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natita [175]
3 years ago
15

You stand 1.40 m in front of a wall and gaze downward at a small vertical mirror mounted on it. In this mirror you can see the r

eflection of your shoes. If your eyes are 1.70 m above your feet, through what angle should the mirror be tilted for you to see your eyes reflected in the mirror? (The location of the mirror remains the same, only its angle to the vertical is changed.)
Physics
1 answer:
Bess [88]3 years ago
7 0

Answer:

The angle is 31.26°.

Explanation:

Given that,

Distance = 1.40 m

Height = 1.70 m

We need to calculate the angle

Using formula of angle

tan\theta=\dfrac{y}{d}

Here, y = \dfrac{h}{2}

y=\dfrac{1.70}{2}=0.85\ m

Put the value into the formula

\tan\theta =\dfrac{0.85}{1.40}

\theta=\tan^{-1}0.6071

\theta=31.26^{\circ}

Hence, The angle is 31.26°.

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Based on your observations of the six collisions, describe the physical difference between elastic and inelastic collisions.
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Answer:

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In elastic collision, the total kinetic energy, momentum are conserved, and there is no wasting of energy occurs. Swinging balls is the good example of elastic collision. In inelastic collision, the energy is not conserved it changes from one form to another for example thermal energy or sound energy. Automobile collision is good example, of inelastic collision.

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2 years ago
A 1350 kg uniform boom is supported by a cable. The length of the boom is l. The cable is connected 1/4 the
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Answer:

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Explanation:

Tension in the cable is calculated by:

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FTorque(3/4L)= FBcostheta(1/2L)+ FWcostheta(L)

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Ftorque= 2/3FBcostheta+ 4/3FWcostheta

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b) components of Normal force

Efx=FNx-FTcos(90-theta)=0 static equilibrium

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Fy=FNy+ FTsin(90-theta)-FB-FW=0

FNy= -FTsin(90-55)+FB+FW

FNy= -21900sin(35)+(1350+2250)×9.81=22700N

The Normal force

FN=sqrt(17900^2+22700^2)

FN= 28.900N

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