Answer:
1.635×10^-3m
Explanation:
Young modulus is the ratio of the tensile stress of a material to its tensile strain.
Young modulus = Tensile stress/tensile strain
Tensile stress = Force/Area
Given force = 130N
Area = Πr² = Π×(1.55×10^-3)²
Area = 4.87×10^-6m²
Tensile stress = 130/4.87×10^-6 = 8.39×10^7N/m²
Tensile strain = extension/original length
Tensile strain = e/3.9
Substituting in the young modulus formula given young modulus to be 2×10¹¹N/m²
2×10¹¹N/m² = 8.39×10^7/{e/3.9)}
2×10¹¹ = (8.39×10^7×3.9)/e
2×10¹¹e = 3.27×10^8
e = 3.27×10^8/2×10¹¹
e = 1.635×10^-3m
The stretch of the steel wire will be
1.635×10^-3m
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
Answer:
<h3>1.43m/s²</h3>
Explanation:
According to newtons second law.
F = mass * acceleration
If the doll has a mass of 0.2 kg, and the robot has a mass of 0.5 kg, the resulting mass will be 0.7kg
Force applied = 1N
acceleration = Force/mass
Substitute the values and get acceleration
acceleration = 1/0.7
acceleration = 1.43m/s²
Hence the magnitude of the acceleration of the robot is 1.43m/s²
Answer:
Explanation:
It is given that,
A planar electromagnetic wave is propagating in the +x direction.The electric field at a certain point is, E = 0.082 V/m
We need to find the magnetic vector of the wave at the point P at that instant.
The relation between electric field and magnetic field is given by :
c is speed of light
B is magnetic field
So, the magnetic vector at point P at that instant is 
.
Hey there,
<span>They interfere essentially like any other form of wave.
</span>
Hope this helps :))
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