Make a general statement concerning how large bodies of water affect the climate of nearby coastal communities.

A solenoid of length 0.35 m and diameter 0.040 m carries a current of 5.0 A through its windings. If the magnetic field in the c

puteri [66]

Correct question:

A solenoid of length 0.35 m and diameter 0.040 m carries a current of 5.0 A through its windings. If the magnetic field in the center of the solenoid is 2.8 x 10⁻² T, what is the number of turns per meter for this solenoid?

Answer:

the number of turns per meter for the solenoid is 4.5 x 10³ turns/m.

Explanation:

Given;

length of solenoid, L= 0.35 m

diameter of the solenoid, d = 0.04 m

current through the solenoid, I = 5.0 A

magnetic field in the center of the solenoid, 2.8 x 10⁻² T

The number of turns per meter for the solenoid is calculated as follows;

$B =\mu_o I(\frac{N}{L} )\\\\B = \mu_o I(n)\\\\n = \frac{B}{\mu_o I} \\\\n = \frac{2.8 \times 10^{-2}}{4 \pi \times 10^{-7} \times 5.0} \\\\n = 4.5 \times 10^3 \ turns/m$

Therefore, the number of turns per meter for the solenoid is 4.5 x 10³ turns/m.

3 0

3 years ago

The owner of a company that manufactures drinking cups decides it would be impressive to build an inground swimming pool that is

garri49 [273]

Answer:

The depth is 5.15 m.

Explanation:

Lets take the depth of the pool = h m

The atmospheric pressure ,P = 101235 N/m²

The area of the top = A m²

The area of the bottom = a m²

Given that A= 1.5 a

The force on the top of the pool = P A

The total pressure on the bottom = P + ρ g h

ρ =Density of the water = 1000 kg/m³

The total pressure at the bottom of the pool = (P + ρ g h) a

The bottom and the top force is same

(P + ρ g h) a = P A

P a +ρ g h a = P A

ρ g h a = P A - P a

$h=\dfrac{P ( A-a)}{\rho g a}$

$h=\dfrac{P ( 1.5 a-a)}{\rho g a}$

$h=\dfrac{P ( 1.5- 1)}{\rho g}$

$h=\dfrac{101235 ( 1.5- 1)}{1000\times 9.81}\ m$

h=5.15 m

The depth is 5.15 m.

7 0

3 years ago

A record player turntable initially rotating at 3313 rev/min is braked to a stop at a constant rotational acceleration. The turn

Rus_ich [418]

Answer:

(A) It will take 22 sec to come in rest

(b) Work done for coming in rest will be 0.2131 J

Explanation:

We have given the player turntable initially rotating at speed of $33\frac{1}{3}rpm=33.333rpm=\frac{2\times 3.14\times 33.333}{60}=3.49rad/sec$

Now speed is reduced by 75 %

So final speed $\frac{3.49\times 75}{100}=2.6175rad/sec$

Time t = 5.5 sec

From first equation of motion we know that '

$\alpha =\frac{\omega -\omega _0}{t}=\frac{2.6175-3.49}{4}=-0.158rad/sec^2$

(a) Now final velocity $\omega =0rad/sec$

So time t to come in rest $t=\frac{0-3.49}{-0.158}=22sec$

(b) The work done in coming rest is given by

$\frac{1}{2}I\left ( \omega ^2-\omega _0^2 \right )=\frac{1}{2}\times 0.035\times (0^2-3.49^2)=0.2131J$

4 0

3 years ago

What factor has the greatest impact on flexibility?

hjlf

That would be c. :) :) :)

4 0

3 years ago

A car travelling at 5.0 m/s starts to speed up. After 3.0 s its velocity has increased to 11 m/s. a) What is its acceleration? (

lions [1.4K]

initially, the car is traveling at 5.0 m/s.

so, we know acceleration for changing velocity is :

a = (v-v $_{o}$ )/t ..........(i)

where v is the final velocity

v $_{o}$ is the initial velocity

t is the time taken to change velocity

Now, as per the question :

initial velocity, v $_{o}$ =5.0 m/s

final velocity, v =11 m/s

time taken, t = 3 s

putting the values in equation (i),

a = ( 11-5 )/3

a = 2 m/s²

Therefore, a, after 3 s, is <em>2 m/s².</em>

4 0

1 year ago