Should be 250m. check with your teacher and let me know
Answer:
v_2 = 2*v
Explanation:
Given:
- Mass of both charges = m
- Charge 1 = Q_1
- Speed of particle 1 = v
- Charge 2 = 4*Q_1
- Potential difference p.d = 10 V
Find:
What speed does particle #2 attain?
Solution:
- The force on a charged particle in an electric field is given by:
F = Q*V / r
Where, r is the distance from one end to another.
- The Net force acting on a charge accelerates it according to the Newton's second equation of motion:
F_net = m*a
- Equate the two expressions:
a = Q*V / m*r
- The speed of the particle in an electric field is given by third kinetic equation of motion.
v_f^2 - v_i^2 = 2*a*r
Where, v_f is the final velocity,
v_i is the initial velocity = 0
v_f^2 - 0 = 2*a*r
Substitute the expression for acceleration in equation of motion:
v_f^2 = 2*(Q*V / m*r)*r
v_f^2 = 2*Q*V / m
v_f = sqrt (2*Q*V / m)
- The velocity of first particle is v:
v = sqrt (20*Q / m)
- The velocity of second particle Q = 4Q
v_2 = sqrt (20*4*Q / m)
v_2 = 2*sqrt (20*Q / m)
v_2 = 2*v
<span>law of conservation of </span>energy<span> is </span><span><span>states that energy of the universe remains constant cant be created nor destroyed and conserving energy is not using as much power as you was like trying to make power bill lower while law of conservation is constant </span> </span>
Answer:
Flow rate 2.34 m3/s
Diameter 0.754 m
Explanation:
Assuming steady flow, the volume flow rate along the pipe will always be constant, and equals to the product of flow speed and cross-section area.
The area at the well head is
So the volume flow rate along the pipe is
We can use the similar logic to find the cross-section area at the refinery
The radius of the pipe at the refinery is:
So the diameter is twice the radius = 0.38*2 = 0.754m
