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AVprozaik [17]
3 years ago
15

A car travels due east with a speed of 52.0 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 66.0° with the vertical. Find the velocity of the rain with respect to the following reference frames. (Enter the magnitude of the velocity.)
(a) the car
m/s
(b) the Earth
m/s
Physics
1 answer:
serg [7]3 years ago
6 0

Answer:

a) v = 6.43 m/s

b) v = 15.8 m/s

Explanation:

Speed of car = 56 km/h

56 km/h = 14.4 m/s

Angle rain makes on the glass to the vertical = 66°

Thus knowing that the opposite side of the angle is the distance moved by the car, and the adjacent side is the distance traveled by the rain in the same time

both of which are directly proportional to their velocities

Then

tan(66°) = 14.44m/s ÷ x

or x = 14.44/tan(66°)

Which is the vertical raindrop velocity of the relative to earth

v = 6.43 m/s vertically towards earth

For v relative to the car is we have vector sum of both velocities

v = √(14.44^2 + 6.43^2) = 15.8 m/s which is the velocity relative to car

= 15.8 m/s

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A remote controlled toy car starts from rest and begins to accelerate in a straight line. The figure below represents "snapshots
Nikolay [14]

(a). The car's average velocity between t = 1.0s to t = 1.5s will be - 1\;m/s

(b). The car's acceleration at t = 1.5s will be - 0.4\;m/s^{2}

(c). Car's speed is increasing with time.

We have a a remote controlled toy car that starts from rest and begins to accelerate in a straight line.

We have to determine -

  • The car's average velocity (in m/s) in the interval between -

        t = 1.0 s  to  t = 1.5 s.

  • The car's acceleration at t = 1.5 s.
  • Determining whether car's speed increasing or decreasing with time.

<h3>What is Acceleration?</h3>

The rate of change of velocity with respect to time is called Acceleration. Mathematically -

$a=\frac{dv}{dt}

According to the question, we have the following data for the Car -

t = 0s → x = 0m

t = 0.5s → x = 0.1m

t = 1.0s → x = 0.4m

t = 1.5s → x = 0.9m

t = 2.0s → x = 1.6m

PART - A

The car's average velocity between t = 1.0s to t = 1.5s will be -

$v_{avg} = \frac{0.9-0.4}{1.5-1}= 1 m/s

PART - B

Velocity at t = 1.5 s will be -

$v(1.5)=\frac{0.9}{1.5}= 0.6\;m/s

The car's acceleration at t = 1.5s will be -

$a(1.5) = \frac{v}{t} = \frac{0.6}{1.5} = 0.4\;m/s^{2}

PART - C

Since, the acceleration of the car is positive, this means that the car is accelerating in the forward direction. Hence, its speed is increasing with time.

[ The following data was missing in your answer. The complete question would include this data also -

t = 0s → x = 0m

t = 0.5s → x = 0.1m

t = 1.0s → x = 0.4m

t = 1.5s → x = 0.9m

t = 2.0s → x = 1.6m ]

To solve more questions on Kinematics, visit the link below-

brainly.com/question/17272824

#SPJ2

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1 year ago
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A light bulb radiates 110 nW of single-frequency sinusoidal electromagnetic waves uniformly in all directions. Calculate the ave
krek1111 [17]

Answer:

The intensity of the light from the bulb would be

3.501 x 10^{-6} W/m^{2}

Explanation:

Given

The Power = 110 n W = 110 x 10^{-9} W

the distance r = 50 mm = 50 /1000 = 0.05 m

The intensity can be obtained with the relationship below;

I = Power/area ......1

The area of the sphere would be used in this case since the bulb is spherical;    A=4πr^{2}

Putting it into equation 1, we have;

I = P/ 4πr^{2}

I =  110 x 10^{-9} / 4 x π x 0.05^{2}

I = 3.501 x 10^{-6} W/m^{2}

Therefore the intensity of the light from the bulb would be

3.501 x 10^{-6} W/m^{2}

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An airplane flies with a constant speed of 560 miles per hour. How far can it travel in 1 1/2 hours?
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You would multiply the speed by the time. So the answer would be 840 miles.
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Answer is miles sir your welcome it was simple
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Answer:

D

Explanation:

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