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koban [17]
3 years ago
15

The speed of a 180-g toy car at the bottom of a vertical circular portion of track is 7.75 m/s. If the radius of curvature of th

is portion of the track is 57.5 cm, what are the magnitude and direction of the force the track exerts on the car
Physics
1 answer:
Bezzdna [24]3 years ago
7 0

Answer:

11.28 N toward the center of the track

Explanation:

Centripetal force: This is the force that tend to draw a body close to the center of a circle, during circular motion.

The formula for centripetal force is given as,

F = mv²/r................................ Equation 1

Where F = force, m = mass of the toy car, v = velocity, r = radius

Given: m = 108 g = 0.108 kg, v = 7.75 m/s, r = 57.5 cm = 0.575 m

Substitute into equation 1

F = 0.108(7.75²)/0.575

F = 11.28 N

Hence the magnitude and direction of the force = 11.28 N toward the center of the track

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A biker goes up hill with a constant speed of 10km/h. Then he goes downhill with a constant speed of 50km/h. What's his average
Sladkaya [172]

This is not as simple as it looks.  

His average speed is NOT (10km/hr + 50km/hr)/2 = 30 km/hr.

You have to use the definition of speed:

Speed = (total distance covered) / (time to cover the distance).

Let's say the distance up (and down) the hill is 'd' .

Then the time it takes to go up the hill is (d/10) hours.

And the time it takes to come down the hill is (d/50) hours.

Total distance = 2d km

Total time = (d/10) + (d/50) = (5d/50) + (d/50) = 6d/50

Speed = distance/time = 2d/(6d/50) = 100d/6d

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3 years ago
In the first stage of a two-stage Carnot engine, energy is absorbed as heat Q1 at temperature T1 = 500 K, work W1 is done, and e
Nataly [62]

Answer:

Efficiency = 52%

Explanation:

Given:

First stage

heat absorbed, Q₁ at temperature T₁ = 500 K

Heat released, Q₂ at temperature T₂ = 430 K

and the work done is W₁

Second stage

Heat released, Q₂ at temperature T₂ = 430 K

Heat released, Q₃ at temperature T₃ = 240 K

and the work done is W₂

Total work done, W = W₁ + W₂

Now,

The efficiency is given as:

\eta=\frac{\textup{Total\ work\ done}}{\textup{Energy\ provided}}

or

Work done = change in heat

thus,

W₁ = Q₁ - Q₂

W₂ = Q₂ - Q₃

Thus,

\eta=\frac{(Q_1-Q_2)\ +\ (Q_2-Q_3)}{Q_1}}

or

\eta=1-\frac{(Q_1-Q_3)}{Q_1}}

or

\eta=1-\frac{(Q_3)}{Q_1}}

also,

\frac{Q_1}{T_1}=\frac{Q_2}{T_2}=\frac{Q_3}{T_3}

or

\frac{T_3}{T_1}=\frac{Q_3}{Q_1}

thus,

\eta=1-\frac{(T_3)}{T_1}}

thus,

\eta=1-\frac{(240\ K)}{500\ K}}

or

\eta=0.52

or

Efficiency = 52%

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matrenka [14]
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marysya [2.9K]

Answer:

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Explanation:

Density is used to describe how much space an object or substance takes up in relation to the amount of matter in that object or substance (its mass).

Another way to put it is that, density of a substance is the amount of mass of that substance per unit volume. If an object is heavy and compact, it has a high density.

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EleoNora [17]

Answer:

Yes, its true

Explanation:

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