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3 years ago

12

A rock climber stands on top of a 50 m -high cliff overhanging a pool of water. He throws two stones vertically downward 1.0 s a

part and observes that they cause a single splash. The initial speed of the first stone was 2.4 m/s .
a) How long after the release of the first stone does the second stone hit the water?
b) What was the initial speed of the second stone?
c) What is the speed of the first stone as it hits the water?
d) What is the speed of the second stone as it hits the water?

1 answer:

bagirrra123 [75]3 years ago

3 0

<h2><em>Answer: b) What was the initial speed of the second stone?</em></h2>

Explanation:

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After school, several students skateboard at the nearby park. Your best friend, Erwin, is a good skateboarder and happens to fal

Alchen [17]

Answer:

A person who never gives up.

Explanation:

due to his passion for skateboarding he try's his never gives up until he will finally learns the trick.

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2 years ago

To teach you how to find the parameters characterizing an object in a circular orbit around a much heavier body like the earth.

Karolina [17]

Answer:

a) T² = ( $\frac{4\pi ^2}{GM}$ ) r³

b) veloicity the dependency is the inverse of the root of the distance

kinetic energy depends on the inverse of the distance

potential energy dependency is the inverse of distance

angular momentum depends directly on the root of the distance

Explanation:

1) for this exercise we will use Newton's second law

F = ma

in this case the acceleration is centripetal

a = v² / r

the linear and angular variable are related

v = w r

we substitute

a = w² r

force is the universal force of attraction

F = $G \frac{m M}{r^2}$

we substitute

$G \frac{m M}{r^2} = m w^2 r$

w² = $\frac{GM}{r^3}$

angular velocity is related to frequency and period

w = 2π f = 2π / T

we substitute

$( \frac{2\pi }{T} ) = \frac{GM}{r^3}$

the final equation is

T² = () r³

b) the speed of the orbit can be found

v = w r

v = $\sqrt{\frac{GM}{r^3} } \ r$

v = $\sqrt{\frac{GM}{r} }$

in this case the dependency is the inverse of the root of the distance

Kinetic energy

K = ½ M v²

K = ½ M GM / r

K = ½ GM² 1 / r

the kinetic energy depends on the inverse of the distance

Potential energy

U =

U = -G mM / r

dependency is the inverse of distance

Angular momentum

L = r x p

for a circular orbit

L = r p = r Mv

L =

L =

The angular momentum depends directly on the root of the distance

8 0

2 years ago

A compact car, mass 660 kg, is moving at 1.00 ✕ 102 km/h toward the east. The driver of the compact car suddenly applies the bra

Maslowich

Answer: $59.43\times 10^3 kg-m/s$

Explanation:

Given

mass of car $m=660 kg$

Initial velocity of car $=102 km/h\approx 28.33 m/s$ towards east

Time taken to stop $t=2.1 s$

Force exerted $F_{avg}=4.8\times 10^3 N$

change in momentum is given by impulse imparted to the car

$Impulse(J)=-F\cdot t$

$J=-4.8\times 10^3\times 2.1$

$J=-59.49\times 10^3 kg-m/s$

negative Sign indicates that impulse is imparted opposite to the direction of motion

magnitude of momentum $J=59.49\times 10^3$

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3 years ago

Read 2 more answers

A uniform electric field of strength E points to the right. An electron is fired with a velocity v0 to the right and travels a d

zvonat [6]

Answer:

<u /> $D_l=d$ <u />

Explanation:

From the question we are told that:

The Electric field of strength direction =Right

The Velocity of The First Electron=V_0

The Velocity of The Second Electron=V_0

Therefore

$V_{e1}=V_{e2}$

Generally, the equation for the Horizontal Displacement of electron is mathematically given by

$D=\frac{at^2}{2}$

Where

Acceleration is given as

$a=\frac{V_o}{2d}$

And

Time

$T=\frac{d}{v_0}$

Therefore horizontal displacement towards the left is

$D_l=\frac{(\frac{V_o}{2d})(\frac{d}{v_0})^2}{2}$

<u /> $D_l=d$ <u />

5 0

2 years ago

Find the intensity of a 55 dB sound given I 0=10^-12W/m^2

MakcuM [25]

Answer:

3.16 × $10^{-7}$ W/ $m^{2}$

Explanation:

β(dB)=10 × $log_{10}(\frac{I}{I_{0} })$

$I_{0}$ = $10^{-12}$ W/ $m^{2}$

β=55 dB

Therefore plugging into the equation the values,

55=10 $log_{10}(\frac{I}{ [tex]10^{-12}$ })[/tex]

5.5= $log_{10}(\frac{I}{ [tex]10^{-12}$ })[/tex]

$10^{5.5}$ = $\frac{I}{10^{-12} }$

316227.76× $10^{-12}$ = I

I= 3.16 × $10^{-7}$ W/ $m^{2}$

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2 years ago