mg sin \theta), so we can find what is the applied force:

F=F_a - mg sin \theta\\F_a = F+mg sin \theta = 96+(24)(9.8)(sin 37^{\circ})=237.5 N

where

m = 24 kg is the mass of the object

g=9.8 m/s^2 is the acceleration of gravity

Now we can finally find what is the work done by the applied force, which is parallel to the ramp, therefore:

W=F_a s = (237.6)(18)=4276 J

where s = 18 m is the displacement.

Therefore the average power needed is:

P=\frac{W}{t}=\frac{4276}{3}=1425 W

b)

The instantaneous power at any point of the motion is given by

P=F_av

where

F_a is the force applied

v is the velocity of the object

We already calculated the applied force:

F_a=237.5 N

While since this is a uniformly accelerated motion, we can find the velocity at the end of the 3.0 seconds using the suvat equation:

v=u+at=0+(4)(3.0)=12.0 m/s

And therefore, the instantaeous power at 3.0 sec is:

P=Fv=(237.5)(12)=2850 W

Learn more about power:

brainly.com/question/7956557

#LearnwithBrainly

8 0
2 years ago
At the equator, the earth’s field is essentially horizontal; near the north pole, it is nearly vertical. In between, the angle v
Yanka [14]

Answer:

A current of 0.358A is required

Explanation:

Pls refer to attached document for more details

7 0
2 years ago
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