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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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A chemical change means a new substance with new properties was<br> formed.<br> True<br> False
Contact [7]

Answer:

False

Explanation:

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What is the meaning of the saying "the power of a lens is 1 dioptre "​
timama [110]

Answer: 1 dioptre of power of a lens is defined as the unit of measuring the power of optical lens Or curved mirror which is equal to the reciprocal of the focal length. The focal length is measured in the meter. 1 dioptre is equal to 1/m where m is the focal length. Basically dioptre is the SI unit of optical power of the lens.

7 0
3 years ago
A 8.34 × 103-kg lunar landing craft is about to touch down on the surface of the moon, where the acceleration due to gravity is
Mnenie [13.5K]

Answer:

21870.3156 N

Explanation:

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 1.6 m/s²

Equation of motion

v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-18.2^2}{2\times 162}\\\Rightarrow a=-1.02234\ m/s^2

The acceleration of the craft should be 1.02234 m/s²

F=ma\\\Rightarrow F=8.34\times 10^3\times 1.02234\\\Rightarrow F=8526.3156\ N

Weight of the craft

W=mg\\\Rightarrow W=8.34\times 10^3\times 1.6\\\Rightarrow W=13344\ N

Thrust

F_t=F+W\\\Rightarrow F_t=8526.3156+13344\\\Rightarrow F_t=21870.3156\ N

The thrust needed to reduce the velocity to zero at the instant when the craft touches the lunar surface is 21870.3156 N

3 0
3 years ago
A 4 kg rock is dropped from 5 m. There is no friction. What kind of energy does is have before? What kind of energy does it have
denpristay [2]
1) The total mechanical energy of the rock is:
E=U+K
where U is the gravitational potential energy and K the kinetic energy.

Initially, the kinetic energy is zero (because the rock starts from rest, so its speed is zero), and the total mechanical energy of the rock is just gravitational potential energy. This is equal to
E_i=U=mgh
where m=4 kg is the mass, g=9.81 m/s^2 is the gravitational acceleration and h=5 m is the height.
Putting the numbers in, we find the potential energy
U=mgh=(4 kg)(9.81 m/s^2)(5 m)=196.2 J

2) Just before hitting the ground, the potential energy U is zero (because now h=0), and all the potential energy of the rock converted into kinetic energy, which is equal to:
E_f=K= \frac{1}{2}mv^2
where v is the speed of the rock just before hitting the ground. Since the mechanical energy of the rock must be conserved, then the kinetic energy K before hitting the ground must be equal to the initial potential energy U of the rock:
K=U=196.2 J

3) For the work-energy theorem, the work W done by the gravitational force on the rock is equal to the variation of kinetic energy of the rock, which is:
W=196.2 J-0 J=196.2 J
6 0
3 years ago
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