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ValentinkaMS [17]
3 years ago
6

A car travels due east with a speed of 38.0 km/h. Raindrops are falling at a constant speed vertically with respect to the Earth

. The traces of the rain on the side windows of the car make an angle of 72.0° with the vertical. Find the velocity of the rain with respect to the following reference frames. (Enter the magnitude of the velocity.)
Physics
1 answer:
DiKsa [7]3 years ago
3 0

Answer: 116.926 km/h

Explanation:

To solve this we need to analise the relation between the car and the Raindrops. The cars moves on the horizontal plane with a constant velocity.

Car's Velocity (Vc) = 38 km/h

The rain is falling perpedincular to the horizontal on the Y-axis. We dont know the velocity.

However, the rain's traces on the side windows makes an angle of 72.0° degrees. ∅ = 72°

There is a relation between this angle and the two velocities. If the car was on rest, we will see that the angle is equal to 90° because the rain is falling perpendicular. In the other end, a static object next to a moving car shows a horizontal trace, so we can use a trigonometric relation on this case.

The following equation can be use to relate the angle and the two vectors.

Tangent (∅) = Opposite (o) / adjacent (a)

Where the Opposite will be the Rain's Vector that define its velocity and the adjacent will be the Car's Velocity Vector.

Tan(72°) = Rain's Velocity / Car's Velocity

We can searching for the Rain's Velocity

Tan(72°) * Vc = Rain's Velocity

Rain's Velocity = 116.926 km/h

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Answer: The energy incident on the solar panel during that day is 9.24 \times 10^{7} J.

Explanation:

Given: Mass = 250 kg

Initial temperature = 16^{o}C

Final temperature = 38^{o}C

Specific heat capacity = 4200 J/kg^{o}C

Formula used to calculate the energy is as follows.

q = m \times C \times (T_{2} - T_{1})

where,

q = heat energy

m = mass of substance

C = specific heat capacity

T_{1} = initial temperature

T_{2} = final temperature

Substitute the values into above formula as follows.

q = 250 kg \times 4200 J/kg^{o}C \times (38 - 16)^{o}C\\= 250 kg \times 4200 J/kg^{o}C \times 22^{o}C

As it is given that water absorbs 25% of the energy incident on the solar panel. Hence, energy incident on the solar panel can be calculated as follows.

\frac{25}{100} \times q = 250 kg \times 4200 J/kg^{o}C \times 22^{o}C\\q = 9.24 \times 10^{7} J

Thus, we can conclude that the energy incident on the solar panel during that day is 9.24 \times 10^{7} J.

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Vadim26 [7]

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In this example (a flashlight being turned on), we have a conversion of energy from chemical energy to electrical energy. In fact:

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