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Answer:
a)V= 0.0827 m³
b)P=181.11 x 10² N/m²
Explanation:
Given that
m = 81.5 kg
Density ,ρ = 985 kg/m³
As we know that
Mass = Volume x Density
81.5 = V x 985
V= 0.0827 m³
The force exerted by weight = m g
F= m g= 81.5 x 10 = 815 N ( Take ,g= 10 m/s²)
Area ,A= 4.5 x 10⁻² m²
The Pressure P
P=181.11 x 10² N/m²
Answer:
The International Space Station move at 7.22 km/s.
Explanation:
Orbital speed of satellite is given by , where G is gravitational constant, M is mass of Earth and r is the distance to satellite from centre of Earth.
r = R + h = 6350 + 1400 = 7750 km = 7.75 x 10⁶ m
G = 6.673 x 10⁻¹¹ Nm²/kg²
M = 5.98 x 10²⁴ kg
Substituting
The International Space Station move at 7.22 km/s.
Answer:
(a) 1.939 m/h
(b) 0.926 m/h
(c) -0.315 m/h
(d) -1.21 m/h
Explanation:
Here, we have the water depth given by the function of time;
D(t) = 7 + 5·cos[0.503(t-6.75)]
Therefore, to find the velocity of the depth displacement with time, we differentiate the given expression with respect to time as follows;
D'(t) =
= 5×(-sin(0.503(t-6.75))×0.503
= -2.515×(-sin(0.503(t-6.75))
= -2.515×(-sin(0.503×t-3.395))
Therefore we have;
(a) at 5:00 AM = 5 - 0:00 = 5
D'(5) = -2.515×(-sin(0.503×5-3.395)) = 1.939 m/h
(b) at 6:00 AM = 6 - 0:00 = 6
D'(5) = -2.515×(-sin(0.503×6-3.395)) = 0.926 m/h
(c) at 7:00 AM = 7 - 0:00 = 7
D'(5) = -2.515×(-sin(0.503×7-3.395)) = -0.315 m/h
(d) at Noon 12:00 PM = 12 - 0:00 = 12
D'(5) = -2.515×(-sin(0.503×12-3.395)) = -1.21 m/h.