The relationship between electrical and charge force is that electrical force strengthens and increases the energy pulses and reflects off an object and makes the force about two times greater with charge.
Answer:
L2 = 1.1994 m
the length of the pendulum rod when the temperature drops to 0.0°C is 1.1994 m
Explanation:
Given;
Initial length L1 = 1.2m
Initial temperature T1 = 27°C
Final temperature T2 = 0.0°C
Linear expansion coefficient of brass x = 1.9 × 10^-5 /°C
The change i length ∆L;
∆L = L2 - L1
L2 = L1 + ∆L ...........1
∆L = xL1(∆T)
∆L = xL1(T2 - T1) ......2
Substituting the given values into equation 2;
∆L = 1.9 × 10^-5 /°C × 1.2m × (0 - 27)
∆L = 1.9 × 10^-5 /°C × 1.2m × (- 27)
∆L = -6.156 × 10^-4 m
From equation 1;
L2 = L1 + ∆L
Substituting the values;
L2 = 1.2 m + (- 6.156 × 10^-4 m)
L2 = 1.2 m - 6.156 × 10^-4 m
L2 = 1.1993844 m
L2 = 1.1994 m
the length of the pendulum rod when the temperature drops to 0.0°C is 1.1994 m
Answer:
<h3>Because one Coulomb of charge is an abnormally large quantity of charge, the units of microCoulombs (µC) or nanoCoulombs (nC) are more commonly used as the unit of measurement of charge. To illustrate the magnitude of 1 Coulomb, an object would need an excess of 6.25 x 1018 electrons to have a total charge of -1 C.</h3>
Explanation:
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Answer:
D is the answer
Explanation:
D is the most highest one so
the answer is D
Answer:
2.2nC
Explanation:
Call the amount by which the spring’s unstretched length L,
the amount it stretches while hanging x1
and the amount it stretches while on the table x2.
Combining Hooke’s law with Newton’s second law, given that the stretched spring is not accelerating,
we have mg−kx1 =0, or k = mg /x1 , where k is the spring constant. On the other hand,
applying Coulomb’s law to the second part tells us ke q2/ (L+x2)2 − kx2 = 0 or q2 = kx2(L+x2)2/ke,
where ke is the Coulomb constant. Combining these,
we get q = √(mgx2(L+x2)²/x1ke =2.2nC