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vitfil [10]
3 years ago
6

The electric potential, when measured at a point equidistant from two particles that have charges equal in magnitude but of oppo

site sign, is
Physics
1 answer:
olga2289 [7]3 years ago
4 0

Answer:

V = 0

Explanation:

To find the potential at the middle of the two charges with opposite signs, you use the following formula:

V_T=V_1+V_2=k\frac{q}{r}-k\frac{q}{r}=0V      (1)

where you have used the fact that the charges are the same and the distances are the same.

The electric potential is zero at the point in the middle of the two charges

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Susan, driving north at 53 mphmph , and Shawn, driving east at 63 mphmph , are approaching an intersection. Part A What is Shawn
mafiozo [28]

Answer:

Shawn's speed relative to Susan's speed = 10 mph

Resultant velocity = 82.32 mph

Explanation:

The given data :-

i) Susan driving in north and speed of Susan is ( v₁ ) = 53 mph.

ii) Shawn driving in east and speed of Shawn is ( v₂ ) = 63 mph.

iii) The speed of both Susan and Shawn is relative to earth.

iv) The angle between Susan in north and Shawn in east is 90°.

We have to find Shawn's speed relative to Susan's speed.

v₂₁ = v₂ - v₁   = 63 - 53 = 10 mph

Resultant velocity,

v = \sqrt{v_{2} ^{2}+ v_{1} ^{2}  }  =\sqrt{63^{2} +53^{2} }

v = 82.32 mph

5 0
3 years ago
Differentiate scalar & vector quantity?
Keith_Richards [23]
\textbf{Hello Friend}


Scalar Quantity :-

→ These are the quantities with magnitude only . These quantities doesn't have to be mentioned with direction

eg.)=> Mass , Temprature .



Vector Quantity :-

→ These quantities are described with both Magnitude and Direction . These quantities follow special type of algebra called Vector algebra .


eg.)=> Force , Displacement






_______________________________



Hope It Helps You. ☺
5 0
3 years ago
A truck ttaveling with an initial velocity of 60.0m/s comes to a stop in 12.93 secs. What is the accelerationof the truck?
anastassius [24]

When a car travelling at an initial velocity of 10 m/s applies the brakes and bring ... accelerates from rest for a time of 8 seconds with an acceleration of 3.2m/s^2?


3 0
3 years ago
A 645 g block is released from rest at height h0 above a vertical spring with spring constant k = 530 N/m and negligible mass. T
Nina [5.8K]

Answer:

a)5.88J

b)-5.88J

c)0.78m

d)0.24m

Explanation:

a) W by the block on spring is given by

W= \frac{1}{2}kx² = \frac{1}{2}(530)(0.149)² =  5.88 J

b)  Workdone by the spring = - Workdone by the block = -5.88J

c) Taking x = 0 at the contact point we have U top = U bottom

So, mgh_o = \frac{1}{2}kx² - mgx

And, h_o= ( \frac{1}{2}kx² - mgx )/(mg) = [\frac{1}{2} (530)(0.149^2)-(0.645)(9.8)(0.149)]/(0.645x9.8)    

   h_o=   0.78m            

d) Now, if the initial initial height of block is 3h_o

h_o = 3 x 0.78 = 2.34m

then, \frac{1}{2}kx² - mgx - mgh_o =0

 

\frac{1}{2}(530)x²  - [(0.645)(9.8)x] - [(0.645)(9.8)(2.34) = 0

265x² - 6.321x - 14.8 = 0  

a=265

b=-6.321

c=-14.8

By using quadratic eq. formula, we'll have the roots

x= 0.24 or x=-0.225

Considering only positive root:

x= 0.24m (maximum  compression of the spring)

4 0
2 years ago
One property that makes electromagnetic waves differ from other types of waves is that they can
Phoenix [80]
That they travel in a vacuum.  All other waves require a medium in which they wave.
4 0
3 years ago
Read 2 more answers
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