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3 years ago

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20) A 5 Kg watermelon is dropped off a 10 meter balcony. What will the watermelon's velocity be right before it hits the ground?

1 answer:

Ulleksa [173]3 years ago

3 0

Answer:

14m/s

Explanation:

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the refractive index is the blank of the speed of light in a medium to the speed of light in a vacuum

andrew11 [14]

Answer:ratio

Explanation:

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3 years ago

Read 2 more answers

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

The pull of what is weaker when 2 objects are farther apart

9966 [12]

The gravitational pull is weaker.

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3 years ago

A 12 volt battery in a motor vehicle is capable of supplying the starter motor with 150 A. It is noticed that the terminal volta

qwelly [4]

Answer: $0.0133 \Omega$

Explanation:

Given

Voltage=12 V

Current(I)=150 A

$V_{terminal}=10 V$

$r_{internal}=\frac{\Delta V}{I}$

$r_{internal}=\frac{12-10}{150}$

$r_{internal}=\frac{2}{150}=0.0133 \Omega$

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What is the relationship between the horizontal and vertical components of velocity for a projectile launched at an angle betwee

slavikrds [6]

Each is independent of the other -Each is dependent on each other -The horizontal component is dependent on gravitational acceleration -The horizontal component decreases with an increase in the vertical component

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3 years ago