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tekilochka [14]
4 years ago
7

Suzy drops a rock from the roof of her house. Mary sees the rock pass her 2.9 m tall window in 0.134 sec. From how high above th

e top of the window was the rock dropped? The acceleration of gravity is 9.8 m/s 2 . Answer in units of m.
Physics
1 answer:
timama [110]4 years ago
8 0

Answer:

Explanation:

Given

length of window h=2.9\ m

time Frame for which rock can be seen is \Delta t=0.134\ s

Suppose h is height above which rock is dropped

Time taken to cover h+2.9 is t_1

so using equation of motion

y=ut+\frac{1}{2}at^2

where  y=displacement

u=initial velocity

a=acceleration

t=time

time taken to travel h  is

h=0+0.5\times g\times (t_2)^2---2

Subtract 1 and 2 we get

2.9=0.5g(t_1^2-t_2^2)

5.8=g(t_1+t_2)(t_1-t_2))

and from equation t_1-t_2=0.134\ s

so t_1+t_2=\frac{5.8}{9.8\times 0.134}

t_1+t_2=4.416\ s

and t_1=t_2+\Delta t

so t_2+\Delta t+t_2=4.416

2t_2+0.134=4.416

t_2=0.5\times 4.282

t_2=2.141\ s

substitute the value of t_2 in equation 2

h=0.5\times 9.8\times (2.141)^2

h=22.46\ m

                                                     

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A ball is dropped from a rooftop 60m high. <br> How long is the ball in the air?
alina1380 [7]

Answer: 3.49 s

Explanation:

We can solve this problem with the following equation of motion:

y=y_{o}+V_{o}t-\frac{1}{2}gt^{2} (1)

Where:

y=0 m is the final height of the ball

y_{o}=60 m is the initial height of the ball

V_{o}=0 m/s is the initial velocity (the ball was dropped)

g=9.8 m/s^{2} is the acceleratio due gravity

t is the time

Isolating t:

t=\sqrt{\frac{2 y_{o}}{g}} (2)

t=\sqrt{\frac{2 (60 m)}{9.8 m/s^{2}}} (3)

Finally we find the time the ball is in the air:

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7 0
4 years ago
The momentum of light, as it is for particles, is exactly reversed when a photon is reflected straight back from a mirror, assum
LenKa [72]

Answer:

a)   E = 2.00 10³ J , b)   I = 6.66 10⁻⁶ N s , c)   F = 1.66 10⁻⁶ N

Explanation:

a) The intensity is defined as the power per unit area

          I = P / A

          P = I A

Power is energy for time

         P = E / t

We replace

        E / t = I A

        E = I A t

        E = 1.0 10³ 2.0 1.00

        E = 2.00 10³ J

b) The moment is

       p = U / c

In the case of a reflection the speed is reversed, so the moment

      Δp = 2 U / c

       I = Δp

       I = 2 U / c

       I = 2.00 10³/3 10⁸

       I = 6.66 10⁻⁶ N s

c) The defined impulse is

        I = F t

       F = I / t

For a time of 1 s

       F = 6.66 10⁻⁶ / 1

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d) Suppose n small mass mirror m = 10 10⁻³ kg, we write Newton's second law

        F = ma

        a = F / m

        a = 1.66 10⁻⁶ / 10 10⁻³

         a = 1.66 10⁻⁴ m / s

We see that the acceleration is very small and attended to increase the mass of the mirror will be less and less, so the assumption of no twisting of the mirror is very reasonable

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3 years ago
I require assistance, please. I would highly appreciate your response. It is a question on the states of matter. I forgot some o
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icang [17]

Answer:

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