Answer:
5.88×10⁸ W
Explanation:
Power = change in energy / time
P = mgh / t
P = (m/t) gh
P = (1.2×10⁶ kg/s) (9.8 m/s²) (50.0 m)
P = 5.88×10⁸ W
Answer:
8
Explanation:
Ca(ClO2)2 - 2*2 = 4 Oxygen atoms
2 Ca(ClO2)2 - 2*4 = 8 Oxygen atoms
Answer:
Explanation:
Expression for relative velocity
=
= (.54 + .82 )c/
= 1.36 c / 1.4428
= .94 c
β = .94
By using drift velocity of the electron, the current flow is 7.20 ampere.
We need to know about drift velocity of electrons to solve this problem. The drift velocity can be determined as
v = I / (n . A . q)
where v is drift velocity, I is current, n is atom number density, A is surface area and q is the charge.
From the question above, we know that
d = 2.097 mm
r = (0.002097 / 2) m
v = 1.54 mm/s = 0.00154 m/s
ρ = 8.92 x 10³ kg/m³
q = e = 1.6 x 10¯¹⁹C
Find the atom density
n = Na x ρ / Mr
where Na is Avogadro's number (6.022 x 10²³), Mr is the atomic weight of copper (63.5 g/mol = 0.635 kg/mol).
n = 6.022 x 10²³ x 8.92 x 10³ / 0.635
n = 8.46 x 10²⁷ /m³
Find the current flows
v = I / (n . A . q)
0.00154 = I / (8.46 x 10²⁷ . πr² . 1.6 x 10¯¹⁹)
0.00154 = I / (8.46 x 10²⁷ . π(0.002097 / 2)² . 1.6 x 10¯¹⁹)
I = 7.20 ampere
For more on drift velocity at: brainly.com/question/25700682
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Answer:
The answer to your question is: F = 0.4375 N. The force will be 16 times lower than with the first conditions.
Explanation:
Data
F = 7 N
F = ? if the masses is quartered
Formula
Process
Normal conditions F = Km₁m₂/r² = 7
When masses quartered F = K(m₁/4)(m₂/4)/r² = ?
F = K(m₁m₂/16)/r²
F = K(m₁m₂/16r² = 7/16 = 0.4375 N