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ivanzaharov [21]
3 years ago
11

The question is on the image below.

Physics
1 answer:
jenyasd209 [6]3 years ago
5 0
The sound bounces from wall to wall, so you draw arrows down the length of the canyon in both directions.
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A machine is rolling a metal cylinder under pressure. The radius of the cylinder is decreasing at a constant rate of 0.05 inches
Luden [163]
Yes, the volume of the cylinder will remain constant. As the radius decreases, the height will increase to make sure that the volume is kept the same.
We have been given a value of dr/dt and are required to find dh/dt
Because the volume is constant, we can plug it into the formula for the volume of the cylinder and rearrange it to make h the subject:
128 = πr²h
h = 128/πr²
Now we differentiate both sides:
dh/dr = -256/πr³
Applying the chain rule:
dh/dt = dh/dr x dr/dt
dh/dt = (-256/πr³) x -0.05
dh/dt = 64/5πr³; substituting the value of r
dh/dt = 64/5π(1.5)³
dh/dt = 1.21 in/sec
4 0
3 years ago
Find the net charge of a system consisting of 6.25×10^6 electrons and 7.75×10^6 protons. Express your answer using three signifi
jok3333 [9.3K]

Answer:

2.40 x 10⁻¹³ C

Explanation:

n_{e} = number of electrons = 6.25 x 10⁶

q_{e} = charge on electron = - 1.6 x 10⁻¹⁹ C

n_{p} = number of protons = 7.75 x 10⁶

q_{p} = charge on proton =  1.6 x 10⁻¹⁹ C

Net charge is given as

Q = n_{e} q_{e} + n_{p} q_{p}

Q = (- 1.6 x 10⁻¹⁹) (6.25 x 10⁶) + (1.6 x 10⁻¹⁹) (7.75 x 10⁶)

Q = 2.40 x 10⁻¹³ C

6 0
3 years ago
Shrinking Loop. A circular loop of flexible iron wire has an initial circumference of 168 cm , but its circumference is decreasi
DedPeter [7]

Answer:

103.1 V

Explanation:

We are given that

Initial circumference=C=168 cm

\frac{dC}{dt}=-15cm/s

Magnetic field,B=0.9 T

We have to find the magnitude of the  emf induced in the loop after exactly time 8 s has passed since the circumference of the loop started to decrease.

Magnetic flux=\phi=BA=B(\pi r^2)

Circumference,C=2\pi r

r=\frac{C}{2\pi}

r=\frac{168}{2\pi} cm

\frac{dr}{dt}=\frac{1}{2\pi}\frac{dC}{dt}=\frac{1}{2\pi}(-15)=-\frac{15}{2\pi} cm/s

\int dr=-\int \frac{15}{2\pi}dt

r=-\frac{15}{2\pi}t+C

When t=0

r=\frac{168}{2\pi}

\frac{168}{2\pi}=C

r=-\frac{15}{2\pi}t+\frac{168}{2\pi}

E=-\frac{d\phi}{dt}=-\frac{d(B\pi r^2)}{dt}=-2\pi rB\frac{dr}{dt}

E=-2\pi(-\frac{5}{2\pi}t+\frac{168}{2\pi})B\times -\frac{15}{2\pi}

t=8 s

B=0.9

E=2\pi\times \frac{15}{2\pi}\times 0.9(-\frac{15}{2\pi}(8)+\frac{168}{2\pi})

E=103.1 V

6 0
3 years ago
An example of a Natural electric field is
liberstina [14]

Answer:

Lightning

Explanation

6 0
3 years ago
Read 2 more answers
A coil of wire containing N turns is in an external magnetic field that is perpendicular to the plane of the coil and it steadil
krok68 [10]

Answer:

The Resultant Induced Emf in coil is 4∈.

Explanation:

Given that,

A coil of wire containing having N turns in an External magnetic Field that is perpendicular to the plane of the coil which is steadily changing. An Emf (∈) is induced in the coil.

To find :-

find the induced Emf if rate of change of the magnetic field and the number of turns in the coil are Doubled (but nothing else changes).

So,

   Emf induced in the coil represented by formula

                          ∈  =   -N\frac{d\phi}{dt}                                  ...................(1)

                                          Where:

                                                    .   \phi = BAcos\theta     { B is magnetic field }

                                                                                 {A is cross-sectional area}

                                                    .  N = No. of turns in coil.

                                                    .  \frac{d\phi}{dt} = Rate change of induced Emf.

Here,

Considering the case :-

                                    N1 = 2N  &      \frac{d\phi1}{dt} = 2\frac{d\phi}{dt}

Putting these value in the equation (1) and finding the  new emf induced (∈1)

                           

                                      ∈1 =-N1\times\frac{d\phi1}{dt}

                                      ∈1 =-2N\times2\frac{d\phi}{dt}

                                       ∈1 =4 [-N\times\frac{d\phi}{dt}]

                                        ∈1 = 4∈             ...............{from Equation (1)}      

Hence,

The Resultant Induced Emf in coil is 4∈.        

                           

8 0
3 years ago
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