It is 1,000kg/m³ i hope this helps you with something
Answer:
Q=1670J
Explanation:
Mass of ice: m=5g=0.005kg
Latent heat: lambda=3.34×10⁵J/kg
Heat received by ice: Q=m×lambda
Q=0.005×3.34×10⁵=5×334=1670J
Answer:
The total work done by the two tugboats on the supertanker is 3.44 *10^9 J
Explanation:
The force by the tugboats acting on the supertanker is constant and the displacement of the supertanker is along a straight line.
The angle between the 2 forces and displacement is ∅ = 15°.
First we have to calculate the work done by the individual force and then we can calculate the total work.
The work done on a particle by a constant force F during a straight line displacement s is given by following formula:
W = F*s
W = F*s*cos∅
With ∅ = the angles between F and s
The magnitude of the force acting on the supertanker is F of tugboat1 = F of tugboat 2 = F = 2.2 * 10^6 N
The total work done can be calculated as followed:
Wtotal = Ftugboat1 s * cos ∅1 + Ftugboat2 s* cos ∅2
Wtotal = 2Fs*cos∅
Wtotal = 2*2.2*10^6 N * 0.81 *10³ m s *cos15°
Wtotal = 3.44*10^9 Nm = <u>3.44 *10^9 J</u>
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The total work done by the two tugboats on the supertanker is 3.44 *10^9 J
Answer:
Explanation:
Let fuel is released at the rate of dm / dt where m is mass of the fuel
thrust created on rocket
= d ( mv ) / dt
= v dm / dt
this is equal to force created on the rocket
= 220 dv / dt
so applying newton's law
v dm / dt = 220 dv / dt
v dm = 220 dv
dv / v = dm / 220
integrating on both sides
∫ dv / v = ∫ dm / 220
lnv = ( m₂ - m₁ ) / 220
ln4000 - ln 2500 = ( m₂ - m₁ ) / 220
( m₂ - m₁ ) = 220 x ( ln4000 - ln 2500 )
( m₂ - m₁ ) = 220 x ( 8.29 - 7.82 )
= 103.4 kg .