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Murrr4er [49]
3 years ago
14

By what factor is the self-inductance of an air solenoid changed if its length and number of coil turns are both tripled

Physics
1 answer:
fredd [130]3 years ago
4 0

Answer:

The new self inductance is 3 times of the initial self inductance.

Explanation:

The self inductance of a solenoid is given by :

L=\dfrac{\mu_oN^2 A}{L}

Where

N is number of turns per unit length

A is area of cross section

l is length of solenoid

If length and number of coil turns are both tripled,

l' = 3l and N' = 3N

New self inductance is given by :

L'=\dfrac{\mu_oN'^2 A}{L'}\\\\=\dfrac{\mu_o(3N)^2 A}{3L}\\\\=3\dfrac{\mu_oN^2 A}{L}\\\\=3L

So, the new self inductance is 3 times of the initial self inductance.

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Determine the launch speed of a horizontally launched cannonball that lands 26.3
nlexa [21]

Answer:

The cannon has an initial speed of 13.25 m/s.

Explanation:

The launched cannonball is an example of a projectile. Thus, its launch speed can be determined by the application of the formula;

R = u\sqrt{\frac{2H}{g} }

Where: R is the range of the projectile, u is its initial speed, H is the height of the cliff and g is the gravitaty.

R = 26.3 m, H = 19.3 m, g = 9.8 m/s^{2}.

So that:

26.3 = u\sqrt{\frac{2*19.3}{9.8} }

(26.3)^{2} = u^{2} x \frac{38.6}{9.8}

691.69 =  u^{2} x \frac{38.6}{9.8}

u^{2} = \frac{691.69*9.8}{38.6}

   = \frac{6778.562}{38.6}

u^{2} = 175.6104

⇒ u = \sqrt{175.6104}

  = 13.2518

u = 13.25 m/s

The initial speed of the cannon is 13.25 m/s.

4 0
3 years ago
How do I do number 10? Please show work!!
mafiozo [28]
The answer would be 40 cause 10*4=40
6 0
3 years ago
A projectile is fired straight up from ground level. After t seconds it's height is above the ground is h feet where h=-16t^2+14
Rainbow [258]
<span>Plug in 288 for h, move it over to the right side and do the quadratic formula to solve for t. You will get 2 times, in between and including those times will give you the period it is at least 288 ft off the ground.

</span>You can simplify this and not need to use the quadratic. <span>288=−16<span>t^2</span>+144t
</span><span>Divide through by 16 getting
 18=-t^2 + 9t
</span><span><span>t^2</span>−9t+18=0</span><span> Is what you would get after rearranging the equation Now you have something you can easily factor</span><span>


</span>
5 0
3 years ago
Read 2 more answers
What is the period and group of gold
algol13

Answer:

Chemical elements, a dense lustrous yellow precious metal of group.

4 0
4 years ago
Using the superposition method, calculate the current through R5 in Figure 8-71
Natasha2012 [34]

by superposition method we can find current in R5

here first let say only 2V battery is present in the circuit

now the equivalent resistance to be found for which we can say

2.2 k ohm and 1 k ohm is connected in parallel

r_1 = \frac{2.2 * 1}{2.2 + 1}

r_1 = 0.6875 k ohm

now it is in series with 1 k ohm and then that part is in parallel with 2.2 k ohm

r_2 = \frac{2.2* (1+0.6875)}{2.2 + (1+0.6875)}

r_2 = 0.95 k ohm

now the current flowing through the battery is

i = \frac{2}{1 + 0.95} = 1.02 mA

now this will divide into R3 and R2 so current flowing in R3 will be

i_1 = \frac{2.2}{2.2+1.6875}*1.02 = 0.58 mA

now this will again divide in R4 and R5

so current in R5 will be

i_5 = \frac{R_4}{R_4 + R_5}* i_1

i_5 = 0.18 mA

now when only 3 V battery is present in the circuit

R1 and R2 is in parallel and then it is in series with R3

so parallel combination will be

r_1 = \frac{1*2.2}{2.2 +1} = 0.6875k ohm

also after its series with R3

r_2 = 1 + 0.6875 = 1.6875 k ohm

now it is in parallel with R5 on other side

r_3 = \frac{1.6875 * 2.2}{1.6875 + 2.2} = 0.95 k ohm

now current through the battery will be given as

i = \frac{3}{1 + 0.95} = 1.53 mA

now it is divide in r2 and R5

so current in R5 is given as

i_5 = \frac{r_2}{r_2 + R_5}*i

i_5 = \frac{1.6875}{2.2 + 1.6875} * 1.53

i_5 = 0.67 mA

now the total current in R5 will be given by super position which is

i = 0.67 + 0.18 = 0.85 mA

so there is 0.85 mA current through R5 resistance

8 0
3 years ago
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