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3 years ago

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A whistle of frequency 516 Hz moves in a circle of radius 64.3 cm at an angular speed of 17.9 rad/s. What are (a) the lowest and

(b) the highest frequencies heard by a listener a long distance away, at rest with respect to the center of the circle

1 answer:

Arisa [49]3 years ago

5 0

Answer:

(a) 498.6 Hz

(b) 534.6 Hz

Explanation: Please see the attachments below

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What is the acceleration experienced by a car that takes 10s to reach 27m/s from rest

Lerok [7]

|acceleration| = (change in speed) / (time for the change)

Change in the car's speed = (27 - 0) = 27 m/s
Time for the change = 10 sec

|acceleration| = (27 m/s) / (10 s) = 2.7 m/s² .

That's the magnitude of the car's acceleration.
We don't know anything about its direction.

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3 years ago

A robot is exploring charon, the dwarf planet pluto's largest moon. Gravity on charon is 0.278 meters per second [m divided b

DochEvi [55]

In order to find the efficiency first we will find the Change in Potential energy of the small stone that robot picked up

First we will find the mass of the stone

As it is given that stone is spherical in shape so first we will find its volume

$V = \frac{4}{3}\pi r^3$

$V = \frac{4}{3}\pi *(\frac{0.06}{2})^3$

$V = 1.13 * 10^{-4} m^3$

Now it is given that it's specific gravity is 10.8

So density of rock is

$\rho = 10.8 * 10^3 kg/m^3$

mass of the stone will be

$m = \rho V$

$m = 10.8* 10^3 * 1.13 * 10^{-4}$

$m = 1.22 kg$

now change in potential energy is given as

$\Delta U = mgH$

here

g = gravity on planet = 0.278 m/s^2

H = height lifted upwards = 15 cm

$\Delta U = 1.22* 0.278 * 0.15$

$\Delta U = 0.051 J$

Now energy supplied by internal circuit of robot is given by

$E = Vit$

V = voltage supplied = 10 V

i = current = 1.83 mA

t = time = 12 s

$E = 10* 1.83 * 10^{-3} * 12$

$E = 0.22 J$

Now efficiency is defined as the ratio of output work with given amount of energy used

$\eta = \frac{\Delta U}{E}*100$

$\eta = \frac{0.051}{0.22} = 0.23$

so efficiency will be 23 %

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3 years ago

Suppose you want to to double a copper wire's resistance. To what temperature, in degrees Celsius, must you raise it if it is or

sleet_krkn [62]

Answer:

T=280.41 °C

Explanation:

Given that

At T= 24°C Resistance =Ro

Lets take at temperature T resistance is 2Ro

We know that resistance R given as

R= Ro(1+αΔT)

R-Ro=Ro αΔT

For copper wire

α(coefficient of Resistance) = 3.9 x 10⁻³ /°C

Given that at temperature T

R= 2Ro

Now by putting the values

R-Ro=Ro αΔT

2Ro-Ro=Ro αΔT

1 = αΔT

1 = 3.9 x 10⁻³ x ΔT

ΔT = 256.41 °C

T- 24 = 256.41 °C

T=280.41 °C

So the final temperature is 280.41 °C.

8 0

3 years ago

needhelpp101 Why is the gravitational potential energy of an object 1 meter above the moon’s surface less than its potential ene

Vlad1618 [11]

If you remember the formula for potential energy,
then this question is a piece-o-cake.

<em>Potential energy = (mass) x (<u>acceleration of gravity</u>) x (height) .</em>

-- The object's mass is the same everywhere.
-- You said that the height is the same both times.
-- How about the acceleration of gravity ?

Compared to gravity on Earth, it's only 16.5 percent as much on the Moon.
So naturally, from the formula, you'd expect the Potential Energy to be less
on the Moon.

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3 years ago

A 1980-kg car is traveling with a speed of 15.5 m/s. What is the magnitude of the horizontal net force that is required to bring

Dennis_Churaev [7]

Answer: 6067.5 N

Explanation:

Work = Change in Energy. To start, all of the energy is kinetic energy, so find the total KE using: KE = 1/2(m)(v^2). Plug in 1980 kg for m and 15.5 m/s for v and get KE = 237847.5 J.

Now, plug this in for work: Work = Force * Distance; so, divide work by distance to get 6067.5 N.

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3 years ago