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2 months ago

A car accelerates uniformly from rest to a speed of 6.6 m/s in 6.5 s. Find the distance the car travels during this time.

Physics

2 answers:

Maru [3.3K]2 months ago

6 0

It covered a distance of 39.6 meters.

ValentinkaMS [3.4K]2 months ago

3 0

Answer:

The total distance the car covers is 21.56 meters.

Explanation:

Given that the car starts from rest (initial speed, u = 0), accelerates to 6.6 m/s (final speed, v), over a period of 6.5 seconds (time, t), we first compute the acceleration using the motion formulas.

$v=u+at$

$6.6=0+a\times 6.5$

$a=1.01\ m/s^2$

Next, apply the third equation of motion to calculate distance (s):

$s=\dfrac{v^2-u^2}{2a}$

$s=\dfrac{(6.6)^2}{2\times 1.01}$

Calculating gives s = 21.56 meters.

Therefore, the car travels 21.56 meters within 6.5 seconds, which is the answer you need.

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Explanation:

In this problem, we apply the connection between mechanical energy, kinetic energy, and gravitational potential energy.

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1 month ago

What happens to the particles of a liquid when energy is removed from them?

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Clarification:

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2 months ago

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1 month ago

A merry-go-round with a a radius of R = 1.63 m and moment of inertia I = 196 kg-m2 is spinning with an initial angular speed of

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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

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⇒ 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.

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we can apply the equation

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Complete Question

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Answer:

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