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valkas [14]
2 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]2 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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vladimir2022 [97]

Answer:

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5 0
2 years ago
Car 1 goes around a level curve at a constant speed of 65 km/h . The curve is a circular arc with a radius of 95 m . Car 2 goes
Arte-miy333 [17]

Answer:

The radius of the curve that Car 2 travels on is 380 meters.

Explanation:

Speed of car 1, v_1=65\ km/h

Radius of the circular arc, r_1=95\ m

Car 2 has twice the speed of Car 1, v_2=130\ km/h

We need to find the radius of the curve that Car 2 travels on have to be in order for both cars to have the same centripetal acceleration. We know that the centripetal acceleration is given by :

a=\dfrac{v^2}{r}

According to given condition,

\dfrac{v_1^2}{r_1}=\dfrac{v_2^2}{r_2}

\dfrac{65^2}{95}=\dfrac{130^2}{r_2}

On solving we get :

r_2=380\ m

So, the radius of the curve that Car 2 travels on is 380 meters. Hence, this is the required solution.

4 0
3 years ago
What type of energy does gasoline have? How could you relate it to the movement of a vehicle that uses it?
Licemer1 [7]

Answer:

Chemical potential energy is the energy stored in the chemical bonds of a substance. The various chemicals that make up gasoline contain a large amount of chemical potential energy that is released when the gasoline is burned in a controlled way in the engine of the car.

Explanation:

5 0
3 years ago
The slider of mass m is released from rest in position A and slides without friction along the vertical-plane guide shown. Deter
Anuta_ua [19.1K]

The value of normal force as the slider passes point B is

  • 6 mg

The value of h when the normal force is zero

  • 3R/2

<h3>How to solve for the normal force</h3>

The normal force is calculated using the work energy principle which is applied as below

K₁ + U₁ = K₂

k represents kinetic energy

U represents potential energy

the subscripts 1,2 , and 3 = a, b, and c

for 1 to 2

K₁ + W₁ = K₂

0 + mg(h + R) = 0.5mv²₂

g(h + R) = 0.5v²₂

v²₂ = 2g(1.5R + R)

v²₂ = 2g(2.5R)

v²₂ = 5gR

Using summation of forces at B

Normal force, N  = ma + mg

N = m(a + g)

N = m(v²₂/R + g)

N = m(5gR/R + g)

N = 6mg

for 1 to 3

K₁ + W₁ = K₃ + W₃

0 + mgh = 0.5mv²₃ + mgR

gh = 0.5v²₃ + gR

0.5v²₃ = gh - gR

v²₃ = 2g(h - R)

at C

for normal force to be zero

ma = mg

v²₃/R = g

v²₃ = gR

and v²₃ = 2g(h - R)

gR = 2gh - 2gR

gR + 2gR = 2gh

3gR = 2gh

3R/2 = h

Learn more about normal force at:

brainly.com/question/20432136

#SPJ1

8 0
11 months ago
A 0.500-kg stone is moving in a vertical circular path attached to a string that is 75.0 cm long. The stone is moving around the
Semenov [28]

Answer:

B. 7.07 m/s

Explanation:

The velocity of the stone when it leaves the circular path is its tangential velocity, v, which is given by

v=\omega r

where \omega is the angular speed and r is the radius of the circular path.

\omega is given by

\omega = 2\pi f

where f is the frequency of revolution.

Thus

v=2\pi fr

Using values from the question,

v=2\pi\times 1.50\times0.75

<em>Note the conversion of 75 cm to 0.75 m</em>

v=2\times3.14\times 1.50\times0.75 = 9.42\times0.75 = 7.065=7.07

6 0
3 years ago
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