Q = mc<span>∆t, where:
q = energy flow
m = mass, 120 000 g
c = specific heat capacity, 4.81 J/gC
</span><span>∆t = change in temperature, ~75 (100 - 25, which is room temperature)
Substituting in the values, we get:
q = 120000 x 4.81 x 75 = 43290000 Joules = 43.29 MJ
Hope I helped!! xx
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Answer:
26 m/s
69 m
Explanation:
Given:
v₀ = 20 m/s
a = 2 m/s²
t = 3 s
Find: v and Δx
v = at + v₀
v = (2 m/s²) (3 s) + 20 m/s
v = 26 m/s
Δx = v₀ t + ½ at²
Δx = (20 m/s) (3 s) + ½ (2 m/s²) (3 s)²
Δx = 69 m
The electric field strength is 
Explanation:
The strength of the electric field produced by a single point charge is given by:
where
is the Coulomb's constant
q is the magnitude of the charge
r is the distance from the charge at which the field strength is calculated
For the charge in the problem, we have:
is the charge
Therefore, the electric field strength is
Learn more about electric fields:
brainly.com/question/8960054
brainly.com/question/4273177
#LearnwithBrainly
Answer:
average acceleration = 6 
Explanation:
Recall that the average acceleration 
is defined by the change in velocity from an initial velocity 
, to a final velocity 
over the time (t) it took that change to happen. Then, in mathematical terms this is:
with our information this becomes:
Answer:
Change in Q = 2.1x 10^-3 C
Explanation:
We are given that
The Initialcapacitance C1 = 6.0μF
Initial charge oncapacitor
Q1 = C1 V
= 6.00 x 10^-6 x 100
= 6.00 x 10^-4 C
So the Final capacitance C2 will be
= K x C1 = 4.5 x 6.00 x 10^-6
= 2.7 x 10^ -5 F
So to get Finalcharge
We use Q2 = C2 x V
= 2.7 x 10^ - 5 x 100
= 27 x 10^ -4 C
So Charge flown in thecapacitor is change in Q
Which is = Q2 - Q1
= 27 x 10^-4 - 6.0 x 10^ -4
Change in Q = 2.1x 10^-3 C
