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3 years ago

A roller coaster

Physics

1 answer:

vagabundo [1.1K]3 years ago

5 0

Answer:

the answer is the rollar coster

Explanation:

because the rollar coster is

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The central star of a planetary nebula emits ultraviolet light with wavelength 104nm. This light passes through a diffraction gr

Gala2k [10]

Answer: 31.33 degrees

Explanation:

The diffraction angles $\theta_{n}$ when we have a slit divided into $n$ parts are obtained by the following equation:

$dsin\theta_{n}=n\lambda$ (1)

Where:

$d$ is the width of the slit

$\lambda$ is the wavelength of the light

$n$ is an integer different from zero.

Now, the first-order diffraction angle is given when $n=1$ , hence equation (1) becomes:

$dsin\theta_{1}=\lambda$ (2)

Now we have to find the value of $\theta_{1}$ :

$sin\theta_{1}=\frac{\lambda}{d}$

$\theta_{1}=arcsin(\frac{\lambda}{d})$ (3)

We know:

$\lambda=104nm=104(10)^{-9}m$

In addition we are told the diffraction grating has 5000 slits per mm, this means:

$d=\frac{1mm}{5000}=\frac{1(10)^{-3}m}{5000}$

Substituting the known values in (3):

$\theta_{1}=arcsin(\frac{104(10)^{-9}m}{\frac{1(10)^{-3}m}{5000}})$

$\theta_{1}=arcsin(0.52)$

Finally:

$\theta_{1}=31.33\º$ >>>This is the first-order diffraction angle

4 0

3 years ago

A man 2 meters tall walks at the rate of 2 meters per second toward a streetlight that's 5 meters above the ground. At what rate

GaryK [48]

Answer:

It changes at a rate of 4/3 meter per second

Explanation:

In the given figure below we have

$\Delta OBD\simeq \Delta ABC\\\\\therefore \frac{5}{X+Y}=\frac{2}{Y}\\\\$

Solving for Y given $X=2m/s$ we get

$\frac{5}{2+Y}=\frac{2}{Y}\\\\5Y=4+2Y\\\\Y=\frac{4}{3}m/s$

8 0

3 years ago

A 12 volt car battery is connected to a 3 ohm brake light. What is the current carrying energy to the lights?

zloy xaker [14]

Answer:

4 A

Explanation:

V = IR, where V=voltage, I=current, R=resistance. This is Ohm's Law. (remember that for units V = volts, Ω = ohms, A = amperes.)

V = IR

12 V = I * 3 Ω

12/3 = I

I = 4 A

8 0

3 years ago

Sonar is used to map the ocean floor. if an ultrasonic signal is received 3.3 s after it is emitted, how deep is the ocean floor

Goryan [66]

The speed of sound in fresh water is 1482m/s.
It says ocean floor, so we should a little bit more accurate, and use the fact that the speed of sound in salt water (that has no bubbles) is 1560m/s.
speed = distance / time
Therefore Distance = speed x time = 1560 x 3.3 = 5158m
The sonar wave is sent out by the boat, reflected off the seafloor, and then is received back at the boat on the surface. So the distance 5148m is the distance from the boat to the sea bottom and then back up to the boat again.
So the depth of the water is half this distance Depth of water = 5148/2=2574m

6 0

3 years ago

How much heat is required to warm 122 g of water by 23.0 c?

GaryK [48]

122 g * 4,186 (j/g*°c) * 23°c = 11745.916 j

4 0

3 years ago

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