The final temperature of the seawater-deck system is 990°C.
<h3>What is heat?</h3>
The increment in temperature adds up the thermal energy into the object. This energy is Heat energy.
The deck of a small ship reaches a temperature Ti= 48.17°C seawater on the deck to cool it down. During the cooling, heat Q =3,710,000 J are transferred to the seawater from the deck. Specific heat of seawater= 3,930 J/kg°C.
Suppose for 1 kg of sea water, the heat transferred from the system is given by
3,710,000 = 1 x 3,930 x (T - 48.17)
T = 990°C to the nearest tenth.
The final temperature of the seawater-deck system is 990°C.
Learn more about heat.
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Answer:
E1 = 2996.667N/C E2 = 11237.5N/C
Explanation:
E1 = kQ1/r^2
=8.99 x 10^9 x 30 x 10^-9/(30x10^-2)^2
= 2996.667N/C
E2 = kQ2/r^2
= 8.99 x 10^9 x 50 x 10^-9/(20x10^-2)^2
= 11237.5N/C
The direction are towards the point a
<span>The correct option is D. Soil can best be described as the loose covering of weathered rocks and decaying organic matter. There are different types of soil, the type of soil formed depends majorly on the type of parent rock from which the soil is formed and the amount of organic matter present in the soil.</span>
<span>364N should be your answer.. hope this helps
</span>
Answer:
<em>155.80rad/s</em>
Explanation:
Using the equation of motion to find the angular acceleration:
is the final angular velocity in rad/s
is the initial angular velocity in rad/s
is the angular acceleration
t is the time taken
Given the following
Time = 4.1secs
Convert the angular velocity to rad/s
1rpm = 0.10472rad/s
6100rpm = x
x = 6100 * 0.10472
x = 638.792rad/s
Get the angular acceleration:
Recall that:
638.792 = 0 + ∝(4.1)
4.1∝ = 638.792
∝ = 638.792/4.1
∝ = 155.80rad/s
<em>Hence the angular acceleration as the blades slow down is 155.80rad/s</em>
