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kap26 [50]
3 years ago
10

The rear window in a car is approximately a rectangle, 2.0 m wide and 0.65 m high. The inside rearview mirror is 0.50 m from the

driver's eyes, and 1.80 m from the rear window.What is the minimum length for the rearview mirror if the driver is to be able to see the entire width and height of the rear window in the mirror without moving her head
Physics
1 answer:
Ymorist [56]3 years ago
5 0

Answer:

0.43 m

Explanation:

Angle of incident and angle of reflection is same.

tan Θh = L' / x (eye)

L' = Length of the window

x (eye) = Distance of the mirror from the eye

tan Θh = L / (x (eye) + xw)

xw = Distance of the mirror from the window

L'/ x (eye) = L / ( x (eye) + xw)

L' = L*x (eye) / ( x (eye) + xw)

L' = (2*0.5) / (0.5 + 1.8)

L' = 0.43 m

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To test the performance of its tires, a car
velikii [3]

<u>Answer</u>:

The coefficient of  static friction between the tires and the road is 1.987

<u>Explanation</u>:

<u>Given</u>:

Radius of the track, r =  516 m

Tangential Acceleration a_r=  3.89 m/s^2

Speed,v =  32.8 m/s

<u>To Find:</u>

The coefficient of  static friction between the tires and the road = ?

<u>Solution</u>:

The radial Acceleration is given by,

a_{R = \frac{v^2}{r}

a_{R = \frac{(32.8)^2}{516}

a_{R = \frac{(1075.84)}{516}

a_{R = 2.085 m/s^2

Now the total acceleration is

\text{ total acceleration} = \sqrt{\text{(tangential acceleration)}^2 +{\text{(Radial acceleration)}^2

=>= \sqrt{ (a_r)^2+(a_R)^2}

=>\sqrt{ (3.89 )^2+( 2.085)^2}

=>\sqrt{ (15.1321)+(4.347)^2}

=>19.4791 m/s^2

The frictional force on the car will be f = ma------------(1)

And the force due to gravity is W = mg--------------------(2)

Now the coefficient of  static friction is

\mu =\frac{f}{W}

From (1) and (2)

\mu =\frac{ma}{mg}

\mu =\frac{a}{g}

Substituting the values, we get

\mu =\frac{19.4791}{9.8}

\mu =1.987

8 0
3 years ago
How is newton's first law used in basketball?
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A roller coaster starts from rest at its highest point and then descends on its (frictionless) track. its speed is 40 m/s when i
Alenkinab [10]

Given that,

Initial velocity , Vi = 0

Final velocity , Vf = 40 m/s

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Distance can be calculated as,

2as = Vf² - Vi²

2 * 9.81 *s = 40² - 0²

s = 81.55 m

For half height, that is, s = 40.77m

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2as = Vf² - Vi²

2 * 9.81 * 40.77 = Vf² - 0²

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Therefore, speed of roller coaster when height is half of its starting point will be 28 m/s.  

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