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kotykmax [81]
2 years ago
15

As a science project, you drop a watermelon off the top of the Empire State Building, 320 m above the sidewalk. It so happens th

at Superman flies by at the instant you release the watermelon. Superman is headed straight down with a speed of 30.0 m/s . Part A How fast is the watermelon going when it passes Superman?
Physics
1 answer:
oksano4ka [1.4K]2 years ago
3 0

Answer:

24.21 m/s

Explanation:

Let the time taken by the water melon to fall is t meanwhile in the same time the distance traveled by the superman and watermelon is same.

Let the distance traveled by the super man and the water melon is same which is h.

For water melon:

Use second equation of motion

s = ut +0.5at^2

Here, u = 0, g = - 9.8 m/s^2, time = t

h = 0 - 0.5 x 9.8 x t^2

h =  -4.9 t^2     .... (1)

For superman:

Distance = speed x time

h = -30 x t   .... (2)

By comparing equation (1) and (2), we get

-4.9 t^2 = - 30 t

t = 2.47 second

The speed of teh watermelon is given by first equation of motion

v = u + at

here, u = 0, a = 9.8 m/s^2, t = 2.47 second

So, v = 0 - 9.8 x 2.47

v = - 24.21 m/s

The negative sign shows that the direction of velocity is downwards.

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Answer:

The answer is down below

Explanation:

Las arterias transportan la sangre desde el corazón y se ramifican en vasos más pequeños, formando arteriolas. Las arteriolas distribuyen sangre a los lechos capilares, los lugares de intercambio con los tejidos corporales. Los capilares conducen de regreso a pequeños vasos conocidos como vénulas que fluyen hacia las venas más grandes y finalmente regresan al corazón.

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2 years ago
By what percent does the braking distance of a car decrease, when the speed of the car is reduced by 10.3 percent? Braking dista
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The braking distance is the distance traveled by a car experiencing a braking force until it comes to rest.

Our initial energy is solely kinetic:
E_i = \frac{1}{2}mv^2

And, since the car goes to rest, it is no longer in motion. It will have no kinetic energy.
E_f = 0

Therefore, there was work done by the braking force.

W_B = E_f - E_i = -\frac{1}{2}mv^2

Recall the definition of work:
W = F\cdot \Delta x

Or in this case, since the displacement and breaking force are antiparallel:
W = -F_B\Delta x

This is equivalent to the dissipation of kinetic energy:
W = -F_B\Delta x = -\frac{1}{2}mv^2

Now, to visualize this, let's rearrange the equation to solve for displacement.

\Delta x =\frac{mv^2}{2F_B}

<u>There is a direct, SQUARE relationship between necessary braking distance speed. </u>

If the speed was reduced by 10.3 percent, its new speed is only 89.7% percent of the original, so:
\Delta x' =\frac{m(0.897v)^2}{2F_B}

\Delta x' = 0.8046\Delta x

The reduction by a percentage is:
1 - 0.8046 = 0.1954 \\\\\boxed{= 19.54\%}

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A 90kg man is standing still on frictionless ice. His friend tosses him a 10kg ball, which has a horizontal velocity of 20m/s. A
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Using the conservation of momentum,
ma*va1 + mb*vb1 = ma*va2 + mb*vb2
Let:
ma = mass of the ball
va = velocity of the ball
mb = mass of the man
vb = velocity of the man
The subscript 1 is known as initials while 2 is for finals.
Before the man throws the ball, he starts at rest, meaning the initial velocity of the ball and the initial velocity of the man are zero. So
0 = ma*va2 + mb*vb2
Given ma = 10 kg; va = 20 m/s; mb = 90 kg; vb is unknown, therefore
-(mb*vb2) = ma*va2
vb2 = -(ma*va2)/mb2 = -(10*20)/90 = -2.22 m/s
Notice that his velocity is negative because when he finally throws the ball (say to the right), he moves at the opposite direction (that is to the left) on which he stands on the frictionless surface.

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