Answer:
270Joues
Explanation:
Step one:
given data
Force F= 45N
distance moved d= 6m
Required
The work done in moving the block 6m
Step two:
We know that the expression for the work done is
WD= force* distance
WD= 45*6
WD=270Joues
Answer:
87.1 mph
Explanation:
We are given that
Mass,m=60 kg
Power,P=340 W
Speed,v=5 m/s
Area,
Drag coefficient,
Coefficient of rolling resistance,
Friction force,
Where 
Let speed of cyclist=v'
Drag force,
Density of air,
Power,P=
1 m=0.00062137 miles
1 hour=3600 s
1. no forms of electromagnetic radiation
2. atoms
3. liquid
4. protons
5. gas
6. electrons
7. solid
8. alpha particle
Answer:
16.96 W
Explanation:
Power: This can be defined as the rate at which work is done by an object. The S.I unit of power is Watt(W).
From the question,
P = (F×d)/t....................... Equation 1
Where P = power, F = force, d = distance, t = time.
Given: F = 75 N, d = 42 m, t = 3.1 min = 3.1×60 = 186 s
Substitute these values into equation 1
P = (75×42)/186
P = 16.94 W
Hence the average power delivered by the child = 16.96 W
Answer:
6400 m
Explanation:
You need to use the bulk modulus, K:
K = ρ dP/dρ
where ρ is density and P is pressure
Since ρ is changing by very little, we can say:
K ≈ ρ ΔP/Δρ
Therefore, solving for ΔP:
ΔP = K Δρ / ρ
We can calculate K from Young's modulus (E) and Poisson's ratio (ν):
K = E / (3 (1 - 2ν))
Substituting:
ΔP = E / (3 (1 - 2ν)) (Δρ / ρ)
Before compression:
ρ = m / V
After compression:
ρ+Δρ = m / (V - 0.001 V)
ρ+Δρ = m / (0.999 V)
ρ+Δρ = ρ / 0.999
1 + (Δρ/ρ) = 1 / 0.999
Δρ/ρ = (1 / 0.999) - 1
Δρ/ρ = 0.001 / 0.999
Given:
E = 69 GPa = 69×10⁹ Pa
ν = 0.32
ΔP = 69×10⁹ Pa / (3 (1 - 2×0.32)) (0.001/0.999)
ΔP = 64.0×10⁶ Pa
If we assume seawater density is constant at 1027 kg/m³, then:
ρgh = P
(1027 kg/m³) (9.81 m/s²) h = 64.0×10⁶ Pa
h = 6350 m
Rounded to two sig-figs, the ocean depth at which the sphere's volume is reduced by 0.10% is approximately 6400 m.
