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3 years ago

A merry-go-round rotates at the rate of

Physics

1 answer:

Mila [183]3 years ago

6 0

Explanation:

Angular momentum is conserved.

I₁ ω₁ = I₂ ω₂

(½ Mr² + md²) ω₁ = (½ Mr²) ω₂

(½ (35 kg) (2.3 m)² + (84 kg) (2.3 m)²) (0.28 rev/s) = (½ (35 kg) (2.3 m)²) ω

ω = 1.624 rev/s

ω = 10.2 rad/s

Round as needed.

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A wire carrying a current of 26.9 A is bent into a circular arc with a radius of 0.6 cm that sweeps out 0.900 radians. What is t

melomori [17]

The magnetic field at the center of the arc is 4 × 10^(-4) T.

To find the answer, we need to know about the magnetic field due to a circular arc.

<h3>What's the mathematical expression of magnetic field at the center of a circular arc?</h3>

According to Biot savert's law, magnetic field at the center of a circular arc is

B=(μ₀ I/4π)× (arc/radius²)

As arc is given as angle × radius, so

B=( μ₀I/4π)×(angle/radius)

<h3>What will be the magnetic field at the center of a circular arc, if the arc has current 26.9 A, radius 0.6 cm and angle 0.9 radian?</h3>

B=(μ₀ I/4π)× (0.9/0.006)

= (10^(-7)× 26.9)× (0.9/0.006)

= 4 × 10^(-4) T

Thus, we can conclude that the magnitude of magnetic field at the center of the circular arc is 4 × 10^(-4) T.

Learn more about the magnetic field of a circular arc here:

brainly.com/question/15259752

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5 0

2 years ago

A certain fluid (with properties listed below) is heated from 25 to 75 °C as it moves at 0.2 m/s through a straight thin-walled

dexar [7]

Answer:

not sure. I'll try answering this later

Explanation:

I'm not sure. I'll try answering this later .

3 0

3 years ago

A skier of mass 82.9 kg starts from rest at the top of a frictionless incline of height 20 m. At the bottom of the incline, the

ehidna [41]

Answer: 170.67 N

Explanation:

Given

Mass of skier is $m=82.9\ kg$

Height of the inclination is $h=20\ m$

Here, the potential energy of the skier is converted into kinetic energy which is consumed by the friction force by applying a constant force that does work to stop the skier.

$\Rightarrow mgh=F\cdot x\quad \quad [\text{F=constant friction force}]\\\\\Rightarrow 82.9\times 9.8\times 20=F\cdot 95.2\\\\\Rightarrow F=\dfrac{16,248.4}{95.2}\\\\\Rightarrow F=170.67\ N$

Thus, the horizontal friction force is 170.67 N.

7 0

3 years ago

3b. Grace drives her car with a constant speed

miskamm [114]

Time = (distance covered) / (speed)

Time = (224 mi) / (56 mi/hr)

<em>Time = 4 hours</em>

3 0

3 years ago

Space debris left from old satellites and their launchers is becoming a hazard to other satellites. (a) Calculate the speed of a

Pie

Answer:

Part a)

$v = 7407.1 m/s$

Part b)

$v_{rel} = 1.05 \times 10^4 m/s$

Explanation:

Part a)

As we know that orbital velocity at certain height from the surface of Earth is given as

$v = \sqrt{\frac{GM}{R+h}}$

here we know that

$M = 5.98 \times 10^{24} kg$

$R = 6.37 \times 10^6 m$

$h = 900 km = 9.0 \times 10^5 m$

now we have

$v = \sqrt{\frac{(6.67 \times 10^{-11})(5.98 \times 10^{24})}{6.37 \times 10^6 + 9.0 \times 10^5}}$

$v = 7407.1 m/s$

Part b)

When a loose rivet is moving in same orbit but at 90 degree with the previous orbit path then in that case the relative speed of the rivet with respect to the satellite is given as

$v_{rel} = \sqrt{2} v$

$v_{rel} = 1.05 \times 10^4 m/s$

6 0

2 years ago