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FrozenT [24]
3 years ago
6

While driving north at 21 m/s during a rainstorm you notice that the rain makes an angle of 36° with the vertical. while drivin

g back home moments later at the same speed but in the opposite direction, you see that the rain is falling straight down. from these observations, determine the speed and the angle of the raindrops relative to the ground?

Physics
1 answer:
denis23 [38]3 years ago
8 0
Let the velocity of the vehicle be 
V₁ = (21, 0)
Let the velocity of a raindrop be
V₂ = (a, b)

When returning, the raindrop is vertical. 
Therefore a = 21 m/s

When driving north, a raindrop makes an angle of 36° with the vertical, therefore
tan(36°) = a/b = 21/b
b = 21/tan(36°) = 28.9 m/s

The speed of a raindrop is
√(a² + b²) = √(441 + 835.44) = 35.73 m/s
The angle is 90-36 = 54° with the ground.

Answer: 35.7 m/s at 54° relative to the ground.

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6 0
3 years ago
Just wondering, what do you guys think the 5th dimension is? I've always believed it to be light.
zepelin [54]

Answer: In the 5th dimension, they who claim to know, say that there is only one time, including the past and the future.

8 0
3 years ago
If a 0.08 kg cell phone falls off a table at 15 m/s, then what is its kinetic energy right before it hits the ground?
Mariana [72]

The kinetic energy of the phone right before it hits the ground is 9J.

<h3>Kinetic energy of the phone</h3>

The kinetic energy of the phone right before it hits the ground is calculated as follows;

K.E = ¹/₂mv²

where;

  • m is mass of the phone
  • v is velocity of the phone

K.E = ¹/₂(0.08)(15)²

K.E = 9 J

Thus, the kinetic energy of the phone right before it hits the ground is 9J.

Learn more about kinetic energy here: brainly.com/question/25959744

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7 0
1 year ago
Three equal point charges, each with charge 1.05 μCμC , are placed at the vertices of an equilateral triangle whose sides are of
kirza4 [7]

Answer:

The value is U  = 0.06 \  J

Explanation:

From the question we are told that

The value of charge on each three point charge is

q_1 = q_2 = q_3 =q=  1.05 \mu C  =  1.05 *10^{-6} \  C

The length of the sides of the equilateral triangle is r  =  0.500 \

Generally the total potential energy is mathematically represented as

U  = k *  [ \frac{q_1 *  q_2}{r}  +  \frac{q_2 *  q_3}{r}   + \frac{q_3 *  q_1}{r} ]

=> U  = k * 3 * \frac{q^2}{r}

Here k is coulomb constant with value k = 9*10^{9}\  kg\cdot m^3\cdot s^{-4} \cdot A^{-2}.

=>    U  = 9*10^9 * 3 * \frac{(1.05 *10^{-6})^2}{0.5 }

U  = 0.06 \  J

6 0
3 years ago
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Answer:

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