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1 year ago

The wheels of a car have radius 12 in. and are rotating at 600 rpm. Find the speed of the car in mi/h.

Physics

1 answer:

guajiro [1.7K]1 year ago

7 0

Answer:

21.4 mph

Explanation:

Circumference of tire in FEET = pi * d = pi * 1 ft = pi feet

pi feet x 600 rot/min * 60 min /hr * 1 mile / 5280 feet = 21.4 mph

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Three 20.0 ohm resistors are

V125BC [204]

Answer:

6.67 ohm

Explanation:

From the question given above, the following data were obtained:

Resistor 1 (R₁) =20 ohm

Resistor 2 (R₂) = 20 ohm

Resistor 3 (R₃) = 20 ohm

Equivalent Resistance (R) =?

Since the resistors are arranged in parallel connection, the equivalent resistance can be obtained as follow:

1/R = 1/R₁ + 1/R₂ + 1/R₃

1/R = 1/20 + 1/20 + 1/20

1/R = (1 + 1 + 1) / 20

1/R = 3/20

Invert

R = 20/3

R = 6.67 ohm

Therefore, the equivalent resistance is 6.67 ohm.

5 0

3 years ago

When you stand on tiptoes on a bathroom scale, there is an increase in

pychu [463]

Answer:

B) Pressure on the scale, not registered as weight.

Explanation:

This is because energy (derived from weight) becomes compiled on the tips of your toes, and therefore does not increase your weight, but simply the pressure at a smaller point

3 0

3 years ago

A railroad car is pulled through the distance of 960 m by a train that did 578 kJ of work during this pull.

Wittaler [7]

Answer:

Explanation:

The force supplied by the train can be found by using the formula

$f = \frac{w}{d} \\$

w is the workdone

d is the distance

From the question we have

$f = \frac{578000}{960} \\ = 602.083333...$

We have the final answer as

Hope this helps you

7 0

3 years ago

A 350-g mass is attached to a spring whose spring constant is 64 N/m. Its maximum acceleration is 5.3 m/s2. What is the frequenc

max2010maxim [7]

The frequency of oscillation is 2.153 Hz

What is the frequency of spring?

Spring Frequency is the natural frequency of spring with a weight at the lower end. Spring is fixed from the upper end and the lower end is free.

For the mass-spring system in this problem,

The Frequency of spring is calculated with the equation:

$f = \frac{1}{2\pi } \sqrt{\frac{k}{m} }$

Where,

f = frequency of spring

k = spring constant = 64 N/m

m = mass attached to spring = 350g = 0.350 kg

a = maximum acceleration = 5.3 m/s^2

Substituting the values in the equation,

$f = \frac{1}{2\pi } \sqrt{\frac{64}{0.350} }$

$f = \frac{1}{2\pi } ( 13.522)$

$f = 2.1535 Hz$

Hence,

The frequency of oscillation is 2.153 Hz

Learn more about frequency here:

<u>brainly.com/question/13978015</u>

#SPJ4

6 0

2 years ago

2 m3 of an ideal gas are compressed from 100 kPa to 200 kPa. As a result of the process, the internal energy of the gas increase

rosijanka [135]

Answer:

work done is -150 kJ

Explanation:

given data

volume v1 = 2 m³

pressure p1 = 100 kPa

pressure p2 = 200 kPa

internal energy = 10 kJ

heat is transferred = 150 kJ

solution

we know from 1st law of thermodynamic is

Q = du +W ............1

put here value and we get

-140 = 10 + W

W = -150 kJ

as here work done is -ve so we can say work is being done on system

3 0

3 years ago