Answer:
the gauge pressure at the upper face of the block is 116 Pa
Explanation:
Given the data in the question;
A cubical block of wood, 10.0 cm on a side.
height h = 1.50 cm = ( 1.50 × ( 1 / 100 ) ) m = 0.0150 m
density ρ = 790 kg/m³
Using expression for the gauged pressure;
p-p₀ = ρgh
where, p₀ is atmospheric pressure, ρ is the density of the substance, g is acceleration due to gravity and h is the depth of the fluid.
we know that, acceleration due to gravity g = 9.8 m/s²
so we substitute
p-p₀ = 790 kg/m³gh × 9.8 m/s² × 0.0150 m
= 116.13 ≈ 116 Pa
Therefore, the gauge pressure at the upper face of the block is 116 Pa
Period, T, is 1/frequency
T = 1/340 = 2.941ms
Answer:
I = 0.5 A
Explanation:
Given: P=60 Watts, Voltage supply V = 120 Volts (for primary coil)
Solution:
we have P = V I
⇒ I = P /V = 60 Watts / 120 Volts
I = 0.5 A
Answer:
so 9/3=3 current is 3 amperes
Explanation:
Answer:
F = 1500 [N]
Explanation:
To solve this problem we must use Newton's second law, which tells us that the sum of all forces must be equal to the product of mass by acceleration.
ΣF = m*a
F = 1000*1.5
F = 1500 [N]
