<span>I believe this question has additional detail which stated
that during the 1st half, his speed was 2.01 m/s. From this we can
calculate his speed during the second half, v2, using the formula:</span>
v_ave = (v1 + v2) / 2
2.05 m/s = (2.01 m/s + v2) / 2
<span>v2 = 2.09 m/s</span>
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:
4.18
Explanation:
Givens
The car's initial velocity 
= 0 and covering a distance Δx = 1/4 mi = 402.336 m in a time interval t = 4.43 s.
Knowns
We know that the maximum static friction force is given by:
μ_s*n (1)
Where μ_s is the coefficient of static friction and n is the normal force.
Calculations
(a) First, we calculate the acceleration needed to achieve this goal by substituting the given values into a proper kinematic equation as follows:
Δx=
a=41 m/s
This is the acceleration provided by the engine. Applying Newton's second law on the car, so in equilibrium, when the car is about to move, we find that:
Substituting (3) into (1), we get:
μ_s*m*g
Equating this equation with (4), we get:
ma= μ_s*m*g
μ_s=a/g
=4.18
Answer:
1. 150C.
2. 50sec
3.1.5a
Explanation:
1. I = Q/T
Q= 30x5
=150c
2.applying the formulae, I = Q/T
T= Q/I
=500/10
=50sec.
3. using the formulae i=q/t
i= 120/80
=1.5a.
Answer:
v = 31.84 cm/s or 0.318 m/s
the speed of the water leaving the end of the hose is 31.84 cm/s or 0.318 m/s
Explanation:
Given;
Diameter of hose d = 2.76 cm
Volume filled V = 20.0 L = 20,000 cm^3
Time t = 1.45 min = 105 seconds
The volumetric flow rate of water is;
F = V/t = 20,000cm^3 ÷ 105 seconds
F = 190.48 cm^3/s
The volumetric flow rate is equal the cross sectional area of pipe multiply by the speed of flow.
F = Av
v = F/A
Area A = πd^2/4
Speed v = F/(πd^2/4)
v = 4F/πd^2 ......1
Substituting the given values;
v = (4×190.48)/(π×2.76^2)
v = 31.83767439628 cm/s
v = 31.84 cm/s or 0.318 m/s
the speed of the water leaving the end of the hose is 31.84 cm/s or 0.318 m/s
