You testify as an expert witness in a case involving an accident in which car A slid into the rear of car B, which was stopped a

Answer:

1) L = 299.88 kg-m²/s

2) L = 613.2 kg-m²/s

3) L = 499.758 kg-m²/s

4) ω₁ = 0.769 rad/s

5) Fc = 70.3686 N

6) v = 1.2535 m/s

7) ω₀ = 1.53 rad/s

Explanation:

Given

R = 1.63 m

I₀ = 196 kg-m²

ω₀ = 1.53 rad/s

m = 73 kg

v = 4.2 m/s

1) What is the magnitude of the initial angular momentum of the merry-go-round?

We utilize the formula

L = I₀*ω₀ = 196 kg-m²*1.53 rad/s = 299.88 kg-m²/s

2) What is the angular momentum magnitude of the person 2 meters prior to jumping onto the merry-go-round?

The equation we apply is

L = m*v*Rp = 73 kg*4.2 m/s*2.00 m = 613.2 kg-m²/s

3) What is the angular momentum of the person just before she hops onto the merry-go-round?

We utilize the formula

L = m*v*R = 73 kg*4.2 m/s*1.63 m = 499.758 kg-m²/s

4) What is the angular velocity of the merry-go-round after the individual jumps on?

We can apply the Principle of Conservation of Angular Momentum

L in = L fin

⇒ I₀*ω₀ = I₁*ω₁

where

I₁ = I₀ + m*R²

⇒  I₀*ω₀ = (I₀ + m*R²)*ω₁

At this point, we can determine ω₁

⇒  ω₁ = I₀*ω₀ / (I₀ + m*R²)

⇒  ω₁ = 196 kg-m²*1.53 rad/s / (196 kg-m² + 73 kg*(1.63 m)²)

⇒  ω₁ = 0.769 rad/s

5) Once the merry-go-round moves at this new angular speed, what force must the person exert to hold on?

We must calculate the centripetal force as follows

Fc = m*ω²*R  

⇒  Fc = 73 kg*(0.769 rad/s)²*1.63 m = 70.3686 N

6) When the person is halfway around, they choose to simply release their hold on the merry-go-round to exit the ride.

What is the linear speed of the person the moment they exit the merry-go-round?

we can apply the equation

v = ω₁*R = 0.769 rad/s*1.63 m = 1.2535 m/s

7) What is the angular velocity of the merry-go-round after the individual releases their hold?

ω₀ = 1.53 rad/s

It returns to its original angular speed

Answer:

The work done, W = 19.6 J

Explanation:

It’s provided that

The mass of the block, m = 5 kg

The velocity of the block, v = 10 m/s

The coefficient of kinetic friction between the block and rough surface is 0.2

Distance traveled by the block, d = 2 m

As the block traverses the rough section, it loses energy equal to the work done by the kinetic energy.

W = 19.6 J

Thus, the change in kinetic energy of the block moving through the rough section is 19.6 J. Consequently, this is the required answer.

Answer:

The driveway measures 4.98 m

Explanation:

We aim to find the length of the driveway, thus utilizing the following equations

W=ΔK.E    where W represents work and  ΔK.E   indicates the change in kinetic energy

Moreover,

also

W = F.d  where F is the force and d denotes distance

Given that

m = 2100 Kg  

θ= 20.0°  

V=3.8 m/s representing the car's speed at the bottom of the driveway

W=Δ K.E

As the x component of gravity is

thus

= 7038.77 N

And the Net force is

=

= 7038.77 - 4000 = 3038.77 N

) = 15162/3038.77 = 4.98 m

Answer:

Stars generate energy by the process of nuclear fusion.

They are large entities composed of gaseous elements.

The main constituents of stars are hydrogen and helium.

Explanation:

Stars are colossal objects with extensive gravitational forces causing them to contract, which allows fusion to take place: the atomic nuclei in the star's core are drawn very close together due to gravity and elevated temperatures, leading to the fusion reaction. This fusion serves as the energy output for a star.

Conversely, it is true that stars predominantly consist of hydrogen and helium (two hydrogen nuclei can fuse to become helium), which implies that a star is essentially an enormous ball of gas without a solid surface suitable for standing on.

As for the presence of water on a star, it is simply impossible. The extreme temperatures found in stars are far too high for water to exist in any liquid state on their surfaces.