Answer:
No force acts on it. It will not be pushed.
Explanation:
The proton moving in the -X direction implies its velocity is in the -X direction and the magnetic field is given to be in the +X direction. So the angle between them is 180 degrees. So no force acts on the proton, since the velocity and magnetic field are anti parallel.
Magnetic force is given by the equation F = q v B sin theta where theta is the angle between the velocity and the magnetic field. This force will be a maximum if they are perpendicular to each other.
Explanation:
Elongation of the wire is:
ΔL = F L₀ / (E A)
where F is the force,
L₀ is the initial length,
E is Young's modulus,
and A is the cross sectional area.
ΔL = T (0.5 m) / ((2.0×10¹¹ Pa) (0.02 cm²) (1 m / 100 cm)²)
ΔL = T (1.25×10⁻⁶ m/N)
T = (80,000 N/m) ΔL
Draw a free body diagram of the mass at the bottom of the circle. There are two forces: tension force T pulling up and weight force mg pulling down.
Sum of forces in the centripetal direction:
∑F = ma
T − mg = mv²/r
T − mg = mω²r
T − (15 kg) (9.8 m/s²) = (15 kg) (2 rev/s × 2π rad/rev)² (0.5 m + ΔL)
T − 147 N = (2368.7 N/m) (0.5 m + ΔL)
Substitute:
(80,000 N/m) ΔL − 147 N = (2368.7 N/m) (0.5 m + ΔL)
(80,000 N/m) ΔL − 147 N = 1184.35 N + (2368.7 N/m) ΔL
(797631.3 N/m) ΔL = 1331.35 N
ΔL = 0.00167 m
ΔL = 1.67 mm
Answer:
va = 4.79 m/s
vb = 1.29 m/s
Explanation:
Momentum is conserved:
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
(3.00) (0) + (6.50) (3.50) = (3.00) v₁ + (6.50) v₂
22.75 = 3v₁ + 6.5v₂
For an elastic collision, kinetic energy is conserved.
½ m₁u₁² + ½ m₂u₂² = ½ m₁v₁² + ½ m₂v₂²
m₁u₁² + m₂u₂² = m₁v₁² + m₂v₂²
(3.00) (0)² + (6.50) (3.50)² = (3.00) v₁² + (6.50) v₂²
79.625 = 3v₁² + 6.5v₂²
Two equations, two variables. Solve with substitution:
22.75 = 3v₁ + 6.5v₂
22.75 − 3v₁ = 6.5v₂
v₂ = (22.75 − 3v₁) / 6.5
79.625 = 3v₁² + 6.5v₂²
79.625 = 3v₁² + 6.5 ((22.75 − 3v₁) / 6.5)²
79.625 = 3v₁² + (22.75 − 3v₁)² / 6.5
517.5625 = 19.5v₁² + (22.75 − 3v₁)²
517.5625 = 19.5v₁² + 517.5625 − 136.5v₁ + 9v₁²
0 = 28.5v₁² − 136.5v₁
0 = v₁ (28.5v₁ − 136.5)
v₁ = 0 or 4.79
We know v₁ isn't 0, so v₁ = 4.79 m/s.
Solving for v₂:
v₂ = (22.75 − 3v₁) / 6.5
v₂ = 1.29 m/s
The equal velocity approach for duct size assumes that the air velocity in each duct segment is the same.
How fast is the air moving through a duct?
The most common unit of air velocity (distance traveled in a unit of time) is feet per minute (FPM). The amount of air passing past a location in the duct per period of time may be calculated by multiplying the airflow by the area of the duct. The standard unit for volume flow is cubic feet per minute (CFM).
What happens when the size of ducts changes to the airflow?
- Result for an image The equal velocity technique for duct size makes the assumption that air velocity is constant across the entire duct system.
- The main lesson to be learned from this is that when air goes from a bigger to a narrower duct, its velocity rises. The velocity drops when it transitions from a shorter to a bigger duct. The flow rate or the amount of air passing through the duct in cubic feet per minute is the same in all scenarios.
Learn more about air velocity here:
brainly.com/question/3255148
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