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Margaret [11]
2 years ago
14

Object 1 has a mass of 3m and is moving to the right at a velocity of

Physics
1 answer:
andrew11 [14]2 years ago
6 0

Answer:

In a collision, the velocity change is always computed by subtracting the initial velocity value from the final velocity value. If an object is moving in one direction before a collision and rebounds or somehow changes direction, then its velocity after the collision has the opposite direction as before.

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The electric company charges $0.08/kilowatt-hour. If a 2 kilowatt appliance is run for 3hours, how much will you have to pay to
asambeis [7]

Answer:

$ 0.48

Explanation:

We can calculate this quantity easily using successive products and taking into account the units.

\frac{0.08}{kw*h}*2[kw]*3[hr]\\ \\=0.48

The amount is $ 0.48

3 0
3 years ago
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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Answer: 1. the object is moving away from the origin

4. the object started at 2 meters

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