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valkas [14]
3 years ago
11

A rock of mass m is thrown straight up into the air with initial speed |v0 | and initial position y = 0 and it rises up to a max

imum height of y = h. A second rock with mass 2m (twice the mass of the original) is thrown straight up with an initial speed of 2|v0 |. What maximum height does the second rock reach?
Physics
1 answer:
REY [17]3 years ago
7 0

Answer:

Explanation:

Case 1:

mass = m

initial velocity = vo

final velocity = 0

height = y

Use third equation of motion

v² = u² - 2as

0 = vo² - 2 g y

y = vo² / 2g       ... (1)

Case 2:

mass = 2m

initial velocity = 2vo

final velocity = 0

height = y '

Use third equation of motion

v² = u² - 2as

0 = 4vo² - 2 g y'

y ' = 4vo² / 2g

y' = 4 y

Thus, the second rock reaches the 4 times the distance traveled by the first rock.

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An archer fish launches a droplet of water from the surface of a small lake at an angle of 70° above the horizontal. He is aimin
Norma-Jean [14]

Answer:

a). v=2776 m/s

Explanation:

The speed of the water droplet for the fish be successful is

Taking the distance in axis 'x' and 'y'

x_{tx}=40cm\frac{1m}{100cm}=0.40m\\x_{ty}=23cm\frac{1m}{100cm}=0.23m

The time is the velocity in axis 'x' with the angle 70 so

t=\frac{0.40m}{v_{x}*cos(70)}

Now using the time in terms of velocity the motion in axis 'y' can find the velocity to be the fish successful

x_{yf}=x_{yo}+v_{o}*t+\frac{1}{2}*g*t^{2}\\0.23m=0.40m\frac{vo*sin(70)}{vo*cos(70)} +\frac{1}{2}*9.8*(\frac{0.40}{vo*cos(70)} )^{2}\\0.23m=0.40m*vo*tan(70)+4.9*(\frac{0.16m^{2} }{vo^{2} *cos(70)^{2} }) \\vo^{2}cos(70)^{2}=\frac{0.16m^{2} }{0.177}\\vo=\sqrt{\frac{0.9021}{cos(70)^{2}}} \\vo=2.776 \frac{m}{s}

4 0
3 years ago
Formula One racers speed up much more quickly than normal passenger vehicles, and they also can stop in a much shorter distance.
klasskru [66]

Answer:

The magnitude of the car's acceleration as it slows during braking is 36.81 m/s²

Explanation:

From the question, the given values are as follows:

Initial velocity, u = 90 m/s

final velocity, v = 0 m/s

distance, s = 110 m

acceleration, a = ?

Using the equation of motion, v² = u² + 2as

(90)² + 2 * 110 * a = 0

8100 + 220a = 0

220a = -8100

a = -8100/220

a = -36.81 m/s²

The value for acceleration is negative showing that car is decelerating to a stop. The magnitude of the car's acceleration as it slows during braking is therefore 36.81 m/s²

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3 years ago
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vova2212 [387]
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4 years ago
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statuscvo [17]

Answer:

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alekssr [168]

Answer:

<em>The statement is true</em>

Explanation:

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Answer:

\boxed{\text{The statement is true}}

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