Macronutrients include carbohydrates, proteins, lipids, and ____

Nancy ran a distance of 5km in 30 minutes. What is her speed in meters(m) per hour)h)

Virty [35]

The speed of Nancy who ran a distance of 5 km in 30 minutes is 10000 meters(m) per hour(h).

<h3>What is the rate of speed?</h3>

The rate of speed is the rate at which the total distance is travelled in the time taken. Rate of speed can be given as,

$s=\dfrac{d}{t}$

Here, (d) is the distance travelled by the object and (t) is the time taken but the object to cover that distance.

Nancy ran a distance of 5 km in 30 minutes. There are 60 minutes in one hour. Thus, the time in hour for which Nancy ran is,

$t=\dfrac{30}{60}\\t=0.5\rm\; hr$

The meters in 5 kilometers is,

$d=5\times1000\\d=5000\rm\; m$

She ran 5000 kilometers in 0.5 hours. Thus, the speed of her is,

$s=\dfrac{5000}{0.5}\\s=10000\rm\; km/hr$

Thus, the speed of Nancy who ran a distance of 5 km in 30 minutes is 10000 meters(m) per hour(h).

Learn more about the rate of speed here:

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7 0

2 years ago

A river current has a velocity of 5 km/h relative to the shore. A boat moves in the same direction as the current at 4 km/h rela

AleksAgata [21]

<span>Add the river’s velocity to the boat’s velocity.</span>

7 0

3 years ago

Read 2 more answers

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

kondor19780726 [428]

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 use the equation

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

2) What is the magnitude of the angular momentum of the person 2 meters before she jumps on the merry-go-round?

We use the equation

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

3) What is the magnitude of the angular momentum of the person just before she jumps on to the merry-go-round?

We use the equation

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

4) What is the angular speed of the merry-go-round after the person 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

Now, we can get 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 travels at this new angular speed, with what force does the person need to hold on?

We have to get the centripetal force as follows

Fc = m*ω²*R

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

6) Once the person gets half way around, they decide to simply let go of the merry-go-round to exit the ride.

What is the linear velocity of the person right as they leave the merry-go-round?

we can use the equation

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

7) What is the angular speed of the merry-go-round after the person lets go?

ω₀ = 1.53 rad/s

It comes back to its initial angular speed

8 0

3 years ago

A helium ion of mass 4m and charge 2e is accelerated from rest through a potential difference V in vacuum. Its final speed will

Pavel [41]

Answer:

Final Velocity = √(eV/m)

Explanation:

The Workdone, W, in accelerating a charge, 2e, through a potential difference, V is given as a product of the charge and the potential difference

W = (2e) × V = 2eV

And this work is equal to change in kinetic energy

W = Δ(kinetic energy) = ΔK.E

But since the charge starts from rest, initial velocity = 0 and initial kinetic energy = 0

ΔK.E = ½ × (mass) × (final velocity)²

(Velocity)² = (2×ΔK.E)/(mass)

Velocity = √[(2×ΔK.E)/(mass)]

ΔK.E = W = 2eV

mass = 4m

Final Velocity = √[(2×W)/(4m)]

Final Velocity = √[(2×2eV)/4m]

Final Velocity = √(4eV/4m)

Final Velocity = √(eV/m)

Hope this Helps!!!

8 0

3 years ago

You observe a distant galaxy. You find that a spectral line of hydrogen that is shifted from its normal location in the visible

dimulka [17.4K]

Answer:

The galaxy is moving away from us

Explanation:

Galaxy refers to a system of millions or billions of stars accompanied by gas and dust such that they are held together as a result of gravitational attraction.

When we observe a distant galaxy, we find that a spectral line of hydrogen that is shifted from its normal location in the visible part of the spectrum into the infrared part of the spectrum.

It means that the galaxy is moving away from us.

6 0

3 years ago

Macronutrients include carbohydrates, proteins, lipids, and _____.