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Two point charges of equal magnitude q are held a distance d apart. Consider only points on the line passing through both charge

NemiM [27]

let us consider that the two charges are of opposite nature .hence they will constitute a dipole .the separation distance is given as d and magnitude of each charges is q.

the mathematical formula for potential is $V=\frac{1}{4\pi\epsilon} \frac{q}{d}$

for positive charges the potential is positive and is negative for negative charges.

the formula for electric field is given as- $E=\frac{1}{4\pi\epsilon} \frac{q}{r^2}$

for positive charges,the line filed is away from it and for negative charges the filed is towards it.

we know that on equitorial line the potential is zero.hence all the points situated on the line passing through centre of the dipole and perpendicular to the dipole length is zero.

here the net electric field due to the dipole can not be zero between the two charges,but we can find the points situated on the axial line but outside of charges where the electric field is zero.

now let the two charges of same nature.let these are positively charged.

here we can not find a point between two charges and on the line joining two charges where the potential is zero.

but at the mid point of the line joining two charges the filed is zero.

5 0

3 years ago

The coefficient of linear expansion of ordinary glass is three times that of Pyrex glass. An ordinary glass rod and a Pyrex glas

DochEvi [55]

Answer:

b. 3 times

Explanation:

Lets take

Coefficient for ordinary glass = α₁

Coefficient for pyrex glass = α₂

Given that α₁ = 3 α₂

Initial length of both glasses are equal = L

Change in the temperature is also same .= ΔT

We know that change in the length given as

ΔL = L α ΔT

Therefore

$\dfrac{\Delta L_1}{\Delta L_2}=\dfrac{L\alpha_1\Delta T}{L\alpha_2\Delta T}$

$\dfrac{\Delta L_1}{\Delta L_2}=\dfrac{3\alpha_2}{\alpha_2}$

ΔL₁ = 3ΔL₂

Therefore change in the length of original glass is three time of pyrex glass.

b. 3 times

6 0

3 years ago

A 150kg motorcycle starts from rest and accelerates at a constant rate along a distance of 350m. The applied force is 250N and t

notsponge [240]

A) The net force on the motorbike is 205.9 N

B) The acceleration of the motorbike is $1.37 m/s^2$

C) The final speed is 5.2 m/s

D) The elapsed time is 3.80 s

Explanation:

A)

There are two forces acting on the motorbike:

- The applied force, F = 250 N, forward

- The frictional force, $F_f$ , backward

The frictional force can be written as

$F_f = \mu mg$

where

$\mu=0.03$ is the coefficient of kinetic friction

$m=150 kg$ is the mass of the motorbike

$g=9.8 m/s^2$ is the acceleration of gravity

Therefore the net force is given by

$\sum F = F - F_f = F - \mu mg$

And substituting, we find

$\sum F=250 - (0.03)(150)(9.8)=205.9 N$

2)

The acceleration of the motorbike can be found by using Newton's second law, which states that the net force is equal to the product between mass and acceleration:

$\sum F = ma$

where

m is the mass

a is the acceleration

In this problem, we have

$\sum F = 205.9 N$ is the net force

m = 150 kg is the mass

Solving for a, we find the acceleration:

$a=\frac{\sum F}{m}=\frac{205.9}{150}=1.37 m/s^2$

C)

Since the motion of the motorbike is a uniformly accelerated motion, we can use the following suvat equation:

$v^2-u^2=2as$

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the distance covered

For this motorbike, we have:

u = 0 (it starts from rest)

$a=1.37 m/s^2$

s = 350 m

Solving for v,

$v=\sqrt{u^2+2as}=\sqrt{0+2(1.37)(9.8)}=5.2 m/s$

4)

For this part of the problem, we can use the following suvat equation:

$v=u+at$

where

v is the final velocity

u is the initial velocity

a is the acceleration

t is the elapsed time

Here we have:

v = 5.2 m/s

u = 0

$a=1.37 m/s^2$

Solving for t, we find

$t=\frac{v-u}{a}=\frac{5.2-0}{1.37}=3.80 s$

Learn more about Newton's laws of motion and accelerated motion:

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#LearnwithBrainly

6 0

3 years ago

g 2. The _____ spans the distance from the _____ to the location of the applied force. moment arm; pivot point moment of inertia

Alik [6]

Answer:

The correct answer to the following question will be Option A (moment arm; pivot point).

Explanation:

The moment arm seems to be the duration seen between joint as well as the force section trying to act mostly on the joint. Each joint that is already implicated in the workout seems to have a momentary arm.
The moment arm extends this same distance from either the pivot point to just the position of that same pressure exerted.
The pivotal point seems to be the technical indicators required to fully measure the appropriate demand trends alongside different time-frames.

The other three choices are not related to the given situation. So that option A is the appropriate choice.

7 0

3 years ago

1200 N-m of torque is used to drive a gear (A) of diameter 25 cm, which in turn drives another gear (B) of diameter 52 cm. What

NNADVOKAT [17]

Answer:

2.5kN.m

Explanation:

Torque is directly proportional to pitch diameter

= Ta/Tb= Da/Db

=120/Tb= 0.25/0.5

Tb= 2.469kN.m approx 2.5kN.m

8 0

3 years ago

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Need help on this thanks​

Need help on this thanks