Answer:
The copper wire stretches 6.25 cm and the steel wire stretches 3.75 cm.
Explanation:
Young's modulus is defined as:
E = stress / strain
E = (F / A) / (dL / L)
E = (F L) / (A dL)
Solving for dL:
dL = (F L) / (A E)
The wires have the same force, length, and cross-sectional area. So:
dL₁ + dL₂ = (FL/A) (1/E₁ + 1/E₂)
Given that dL₁ + dL₂ = 0.10 m, E₁ = 20×10¹⁰ N/m², and E₂ = 12×10¹⁰ N/m²:
0.10 = (FL/A) (1/(20×10¹⁰) + 1/(12×10¹⁰))
FL/A = 0.75×10¹⁰ N/m
Solving for dL₁ and dL₂:
dL₁ = (FL/A) / E₁
dL₁ = (0.75×10¹⁰ N/m) / (20×10¹⁰ N/m²)
dL₁ = 0.0375 m
dL₂ = (FL/A) / E₂
dL₂ = (0.75×10¹⁰ N/m) / (12×10¹⁰ N/m²)
dL₂ = 0.0625 m
The copper wire stretches 6.25 cm and the steel wire stretches 3.75 cm.
Answer:
v/c = 0.76
Explanation:
Formula for Length contraction is given by;
L = L_o(√(1 - (v²/c²))
Where;
L is the length of the object at a moving speed v
L_o is the length of the object at rest
v is the speed of the object
c is speed of light
Now, we are given; L = 65%L_o = 0.65L_o, since L_o is the length at rest.
Thus;
0.65L_o = L_o[√(1 - (v²/c²))]
Dividing both sides by L_o gives;
0.65 = √(1 - (v²/c²))
Squaring both sides, we have;
0.65² = (1 - (v²/c²))
v²/c² = 1 - 0.65²
v²/c² = 0.5775
Taking square root of both sides gives;
v/c = 0.76
Answer:The answer is A
Explanation:
To determine the force, we use the equation
Fg= G m1 m2/r2
, where Fg is the force of gravity, G is the gravitational constant,
m1
is the mass of the first body,
m2
is the mass of the second body, and r is the radius. Hence, we have
Fg=6.67×10−11 N×m2/kg2×2.5 kg×2.5 kg/(0.50 m)2=1.7×10−9N
. Since this is a gravitational problem, the force on one body has the same force as the other body, but the directions are opposite.
Answer:
1.To get your team to 25 points each period by serving, spiking, and setting the ball
2. A gymnasium with volleyball court markings, a net, a volleyball
Answer:
Acceleration, 
Explanation:
We are given with a velocity-time graph of an object that starts from origin. y axis is velocity and x-axis is time.
The slope of velocity-time graph gives acceleration of an object. Taking the slope of this graph.
So, the acceleration of the object is 
.
