Answer:
Explanation:
See the attached figure . See the forces acting on man pulling up the box .
Man is stationary so net force acting on man is zero .
T + R = Wman
R is the reaction force of the ground of second floor .
R = Wman - T
Answer:
If an object has a high density then the molicules making up that object are closly packed togeather. Because of this, objects with a higher density will have more mass than objects of the same size that have a lesser density.
Answer:
Time, t = 13.34 seconds.
Explanation:
Given the following data;
Initial velocity, u = 85km/hr to meters per seconds = 85*1000/3600 = 23.61 m/s
Final velocity, v = 45km/hr to meters per seconds = 45*1000/3600 = 12.5 m/s
Acceleration, a = -3 km/hr/sec to meters per seconds square = -3*1000/3600 = -0.833m/s²
To find the time;
Acceleration = (v - u)/t
-0.833 = (12.5 - 23.61)/t
-0.833t = -11.11
t = 11.11/0.833
Time, t = 13.34 seconds.
Answer:
b. The internal resistance must be much smaller than the other resistances in the circuit.
Explanation:
Ammeter is used to measure the current flowing through a circuit. It is connected in series configuration with the load. In such a scenario the resistance of the ammeter should be negligible so as to make sure that the voltage drop across the resistance of ammeter is zero and it shows the correct reading of the current in the circuit.
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.