When placed at a certain point, a 4.20 C charge feels an electric force of 2.55 N. What is the magnitude of the electric field a
t that point?
(Remember, magnitudes are always positive.) (Unit = N/C)
(Remember, magnitudes are always positive.) (Unit = N/C)
2 answers:
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Answer:
0.3659
Explanation:
The power (p) is given as:
P = AeσT⁴
where,
A =Area
e = transmittivity
σ = Stefan-boltzmann constant
T = Temperature
since both the bulbs radiate same power
P₁ = P₂
Where, 1 denotes the bulb 1
2 denotes the bulb 2
thus,
A₁e₁σT₁⁴ = A₂e₂σT₂⁴
Now e₁=e₂
⇒A₁T₁⁴ = A₂T₂⁴
or
substituting the values in the above question we get
or
=0.3659
<h2>Thus the force of friction is 235 N</h2>
Explanation:
When the bear was at the height of 14 m . Its potential energy = m g h
here m is the mass of bear , g is acceleration due to gravity and h is the height .
Thus P.E = 27 x 10 x 14 = 3780 J
The K.E of the bear just before hitting = m v²
= x 27 x ( 6.1 )² = 490 J
The force of friction f = P.E - K.E = 3290 J
Because the work done = Force x Distance
Thus frictional force = = 235 N