Answer:
P = 2439.5 W = 2.439 KW
Explanation:
First, we will find the mass of the water:
Mass = (Density)(Volume)
Mass = m = (1 kg/L)(10 L)
m = 10 kg
Now, we will find the energy required to heat the water between given temperature limits:
E = mCΔT
where,
E = energy = ?
C = specific heat capacity of water = 4182 J/kg.°C
ΔT = change in temperature = 95°C - 25°C = 70°C
Therefore,
E = (10 kg)(4182 J/kg.°C)(70°C)
E = 2.927 x 10⁶ J
Now, the power required will be:
where,
t = time = (20 min)(60 s/1 min) = 1200 s
Therefore,
<u>P = 2439.5 W = 2.439 KW</u>
Answer:
change in height is 1.664 mm
Explanation:
Given data
drops = 3.00 mm
diameter = 5.00 cm = 0.05 mm
decrease = 350 cm^3
temperature = 95°C to 44.0°C
to find out
the decrease in millimeters in level
solution
we will calculate here change in volume so we can find how much level is decrease
change in volume = β v change in temp ...............1
here change in volume = area× height
so = 
/4 × d² h
so we can say change in volume = /4 × d² × change in height .......2
so from equation 1 and 2 we calculate change in height
( β(w) -β(g) )× v× change in temp = /4 × d² × change in height
change in height = 4 × ( β(w) -β(g) ) v× change in temp / /4 × d²
put all value here
change in height = 4 × ( 210 - 27 )(350 ) 
× (95-44) / /4 × 0.05²
change in height is 1.664 mm
<span>The correct answer is B - Light can travel in a vacuum, and its speed is constant if the source is moving or stationary.</span>
Answer:
77%
Explanation:
efficiency= work output/work input X 100%
e = 1,200j/ 1,550 j x100%
e = 1,200/1,550= 0.77
e = 0.77 x 100%
e = 77%
Answer:
0.015
Explanation:
Total volume of water coming out = 1.33
Also volume = Cross sectional area*Length covered
Length covered = Velocity *time
=24.5*3.55
=86.97 meter
Let the cross sectional area be A.
1.33 = 86.97*A
A =0.015
