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mina [271]
3 years ago
14

Determine the force between two long parallel wires, that are separated by a dis- tance of 0.1m the wire on the left has a curre

nt of 10A and the one on the right BT has a current of 15A, both going up. 31十131
Physics
1 answer:
valina [46]3 years ago
3 0

Answer:

3 x 10^-4 N/m

Attractive

Explanation:

r = 0.1 m, i1 = 10 A, i2 = 15 A

The force per unit length between two parallel wires is given by

F = \frac{\mu _{0}}{4\pi }\times \frac{2i_{1}i_{2}}{r}

F = \frac{10^{-7}\times 2\times 10\times 15}{0.1}

F = 3 x 10^-4 N/m

Thus, the force per unit length between two wires is 3 x 10^-4 N/m, the force is attractive in nature because the direction of flow of current in both the wires is same.

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Caleb rides his motorcycle with a constant speed of 72km/h. How far can he Travel in 3 hours 40 minutes
lina2011 [118]
264 km

72 x 3 hr = 216 km
72/60 (min in an hr) =  1.2(km per min) x 40 = 48 km
216 + 48 = 264

5 0
3 years ago
A cube is 4.4 cm on a side, with one corner at the origin. Part 1 (a) What is the unit vector pointing from the origin to the di
Sidana [21]

Answer:

(a) \hat{A} = \frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}}

(b) \theta = 85.44^{\circ}

Solution:

As per the question:

Side of the cube, a = 4.4 cm

Coordinates of the diagonally opposite corner, A = <4.4, 4.4, 4.4> cm

Now,

(a) To calculate the unit vector:

\hat{A} = \frac{\vec{A}}{|A|}

\hat{A} = \frac{4.4\hat{i} + 4.4\hat{j} + 4.4\hat{k}}{\sqrt{()4.4}^{2} + (4.4)^{2} + (4.4)^{2}}

\hat{A} = \frac{4.4(\hat{i} + \hat{j} + \hat{k})}{4.4\sqrt{3}}

\hat{A} = \frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}}

(b) To calculate the angle between the two vectors say A and A' is given by:

\vec{A}\cdot \vec{A'} = \vec{A}\vec{A'}cos\theta                      

\theta = cos^{- 1}(\frac{\vec{A}\cdot \vec{A'}}{\vec{A}\vec{A'}})        (1)

Now,

The coordinates of the diagonally opposite corner, A' is <0, 0, 1> cm

Thus

\vec{A'} = 0\hat{i} + 0\hat{j} + 1\hat{k} = \hat{k}

Now,

Using equation (1) :

\theta = cos^{- 1}(\frac{(\frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}})\cdot \hat{k}}{|A||A'|})

|A||A'| = (\sqrt{4.4^{2} +4.4^{2} + 4.4^{2}})(\sqrt{0^{2} + 0^{2} + 0^{2}}) = 7.261

Thus

\theta = cos^{- 1}(\frac{\frac{1}{\sqrt{3}}}{7.261})

\theta = cos^{- 1}(0.07946) = 85.44^{\circ}

4 0
3 years ago
A temperature of 50F is equal to c
Hitman42 [59]

10.00 °C this is the right answer need more question feel free to post

5 0
3 years ago
How much energy is used when a 110kw appliance is used for 3 hours
Bogdan [553]
We could take the easy way out and just say

(110 kW) x (3 hours) = 330 kilowatt hours .

But that's cheap, and hardly worth even 5 points.
If we want to talk energy, let's use the actual scientific unit of energy.
________________________________________________

" 110 kw " means 110,000 watts = 110,000 joules/second .

(3 hours) x (3600 sec/hour) = 10,800 seconds.

(110,000 joules/second) x (10,800 seconds) = 1.188 x 10⁹ Joules

 That's

==>  1,188,000,000 joules

==>  1,188,000 kilojoules

==>  1,188 megajoules

==>  1.188 gigajoules

Atsa nawfulotta energy ! 
It goes back to that "110 kw appliance" that we started with. 
That's no common ordinary household appliance.  110 kw is something like
147 horsepower.  In order to bring 110 kw into your house, you'd need to
take 458 Amperes through the 240-volt line from the pole.  Most houses
are limited to 100 or 200 Amperes, tops.  And the TRANSFORMER on
the pole, that supplies the whole neighborhood, is probably a 50 kw unit.  
6 0
3 years ago
If you were looking for a metalloid on the periodic table,the best place to look would be?
vovikov84 [41]

Answer:

Long the step line

Explanation:

8 0
3 years ago
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