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3 years ago

9

What net force is required to cause a 1600 kg car to accelerate at a rate of 4.0

1 answer:

balu736 [363]3 years ago

6 0

The correct answer, is Option C, I hope this helps :)

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A wave's _______________ and its frequency have a reciprocal relationship.

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Wavelength and frequency have a reciprocal relationship. If one doubles, the other halves.

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A 5.75 mm high firefly sits on the axis of, and 11.3 cm in front of, the thin lens A, whose focal length is 5.77 cm . Behind len

weeeeeb [17]

Answer

given,

focal length of lens A = 5.77 cm

focal length of lens B= 27.9 cm

flies distance from mirror = 11.3 m

now,

Using lens formula

$\dfrac{1}{f} = \dfrac{1}{p} + \dfrac{1}{q}$

$\dfrac{1}{5.77} = \dfrac{1}{11.3} + \dfrac{1}{q}$

q =11.79 cm

image of lens A is object of lens B

distance of lens = 59.9 - 11.79 = 48.11

now, Again applying lens formula

$\dfrac{1}{f} = \dfrac{1}{p} + \dfrac{1}{q'}$

$\dfrac{1}{27.9} = \dfrac{1}{48.11} + \dfrac{1}{q'}$

q' =66.41 cm

hence, the image distance from the second lens is equal to q' =66.41 cm

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3 years ago

You want to create a device to warm food that is safe. Which type of electromagnetic wave would be most useful to investigate? A

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The air in a kitchen has a mass of 60.0kg and a specific heat of 1505J/(kg°C).

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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):

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

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

<u>Finally:</u>

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

4 0

3 years ago