Answer:
1321 N/C
Explanation:
From the question,
Electric Field (E) = Electric Force(F)/Electric Charge(q)
E = F/q............ Equation 1
Given: F = 7.4 μN = 7.4×10⁻⁶ N, q = 5.6 nC = 5.6×10⁻⁹ C
Substitute these value into equation 1
E = ( 7.4×10⁻⁶)/(5.6×10⁻⁹)
E = 1.321×10³ N/C
E = 1321 N/C
Hence the magnitude of the electric field at that location is 1321 N/C
Potential energy U = mgh
Given h = 123 m,
mg = F = 780 N
Then
U = (123)(780)
= 95940
= 9.59 x 10^4
Answer:
v = 0.4712 m / s In the opposite direction of the turntable
Explanation:
For this exercise the mouse to remain motionless must run at the same speed as the record player point moves,
Let's use rotational kinematics to find the speed of the record player at the point where the mouse is
v = w r
Let's reduce the magnitude to the SI system
w = 45 rpm (2π rad / 1rev) (1 min / 60s) = 4.712 rad / s
r = 10 cm = 0.10 m
Let's calculate
v = 4.712 0.10
v = 0.4712 m / s
In the opposite direction of the turntable
Answer:
15.2 m/s²
Explanation:
Acceleration: This can be defined as the rate of change of velocity. The S.I unit of acceleration is m/s²
From the question,
F-W = ma......................... Equation 1
Where F = Tension on the line, W = weight of the fish, m = mass of the fish, a = acceleration of the fish.
But,
W = mg........................ Equation 2
Where g = acceleration due to gravity.
Substitute equation 2 into equation 1
F-mg = ma
make a the subject of the equation
a = (F-mg)/m.................... Equation 3
Given: F = 150 N, m = 6 kg,
Constant: g = 9.8 m/s²
Substitute into equation 2
a = [150-(6×9.8)]/6
a = (150-58.8)/6
a = 91.2/6
a = 15.2 m/s²
Hence the minimum acceleration of the fish = 15.2 m/s²
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