Answer:
E = 4000 N/C
Explanation:
Given the following data;
Force = 0.080 N.
Charge, q = 20 microcoulomb = 20 * 10^-6 = 2 * 10^-5 Coulombs
To find the electric field strength;
Mathematically, the electric field strength is given by the formula;
Electric field strength = force/charge
Substituting into the formula, we have;
E = 0.080/0.00002
E = 4000 N/C
We have that the Force is mathematically given as
F=170.833N
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Question Parameters:
- The torque required to loosen a nut that is holding a flat tire in place on a car has a magnitude of 41 N · m,
- the end of a 24 cm-long wrench to <em>loosen </em>the nut
Generally the equation for the Force is mathematically given as
Therefore
F=41/0.24
F=170.833N
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Answer:
the average force exerted by seatbelts on the passenger is 5625 N.
Explanation:
Given;
initial velocity of the car, u = 50 m/s
distance traveled by the car, s = 20 m
final velocity of the after coming to rest, v = 0
mass of the passenger, m = 90 kg
Determine the acceleration of the car as it hit the pile of dirt;
v² = u² + 2as
0 = 50² + (2 x 20)a
0 = 2500 + 40a
40a = -2500
a = -2500/40
a = -62.5 m/s²
The deceleration of the car is 62.5 m/s²
The force exerted on the passenger by the backward action of the car is calculated as follows;
F = ma
F = 90 x 62.5
F = 5625 N
Therefore, the average force exerted by seatbelts on the passenger is 5625 N.
Complete Question:
Suppose , where A has the dimensions LT, B has dimensions L²T⁻¹, and C has dimensions LT². Then the exponents n and m have the values
Answer:
The value of n = ¹/₅
The value of m = ³/₅
Explanation:
Given dimensions;
A = LT
B = L²T⁻¹
C = LT²
The values of n and m are calculated as follows;