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3 years ago

9

Twist-on connectors without the spring-steel coils (plastic threads only) are suitable for making branch-circuit connections.

1 answer:

Gnoma [55]3 years ago

6 0

Answer:

if it is a plastic connector it wont work but if there is metal or steel it will work

Explanation:

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In the simple circuit shown, what is the resistance, R? (Remember that V = I*R)

Nookie1986 [14]

B
Bensbsbng f fudge dndjdjd d djdjdbd

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3 years ago

PHYSICS. <br> ……………………………………………….

belka [17]

Answer:

Work, W = F * d, and

Work = change in kinetic energy, so W=deltaKE.

Hence,

deltaKE=F * d

(1/2)*m*v^2 =F * d

d=[(1/2)*m*v^2]/F

d=[(1/2)*0.6*20^2]/5

d=24 m.

Explanation:

Work = change in kinetic energy, so W=deltaKE.

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2 years ago

What are 2 ways that we can express to show our connection to our culture

garik1379 [7]

Answer:

Food, clothes, language, and belief

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3 years ago

A wire is oriented along the x-axis. It is connected to two batteries, and a conventional current of 2.4 A runs through the wire

Deffense [45]

Answer:

$\vec{F}=0.40176 N \hat{k}$

Explanation:

To calculate the force we need to use this equation

$\vec{F}=\int\limits^L_0 {i(\vec{dl}\times\vec{B})}$

where L is the total length of the wire

So in this case the small element of current is

$\vec{dl} = dx \hat{i}$

Because x is the direction of the current flow.

As is said in the problem B is such that

$\vec{B} = B \hat{j} = 0.62\hat{j} [ T]$

so to use the equation above we first calculate the following cross product:

$\vec{dl}\times\vec{B}=dx \hat{i}\times B \hat{j} = Bdx\hat{k}$

so the force:

$F = \int\limits^L_0 {i(\vec{dl}\times\vec{B})}=\int\limits^L_0{iBdx\hat{k}}$

So here we use the fact that B=0 in any point of the x axis that is not $x^{'}=0.27 [m]$ , that means that we only need to do the integration between a very short distant behind the point $x^{'}=0.27 [m]$ and a very short distant after that point, meaning:

$\vec{F}= \lim_{h \to 0}{\int\limits^{x^{'}+h}_{x^{'}-h}{iBdx\hat{k}} }$

so is the same as evaluating $iBx$ at $x=x^{'}$

that is:

$2,4 A * 0,62 T * 0,27 m \hat{k}$

$2,4 A * 0,62 (\frac{Kg}{A s^{2}}) * 0,27 m \hat{k}$

$2,4*0,62*2,7 ( \frac{ kgm }{ s^{2} } ) \hat{k}$

$\vec{F}=0.40176 N \hat{k}$

5 0

3 years ago

Pulsars are neutron stars that emit X rays and other radiation in such a way that we on Earth receive pulses of radiation from t

stira [4]

Answer:

(a) $a_{c} = 5.41\times 10^{9} m/s^{2}$

(b) $a_{t} = 2.99\times 10^{- 5} m/s^{2}$

Given:

Time period of Pulsar, $T_{P} = 33.085 ms == 33.085\times 10^{- 3} s$

Equatorial radius, R = 15 Km = 15000 m

Spinning time, $t_{s} = 9.50\times 10^{10}$

Solution:

(a) To calculate the value of the centripetal acceleration, $a_{c}$ on the surface of the equator, the force acting is given by the centripetal force:

$m\times a_{c} = \frac{mv_{c}^{2}}{R}$

$a_{c} = \frac{v_{c}^{2}}{R}$ (1)

where

$v_{c} = \frac{distance covered(i.e., circumference)}{ T}$

$v_{c} = \frac{2\pi R}{Time period, T}$ (2)

Now, from (1) and (2):

$a_{c} = R\frac({2\pi )^{2}}{T^{2}}$

$a_{c} = 15000\frac{2\pi )^{2}}{(33.085\times 10^{- 3})^{2}}$

$a_{c} = 5.41\times 10^{9} m/s^{2}$

(b) To calculate the tangential acceleration of the object :

The tangential acceleration of the object will remain constant and is given by the equation of motion as:

$v = u + a_{t}t_{s} = 0$

where

u = $v_{c}$

$a_{t} = - \frac{2\pi R}{Tt_{s}}$

$a_{t} = - \frac{2\pi 15000}{33.085\times 10^{- 3}\times 9.50\times 10^{10}}$

$a_{t} = 2.99\times 10^{- 5} m/s^{2}$

7 0

3 years ago