A truck weighing 5 * 10^3 kgf and a cart weighting 500 kgf are moving with the same speed compare their kinetic energies?
1 answer:
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Answer:
The drift speed of the electrons in the wire is 2.12x10⁻⁴ m/s.
Explanation:
We can find the drift speed by using the following equation:
Where:
I: is the current = 4.50 A
n: is the number of electrons
q: is the modulus of the electron's charge = 1.6x10⁻¹⁹ C
A: is the cross-sectional area = 2.20x10⁻⁶ m²
We need to find the number of electrons:
Now, we can find the drift speed:
Therefore, the drift speed of the electrons in the wire is 2.12x10⁻⁴ m/s.
I hope it helps you!
Answer:
A) 31 kJ
B) 1.92 KJ
C) 40 , 2.48
Explanation:
weight of person ( m ) = 79 kg
height of jump ( h ) = 0.510 m
Compression of joint material ( d ) = 1.30 cm ≈ 0.013 m
A) calculate the force
Fd = mgh
F = mgh / d
W = mg
F(net) = W + F = mg ( 1 +
= 79 * 9.81 ( 1 + (0.51 / 0.013) )
= 774.99 ( 40.231 ) ≈ 31 KJ
B) calculate the force when the stopping distance = 0.345 m
d = 0.345 m
Fd = mgh hence F = mgh / d
F(net) = W + F = mg ( 1 +
= 79 * 9.81 ( 1 + (0.51 / 0.345) )
= 774.99 ( 2.478 ) = 1.92 KJ
C) Ratio of force in part a with weight of person
= 31000 / ( 79 * 9.81 ) = 31000 / 774.99 = 40
Ratio of force in part b with weight of person
= 1920 / 774.99 = 2.48