Answer:
Below
Explanation:
First draw the vectors that represent both electric fields.
E1 is the elictric field created by q1, E2 is the one created by q2.
● q1 is negative so E1 will point from P.
● q2 is positive so E2 will point out of P
(Picture below)
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The resulting electric field is equal to the sum of the two fields since both vectors are colinear.
Let E be the total field.
● E = E1 + E2
The formula of the electric field intensity is:
● E = K ×(q/d^2)
-K is Coulomb's constant
-d is the distance between the charge and the object ( here P)
-q is the charge
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● E1 = K × (q1/d1^2)
The distance between q1 and P is the qum of 0.15 m 0.25 m. (0.4 m)
Coulombs constant is 9×10^9 m^2/C^2
● E1 = 9×10^9 ×[-6.39 × 10^(-9)/ 0.4^2]
● E1 = -359.43 N/C
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● E2 = K ×(q2/d^2)
The distance between q2 and P is 0.25 m.
● E2 = 9×10^9×[3.22×10^(-9) /0.25^2]
● E2 = 463.68 N/C
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● E = E1 + E2
● E = -359.43+463.68
● E = 105.25 N/C
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Answer:
The study of charges in motion and their interaction with magnetic fields is known as electromagnetism.
Any material with an unequal number of electrons and or protons could generally be termed as ion.
Explanation:
i hope this will help you :)
Answer:
a) The centripetal acceleration of the car is 0.68 m/s²
b) The force that maintains circular motion is 940.03 N.
c) The minimum coefficient of static friction between the tires and the road is 0.069.
Explanation:
a) The centripetal acceleration of the car can be found using the following equation:

Where:
v: is the velocity of the car = 51.1 km/h
r: is the radius = 2.95x10² m

Hence, the centripetal acceleration of the car is 0.68 m/s².
b) The force that maintains circular motion is the centripetal force:

Where:
m: is the mass of the car
The mass is given by:

Where P is the weight of the car = 13561 N

Now, the centripetal force is:

Then, the force that maintains circular motion is 940.03 N.
c) Since the centripetal force is equal to the coefficient of static friction, this can be calculated as follows:



Therefore, the minimum coefficient of static friction between the tires and the road is 0.069.
I hope it helps you!