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zlopas [31]
3 years ago
7

Light is an electromagnetic wave and travels at a speed of 3.00 × 108 m/s. The human eye is most sensitive to yellow-green light

, which has a wavelength of 5.04 × 10-7 m. What is the frequency of this light?
Physics
1 answer:
Nimfa-mama [501]3 years ago
6 0

Answer:

f = 5.95 \times 10^{14} Hz

Explanation:

As we know that the frequency of the wave is given as

f = \frac{c}{\lambda}

here we know that

c = 3 \times 10^8 m/s

also we know that

\lambda = 5.04 \times 10^{-7} m

now we have

f = \frac{3 \times 10^8}{5.04 \times 10^{-7}}

f = 5.95 \times 10^{14} Hz

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A boy is holding a ball 1 m from the ground with a force of 20 N. He holds it still for 60seconds. How much power in watts is be
Sergio [31]

Answer:

Power = 0.33 Watts

Explanation:

Given the following data;

Distance = 1m

Force = 20N

First of all, we would solve for the work done by the boy.

Workdone = force * distance

Substituting into the equation, we have;

Workdone = 20*1 = 20J

Now to find power;

Power = workdone/time

Power = 20/60

Power = 0.33 Watts.

8 0
3 years ago
a 3.2 kg durian fruit is pushed across the table.If the acceleration of the durian is 3.1 m/s/s to the right,what is the net for
Gala2k [10]

Answer:

<h2>9.92 N</h2>

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 3.2 × 3.1

We have the final answer as

<h3>9.92 N</h3>

Hope this helps you

4 0
3 years ago
Read 2 more answers
Bádminton is played to a score of
Temka [501]
Badminton is played to a score of 21 points
4 0
3 years ago
Read 2 more answers
A cannon fires a 0.2 kg shell with initial velocity vi = 9.2 m/s in the direction θ = 46 ◦ above the horizontal. The shell’s tra
Sedbober [7]

Answer:

∆h = 0.071 m

Explanation:

I rename angle (θ) = angle(α)

First we are going to write two important equations to solve this problem :

Vy(t) and y(t)

We start by decomposing the speed in the direction ''y''

sin(\alpha) = \frac{Vyi}{Vi}

Vyi = Vi.sin(\alpha ) = 9.2 \frac{m}{s} .sin(46) = 6.62 \frac{m}{s}

Vy in this problem will follow this equation =

Vy(t) = Vyi -g.t

where g is the gravity acceleration

Vy(t) = Vyi - g.t= 6.62 \frac{m}{s} - (9.8\frac{m}{s^{2} }) .t

This is equation (1)

For Y(t) :

Y(t)=Yi+Vyi.t-\frac{g.t^{2} }{2}

We suppose yi = 0

Y(t) = Yi +Vyi.t-\frac{g.t^{2} }{2} = 6.62 \frac{m}{s} .t- 4.9\frac{m}{s^{2} } .t^{2}

This is equation (2)

We need the time in which Vy = 0 m/s so we use (1)

Vy (t) = 0\\0=6.62 \frac{m}{s} - 9.8 \frac{m}{s^{2} } .t\\t= 0.675 s

So in t = 0.675 s  → Vy = 0. Now we calculate the y in which this happen using (2)

Y(0.675s) = 6.62\frac{m}{s}.(0.675s)-4.9 \frac{m}{s^{2} }  .(0.675s)^{2} \\Y(0.675s) =2.236 m

2.236 m is the maximum height from the shell (in which Vy=0 m/s)

Let's calculate now the height for t = 0.555 s

Y(0.555s)= 6.62 \frac{m}{s} .(0.555s)-4.9\frac{m}{s^{2} } .(0.555s)^{2} \\Y(0.555s) = 2.165m

The height asked is

∆h = 2.236 m - 2.165 m = 0.071 m

6 0
3 years ago
A star with a large luminosity would have a relatively _____ absolute magnitude.
Mademuasel [1]
A star with large luminosity would have a relatively low absolute magnitude. Absolute magnitude is a number that tells how bright a star is from the Earth. However, this scale is backwards and logarithmic, so having a large absolute magnitude value means that the star is faint.
3 0
3 years ago
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