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2 years ago

What is the wavelength of a wave that has a frequency of 15HZ and a wave speed of 5 m/s?

Physics

1 answer:

stepladder [879]2 years ago

5 0

Answer:

A. .33m

Explanation:

Have a good day.

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What is the thermal energy of an object

Papessa [141]

It's total kinetic energy

6 0

4 years ago

A man has a mass of 66 kg on the earth. What is his weight.

LUCKY_DIMON [66]

Since force = mass X acceleration, then the force he excerts on the earth (aka his weight) equals his mass times the force of gravity.

Therefore
W = (66 kg) X (9.8 m/ss)
W = 646.8 kgm/ss

kg m/ss are also known as Newtons, so your answer is...

646.8 N

8 0

3 years ago

A thin film of cryolite ( nc = 1.34 ) is applied to a camera lens ( ng = 1.58 ). The coating is designed to reflect wavelengths

rosijanka [135]

Answer:

=99.07nm

Explanation:

minimum thickness

2nd = (m - 1/2)λ

d = (m - 1/2)(λ/2n)

refractive index of the thin film, n = 1.34

minimum thickness m = 1

light wavelength λ = 531nm

d = (1 - 1/2) (531 / (2)(1.34)

d = 531/5.36

= 99.07nm

3 0

3 years ago

Suppose that an object is moving along a vertical line. Its vertical position is given by the equation L(t) = 2t3 + t2-5t + 1, w

Tatiana [17]

Answer:

The average velocity is

$266\frac{m}{s},274\frac{m}{s}$ and $117\frac{m}{s}$ respectively.

Explanation:

Let's start writing the vertical position equation :

$L(t)=2t^{3}+t^{2}-5t+1$

Where distance is measured in meters and time in seconds.

The average velocity is equal to the position variation divided by the time variation.

$V_{avg}=\frac{Displacement}{Time}$ = Δx / Δt = $\frac{x2-x1}{t2-t1}$

For the first time interval :

t1 = 5 s → t2 = 8 s

The time variation is :

$t2-t1=8s-5s=3s$

For the position variation we use the vertical position equation :

$x2=L(8s)=2.(8)^{3}+8^{2}-5.8+1=1049m$

$x1=L(5s)=2.(5)^{3}+5^{2}-5.5+1=251m$

Δx = x2 - x1 = 1049 m - 251 m = 798 m

The average velocity for this interval is

$\frac{798m}{3s}=266\frac{m}{s}$

For the second time interval :

t1 = 4 s → t2 = 9 s

$x2=L(9s)=2.(9)^{3}+9^{2}-5.9+1=1495m$

$x1=L(4s)=2.(4)^{3}+4^{2}-5.4+1=125m$

Δx = x2 - x1 = 1495 m - 125 m = 1370 m

And the time variation is t2 - t1 = 9 s - 4 s = 5 s

The average velocity for this interval is :

$\frac{1370m}{5s}=274\frac{m}{s}$

Finally for the third time interval :

t1 = 1 s → t2 = 7 s

The time variation is t2 - t1 = 7 s - 1 s = 6 s

Then

$x2=L(7s)=2.(7)^{3}+7^{2}-5.7+1=701m$

$x1=L(1s)=2.(1)^{3}+1^{2}-5.1+1=-1m$

The position variation is x2 - x1 = 701 m - (-1 m) = 702 m

The average velocity is

$\frac{702m}{6s}=117\frac{m}{s}$

5 0

3 years ago

A ski jumper travels down a slope and

AleksandrR [38]

Answer:

304.86 metres

Explanation:

The x and y cordinates are $dcos\theta$ and $dsin\theta$ respectively

The horizontal distance travelled, $x=v_{ox}t=dcos\theta$

Making t the subject, $t=\frac{dcos\theta}{v_{ox}}$

Since $y=0.5gt^2=dsin\theta$ , we substitute t with the above and obtain

$0.5g(\frac{dcos\theta}{v_{ox}})^2=dsin\theta$

Making d the subject we obtain

$d=\frac{2v_{ox}^2sin\theta}{gcos^2\theta}$

$d=\frac{2*30^2sin48}{9.8cos^248}$

d=304.8584

d=304.86m

5 0

3 years ago