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 .
The formula for this problem that we will be using is:
F * cos α = m * g * μs where:F = 800m = 87g = 9.8
cos α = m*g*μs/F= 87*9.8*0.55/800= 0.59 So solving the alpha, find the arccos above.
α = arccos 0.59 = 54 ° is the largest value of alpha
Density is given by:
D = M/V
D = density, M = mass, V = volume
Given values:
M = 3.7g, V = 4.6cm³
Plug in and solve for D:
D = 3.7/4.6
D = 0.80g/cm³
Your answer will be (B) - intense pressure.
Answer: current I = 0.96 Ampere
Explanation:
Given that the
Resistance R = 60 Ω
Power = 55 W
Power is the product of current and voltage. That is
P = IV ...... (1)
But voltage V = IR. From ohms law.
Substitutes V in equation (1) power is now
P = I^2R
Substitute the above parameters into the formula to get current I
55 = 60 × I^2
Make I^2 the subject of formula
I^2 = 55/60
I^2 = 0.92
I = sqr(0.92)
I = 0.957 A
Therefore, 0.96 A current must be applied.
