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3 years ago

7

a boat sails along the shore. to an observer, the boat appears to move at a speed of 3m/s, and a man on the boat walking foward

appears to have a speed of 4.2 m/s how long does it take the man to move 8 m relative to the boat?

1 answer:

lina2011 [118]3 years ago

6 0

It is 6.7s. you need to slove the promble

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A vertical spring stretches 4.0 cm when a 12-g object is hung from it. The object is replaced with a block of mass 28 g that osc

marin [14]

Answer:

Time period of the osculation will be 0.0671 sec

Explanation:

It is given a vertical spring is stretched by 4 cm

So change in length of the spring x = 4 cm = 0.04 m

Mass which is hung from it m = 12 gram = 0.012 kg

Sprig force will be equal to weight of the mass

So $kx=mg$

$k\times 0.04=0.012\times 9.8$

k = 244.7 N/m

Now new mass is m = 28 gram = 0.028 kg

So time period with new mass will be

$T=2\pi \sqrt{\frac{m}{k}}$

$=2\times 3.14 \sqrt{\frac{0.028}{244.7}}=0.0671sec$

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2 years ago

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

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2 years ago

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Axial Tilt and Sun Energy

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Why might surface mining be less risky for miners than underground mining?

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Surface miners work aboveground. (Apex) ^-^

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