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svetlana [45]
3 years ago
13

How do I do number 10? Please show work!!

Physics
1 answer:
mafiozo [28]3 years ago
6 0
The answer would be 40 cause 10*4=40
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A 460 nm monochromatic light is split into two beams. The two beams arrive at a point on a screen with a path difference of 920n
VladimirAG [237]

Answer:

phase difference \Delta \Phi = 12.566

Explanation:

Phase difference is expressed as \Delta \Phi

\Delta \Phi =\frac {2\pi}{\lambda} \Delta X

Where,

\Delta \Phi= phase difference

\lambda  wavelength = 460nm

\Delta X  path difference = 920 nm

\Delta \Phi =\frac {2\pi}{460} *920 = 4\pi

therefore

phase difference \Delta \Phi = 12.566

7 0
4 years ago
1. A friend measures the length of the school
shusha [124]

Answer:

option A

Explanation:

A soccer field is about 100 meters of length, with the other instruments is possible to make the measure, but it will be very complex and it will take time. Therefore with the 50-m tape, the student will need only two measures.

6 0
3 years ago
An object with a mass of 9 kg weighs 14.4 N on the Moon. What is the
erma4kov [3.2K]

Answer:

1.6 m/s²

Explanation:

Weight equals mass times acceleration due to gravity.

F = mg

14.4 N = (9 kg) g

g = 1.6 m/s²

7 0
4 years ago
Read 2 more answers
A world-class sprinter running a 100 m dash was clocked at 5.4 m/s 1.0 s after starting running and at 9.8 m/s 1.5 s later. In w
cupoosta [38]

Answer:

<em>The output power is greater in the interval from 1.0 s to 2.5 s</em>

Explanation:

<u>Physical Power </u>

It measures the amount of work W an object does in certain time t. The formula needed to compute power is

\displaystyle P=\frac{W}{t}

Work can be computed in several ways since we are given the motion conditions, we'll use this formula, for F= applied force, x=distance parallel to F

W=F.x

The second Newton's law gives us the net force as

F=m.a

being m the mass of the object and a the acceleration it has for a given period of time. In our problem, we have two different behaviors for each interval and we must calculate this force since the acceleration is changing. Let's calculate the acceleration in the first interval. We can use the formula for the final speed vf knowing the initial speed vo (which is 0 because the sprinter starts from rest), the acceleration a, and the time t:

v_f=v_o+at

v_f=at

Solving for a

\displaystyle a=\frac{v_f}{t}={5.4}{1}

a=5.4\ m/s^2

The distance traveled in the interval is given by

\displaystyle x=v_o.t+\frac{a.t^2}{2}

Since vo=0

\displaystyle x=\frac{a.t^2}{2}=\frac{5.4(1)^2}{2}

x=2.7\ m

The force is given by

F=m.a

We don't know the value of m, so the force is

F=2.7m

Computing the work done by the sprinter

W=F.x=2.7m(5.4)

W=14.58m

The power is finally computed

\displaystyle P=\frac{W}{t}=\frac{14.58m}{1}

P=14.58m

During the second interval, from t=1 sec to 1.5 sec, the speed changes from 5.4 m/s to 9.8 m/s. This allows us to compute the second acceleration

\displaystyle a=\frac{v_f-v_o}{t}=\frac{9.8-5.4}{0.5}

a=8.8\ m/s^2

The distance is

\displaystyle x=(5.4).(0.5)+\frac{8.8(0.5)^2}{2}

x=3.8\ m

The net force is

F=m(8.8)=8.8m

The work done by the sprinter is now computed as

W=8.8m(3.8)=33.44m

At last, the output power is

\displaystyle P=\frac{33.44m}{0.5}=66.88m

By comparing both results, and being m the same for both parts, we conclude the output power is greater in the interval from 1.0 s to 2.5 s

6 0
4 years ago
Once cells with nuclei developed, organisms were able to become more complex, which allowed them to adapt to the conditions in m
liq [111]
D). natural selection i am not sure tho
4 0
3 years ago
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