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4 years ago

How does a electron travel through a closed circuit?

1 answer:

3 0

The particles that carry charge through wires in a circuit are mobile electrons. The electric field direction within a circuit is by definition the direction that positive test charges are pushed. Thus, these negatively charged electrons move in the direction opposite the electric field.

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If acceleration of a particle at any time is given by a=2t+5 the velocity after 5 seconds, if it starts from rest is?

kow [346]

The velocity after 5 seconds is 50 m/s.

3 0

3 years ago

Nzvhybprob it is really coll to do it

nika2105 [10]

Answer:

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6 0

3 years ago

A proton is accelerated from rest through a potential difference V0 and gains a speed v0. If it were accelerated instead through

Svet_ta [14]

Answer:

The speed is $\sqrt{2}v_{0}$ .

(a) is correct option.

Explanation:

Given that,

Potential difference $V= V_{0}$

Speed $v = v_{o}$

If it were accelerated instead

Potential difference $V'=2V_{0}$

We need to calculate the speed

Using formula of initial work done on proton

$W = q V$

We know that,

$\Delta W=\Delta K.E$

$q V=\dfrac{1}{2}mv^2$

Put the value into the formula

$q V_{0}=\dfrac{1}{2}mv_{0}^2$

$v_{0}^2=\dfrac{2qV_{0}}{m}$ ....(I)

If it were accelerated instead through a potential difference of $2 V_{0}$ , then it would gain a speed will be given as :

Using an above formula,

$v_{0}'^2=\dfrac{2qV_{0}}{m}$

Put the value of $V_{0}$

$v_{0}'^2=\dfrac{2q\times2V_{0}}{m}$

$v_{0}'=\sqrt{\dfrac{4qV_{0}}{m}}$

$v_{0}'=\sqrt{2}v_{0}$

Hence, The speed is $\sqrt{2}v_{0}$ .

6 0

3 years ago

Two identical blocks are attached to the same massless rope, which is strung around two massless, frictionless pulleys. A massle

Alik [6]

Answer:

The value of mass 1, m1= 6/5m

The value of mass 2, m2= 3/5m

Explanation:

case 1:

here tension and the acceleration will be:

for m1;

mg-T=ma
2mg - 2T = 2ma .....1.

for m2:

2T-mg = ma/2 ..... 2.

adding the both equations,

2mg - 2T + 2T-mg = 2ma + ma/2

a = 2/5 g

putting the value of a into the equation 1.

mg - T = m* (2/5)g

T = 3/5 ( mg )

now

case 2:

The two identical blocks are released from the rest, the tension remains the same as the case 1.

so,

for m1:

2T-m2g=0

for m2:

2m2g - 2T =0

adding both equations we get,

2T-m2g + 2m2g - 2T = 0

m2 = m1 / 2

T = m1*g / 2

here we know that

T (case1) = T (case2)

3/5 ( mg ) = m1*g / 2

m1 = 6/5 m

hence

m2 = 3/5 m

learn more about tension here:

<u>brainly.com/question/23590078</u>

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#SPJ4

4 0

2 years ago

A uniform solid sphere of mass M and radius R is free to rotate about a horizontal axis through its center. A string is wrapped

LenaWriter [7]

Answer:

$a = \frac{mg}{m + \frac{2}{5}M}$

Explanation:

To calculate the Acceleration and the tension of the object, we start by considering the value of the Tension through its moment of Inertia and Acceleration based on the angular velocity

$\tau = I\alpha = Tension(T)*R$