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3 years ago

a coin is dropped from the top of a tall building. determine the coin's (a) velocity and (b) displacement after 1.5 sec.

Physics

1 answer:

Lilit [14]3 years ago

8 0

1) v = gt = 10*1.5 = 15 m/s
2) r = gt^2 /2 = 10*(1.5)^2 / 2 = 11.25 meters

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3 years ago

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Problem One: A beam of red light (656 nm) enters from air into the side of a glass and then into water. wavelength, c. and speed

Ivanshal [37]

Answer:

Part a)

$f_w = f_g = 4.57 \times 10^{14} Hz$

Part b)

$\lambda_w = 492 nm$

$\lambda_g = 437.3 nm$

Part c)

$v_w = 2.25 \times 10^8 m/s$

$v_g = 2.0 \times 10^8 m/s$

Explanation:

Part a)

frequency of light will not change with change in medium but it will depend on the source only

so here frequency of light will remain same in both water and glass and it will be same as that in air

$f = \frac{v}{\lambda}$

$f = \frac{3 \times 10^8}{656 \times 10^{-9}}$

$f = 4.57 \times 10^{14} Hz$

Part b)

As we know that the refractive index of water is given as

$\mu_w = 4/3$

so the wavelength in the water medium is given as

$\lambda_w = \frac{\lambda}{\mu_w}$

$\lambda_w = \frac{656 nm}{4/3}$

$\lambda_w = 492 nm$

Similarly the refractive index of glass is given as

$\mu_w = 3/2$

so the wavelength in the glass medium is given as

$\lambda_g = \frac{\lambda}{\mu_g}$

$\lambda_g = \frac{656 nm}{3/2}$

$\lambda_g = 437.3 nm$

Part c)

Speed of the wave in water is given as

$v_w = \frac{c}{\mu_w}$

$v_w = \frac{3 \times 10^8}{4/3}$

$v_w = 2.25 \times 10^8 m/s$

Speed of the wave in glass is given as

$v_g = \frac{c}{\mu_g}$

$v_g = \frac{3 \times 10^8}{3/2}$

$v_g = 2 \times 10^8 m/s$

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3 years ago

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The formula that you are working with is F = m*a

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The second one is much more difficult to answer because it is basically incomplete. This is one way to interpret it. If you start at a certain speed and increase during a known time period then effectively you are defining acceleration which is "a" in the formula.

Without those modifications, there is no answer.

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