mass of the space probe is given as
m = 1700 lbm = 771.12 kg
each thrust will apply a net force
now we will have
given that we have final speed v = 420 miles/minute
we need to convert it into m/s
now from above equation
<em>so it will require 63.8 weeks to reach the given speed</em>
Answer:
1.75 kW
Explanation:
First, the rate at which work is done is simply termed Power
m = 35kg
v = 5.0 m/s
since the undergoes a vertical motion g = 10 m/s²
F = W = mg = 35×10 = 350 N
Power = Fv = 350×5 = 1750 Watt
BLANK 1: A) Not balanced
BLANK 2: B) Sodium
BLANK 3: B) Not Equal
Answer:
0.00225 N/m
Explanation:
Parameters given:
Current in first wire, I(1) = 15A
Current in second wire, I(2) = 15A
Distance between two wires, R = 1cm = 0.01m
The force per unit length between two current carrying wires is:
F/L = μ₀I(1)I(2)/2πR
μ₀ = 4π * 10^(-7) Tm/A
F/L = [4π * 10^(-7) * 15 * 15] / (2π * 0.01)
F/L = 2.25 * 10^(-3) N/m or 0.00225 * 10^(-3) N/m
Answer:
(a) (18m/s/t₁)m/s²
(b) -12.5m/s²
(c) -20mls²
Explanation:
(a) Let t₁ be the initial time
a = v-u/t
acc = (18m/s/t₁)m/s²
(b) acc = -30m/s/2.4
= -12.5m/s²
(c)The particle was at a speed of 18m/s in the positive x-direction and later after 2.4s ≡Δt, it was at speed of -30m/s in the negative x-direction.
so this imply that the velocity was first v₁ =18m/s and later v₂ = -30m/s.
The average acceleration is then:
Aavg =<u> Δv</u>
Δt
= v₂-v₁/Δt
= -30-18/2.4 = -20mls²