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2 years ago

How can you observe the law of conservation of energy in action at the skatepark?

Physics

1 answer:

ivann1987 [24]2 years ago

3 0

Explanation:

When the skater is dropped onto the ramp from above, the potential energy decreases and the kinetic energy increases.

Every time the skater bounces from the impact, thermal energy is gained, and both potential and kinetic energy are lost.

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7. A 1000 kg car is rolling down the street at 2.5 m/s. How fast would a 2500 kg car have to

babunello [35]

1 m/s

Explanation:

To solve this question we use the following formula:

momentum = mass × velocity

momentum of the first car = 1000 kg × 2.5 m/s

momentum of the second car = 2500 kg × X m/s

To bring the cars at rest the momentum of the first car have to be equal to the momentul of the second car.

momentum of the first car = momentum of the second car

1000 kg × 25 m/s = 2500 kg × X m/s

X (velocity of the second car) = (1000 × 25) / 2500 = 1 m/s

Learn more about:

momentum

brainly.com/question/13378780

#learnwithBrainly

7 0

2 years ago

1) An ice skater with a moment of inertia of 2.2 kg m^2 rotates at a frequency of 0.8 rotations per second. The ice skater tucks

trapecia [35]

Answer:

Explanation:

2.3 kg·m/s²

4 0

2 years ago

The mass of a hot-air balloon and its cargo (not including the air inside) is 170 kg. The air outside is at 10.0°C and 101 kPa.

scoundrel [369]

Answer:

108.37°C

Explanation:

P₁ = Initial pressure = 101 kPa

V₁ = Initial volume = 530 m³

T₁ = Initial temperature = 10°C = 10+273.15 =283.15 K

P₂ = Final pressure = 101 kPa (because it is open to atmosphere)

V₂ = Final volume = 530 m³

P₁V₁ = n₁RT₁

⇒101×530 = n₁RT₁

⇒53530 J = n₁RT₁

P₂V₂ = n₂RT₂

⇒53530 J = n₂RT₂

$\frac{m_1}{m_2}=\frac{\rho V_1}{\rho V_1-170}\\\Rightarrow \frac{m_1}{m_2}=\frac{1.244\times 530}{1.244\times 530-170}=1.347\\\Rightarrow \frac{m_1}{m_2}=1.347\\\Rightarrow \frac{n_1}{n_2}=1.347$

Dividing the first two equations we get

$1=\frac{n_1}{n_2}\frac{T_1}{T_2}\\\Rightarrow 1=1.347\frac{283.15}{T_2}\\\Rightarrow T_2=1.347\times 283.15= 381.52\ K$

∴Temperature must the air in the balloon be warmed before the balloon will lift off is 381.25-273.15 = 108.37°C

8 0

2 years ago

Given the necessary information about the two components of projectile motion, your task is to explain the difference or even de

ivann1987 [24]

The two components of projectile motion include the following:

Horizontal motion

Vertical motion.

<h3>What is a Projectile?</h3>

This is defined as a missile propelled by the application of an external force and allowed to move freely under the influence of gravity and air resistance.

The equation for Horizontal motion Vx = V * cos(α)

Vertical velocity component: Vy = V * sin(α)

Read more about Projectile here brainly.com/question/24216590

8 0

1 year ago

What is the kinetic energy of a 0.25kg ball rolling at a speed of 2.5m/s

Lisa [10]

Answer:

Explanation:

Kinetic Energy formula:

KE = $\frac{1}{2}$ mv²

m=mass

v=speed

Given:

m=0.25kg

v=2.5m/s

Plug the values in:

KE = 1/2(0.25kg)(2.5m/s)²

KE = 0.78125 J (Joules)

4 0

2 years ago