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Salsk061 [2.6K]
3 years ago
14

A 0.10 kg stone is thrown from the edge of an ocean cliff with an initial speed

Physics
1 answer:
AfilCa [17]3 years ago
6 0

Answer:

79.20 joule

Explanation:

1.

A 0.10kg stone is thrown from the edge of an ocean cliff with an initial speed of 21m/s. When it strikes the water below, it is traveling at 45m/s. What is the change in kinetic energy of the stone?

ΔEk = m/2*(V2^2-V1^2) = 0.05*(45^2-21^2) = 79.20 joule

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Determine the kinetic energy of a 2000 kg roller coaster car that is moving at the speed of 10 ms
kondaur [170]

Answer:

\boxed {\boxed {\sf 100,000 \ Joules}}

Explanation:

Kinetic energy is energy due to motion. The formula is half the product of mass and velocity squared.

E_k= \frac{1}{2} mv^2

The mass of the roller coaster car is 2000 kilograms and the car is moving 10 meters per second.

  • m= 2000 kg
  • s= 10 m/s

Substitute these values into the formula.

E_k= \frac{1}{2} (2000 \ kg ) \times (10 \ m/s)^2

Solve the exponent.

  • (10 m/s)²= 10 m/s * 10 m/s= 100 m²/s²

E_k= \frac{1}{2} (2000 \ kg ) \times (100 \ m^2/s^2)

Multiply the first two numbers together.

E_k= 1000 \ kg  \times (100 \ m^2/s^2)

Multiply again.

E_k= 100,000 \ kg*m^2/s^2

  • 1 kilogram square meter per square second is equal to 1 Joule.
  • Our answer of 100,000 kg*m²/s² is equal to 100,000 Joules.

E_k= 100,000 \ J

The roller coaster car has <u>100,000 Joules</u> of kinetic energy.

3 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
if the amplitude of sine curve is 4 and the frequency is 1 with no phase shift, what is the function that descirbes this curve?
Vlada [557]

Answer:

y = 4 Sin (2πt)

Explanation:

Amplitude, A = 4

frequency, f = 1

Wave function is given by

y = A sinωt

where, ω is angular frequency

ω = 2 π f = 2 π x 1 = 2π

So, the desired wave function

y = 4 Sin (2πt)

8 0
3 years ago
One half of a balanced chemical equation is shown.
Zarrin [17]

Answer:

A: 3 Mg, 2 P, 14 O, 12 H

Explanation:

I took the test on Edgenuity

6 0
3 years ago
Read 2 more answers
The experimental apparatus shown in the figure above contains a pendulum consisting of a 0.66 kg ball attached to a string of le
lara31 [8.8K]

The problem is solved and the questions are answered below.

Explanation:

a. To calculate the speed of the 0.66 kg ball just before the collision

V₀ + K₀ = V₁ + K₁

= mgh₀ = 1/2 mv₁²

where, h= r - r cosθ

V = \sqrt{2gh}

 V = 2.42 m/s

b. Calculate the speed of the 0.22 kg ball immediately after the collision

y = y₀ + Vy₀t - 1/2 gt²

0 = 1.2 - 1/2 gt²

t = 0.495 s

x = x₀ + Vx₀t

1.4 = 0 + vx₀ (0.495)

Vx₀ = 2.83 m/s

C. To Calculate the speed of the 0.66 kg ball immediately after the collision

m₁ v₁ = m₁ v₃ + m₂ v₄

(0.66)(2.42) = (0.66) v₃ + (0.22)(2.83)

V₃ = 1.48 m/s

D. To Indicate the direction of motion of the 0.66 kg ball immediately after the collision is to the right.

E. To Calculate the height to which the 0.66 kg ball rises after the collision

V₀ + k₀ = V₁ + k₁

1/2 mv₀² = mgh₁

h₁ = v₀²/2 g

  = 0.112 m

F. Based on your data, No the collision is not elastic.

Δk = 1/2 m₁v₃² =1/2 m₂v₄² - 1/2 m₁v₁²

     = 1/2 (0.66)(1.48)² + 1/2 (0.22)(2.83)² - 1/2 (0.66)(2.42)²

    = - 0.329 J

Hence, kinetic energy is not conserved.

8 0
3 years ago
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