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2 years ago

PLEASE HELP ME TIMED TEST!!!

Physics

2 answers:

Dmitry [639]2 years ago

8 0

Explanation:

$\boxed{\sf{K.E = \dfrac{1}{2} \times m \times v^2}}$

Here, K . E or Kinetic Energy is given as 14 J
Here, mass is 17 kg

Let's Do!

$\boxed{\sf{14 = \dfrac{1}{2} \times 17 \times v^2}}$

$\boxed{\sf{\dfrac{14 \times 2}{17} = v^2}}$

$\boxed{\sf{ \sqrt{\dfrac{14 \times 2}{17}}= v}}$

So, 1.283 m/s is the required answer.

malfutka [58]2 years ago

7 0

Answer:

Explanation:

I just want the points to be completely honest with you.

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3 years ago

All of the visible color light waves together result in____light

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3 years ago

(a) Write an equation describing a sinusoidal transverse wave traveling on a cord in the positive of a y axis with an angular wa

Vedmedyk [2.9K]

Missing data: the wave number

$k=60 cm^{-1}$

(a) $z = 0.003 sin (6000y-31.4 t)$

For a transverse wave travelling in the positive y-direction and with vibration along the z-direction, the equation of the wave is

$z = A sin (ky-\omega t)$

where

A is the amplitude of the wave

k is the wave number

$\omega$ is the angular frequency

t is the time

In this situation:

A = 3.0 mm = 0.003 m is the amplitude

$k = 60 cm^{-1} = 6000 m^{-1}$ is the wave number

$T = 0.20 s$ is the period, so the angular frequency is

$\omega=\frac{2\pi}{T}=\frac{2\pi}{0.20}=31.4 rad/s$

So, the wave equation (in meters) is

$z = 0.003 sin (6000y-31.4 t)$

(b) 0.094 m/s

For a transverse wave, the transverse speed is equal to the derivative of the displacement of the wave, so in this case:

$v_t = z' = -A \omega cos (ky-\omega t)$

So the maximum transverse wave occurs when the cosine term is equal to 1, therefore the maximum transverse speed must be

$v_{t}_{max} =\omega A$

where

$\omega = 31.4 rad/s\\A = 0.003 m$

Substituting,

$v_{t}_{max}=(31.4)(0.003)=0.094 m/s$

(c) 5.24 mm/s

The wave speed is given by

$v=f \lambda$

where

f is the frequency of the wave

$\lambda$ is the wavelength

The frequency can be found from the angular frequency:

$f=\frac{\omega}{2\pi}=\frac{31.4}{2\pi}=5 Hz$

While the wavelength can be found from the wave number:

$\lambda = \frac{2\pi}{k}=\frac{2\pi}{6000}=1.05\cdot 10^{-3} m$

Therefore, the wave speed is

$v=(5)(1.05\cdot 10^{-3} )=5.24 \cdot 10^{-3} m/s = 5.24 mm/s$

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3 years ago

A chimpanzee sitting against his favorite tree gets up and walks 70.9 m due east and 31.9 m due south to reach a termite mound,

DIA [1.3K]

A right triangle is formed by the 70.9 m walked east and 31.9 m walked south.
The legs of this right triangle are 70.9 and 31.9.
The shortest distance between two points is a straaight line. Therefore the hypotenuse of this triangle is going to be that shortest distance.
We can use the Pythagorean Theorem to find the hypotenuse of this triangle.
$a^2+b^2=c^2\\70.9^2+31.9^2=c^2\\5026.81+1017.61=c^2\\6044.42=c^2\\c=\sqrt{6044.42}$

$c=\sqrt{6044.42}\approx\boxed{77.7458681}\ (decimal\ form)\\\\c=\sqrt{6044.42}=\sqrt{\frac{604442}{100}}=\boxed{\frac{\sqrt{604442}}{10}}\ (exact\ form)$

As for the second part of the question, we want to find the angle formed by the hypotenuse and the 31.9 walked east.
We could use any of the three trigonometric ratios here since we know all 3 sides.
sine = opposite / hypotenuse
cosine = adjacent / hypotenuse
tangent = opposite / adjacent
I am going to use tangent, because then I won't have to deal with the hypotenuse and so the answer will be more accurate.

If you haven't already drawn yourself a diagram, now is a good time to.
The side opposite our angle is the 31.9, and the adjacent is 70.9.
Therefore, $\tan(m\angle)=\frac{31.9}{70.9}$ .

We can use inverse trig ratios here to find the measure of our angle.
$\tan^{-1}(\tan(m\angle))=\tan^{-1}(\frac{31.9}{70.9})\\\\m\angle=\tan^{-1}(\frac{31.9}{70.9})\approx\boxed{24.2243851\°\ or\ 0.422795279\ rad}$

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3 years ago