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
Answer:
Pascal's law says that pressure applied to an enclosed fluid will be transmitted without a change in magnitude to every point of the fluid and to the walls of the container.
Explanation:
The pressure at any point in the fluid is equal in all directions.
Answer:
A divergent boundary is the answer
Explanation:
The best name for the ionic bond that forms between them is Beryllium Bromide.
We have been provided with data,
Beryllium charge, q = 2
Bromine charge, q = -1
As we know the valance electron of Be is +2 and the valance electron of bromine is -1. Since one is metallic and the other is non-metallic.
Now, when they combine they exchange valance electron, and bromine change into bromide so they form Beryllium Bromide.
So, the best name for the ionic bond that forms between them is Beryllium Bromide.
Learn more about ionic bonds here:
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Answer:
970 kN
Explanation:
The length of the block = 70 mm
The cross section of the block = 50 mm by 10 mm
The tension force applies to the 50 mm by 10 mm face, F₁ = 60 kN
The compression force applied to the 70 mm by 10 mm face, F₂ = 110 kN
By volumetric stress, we have that for there to be no change in volume, the total pressure applied by the given applied forces should be equal to the pressure removed by the added applied force
The pressure due to the force F₁ = 60 kN/(50 mm × 10 mm) = 120 MPa
The pressure due to the force F₂ = 110 kN/(70 mm × 10 mm) = 157.142857 MPa
The total pressure applied to the block, P = 120 MPa + 157.142857 MPa = 277.142857 MPa
The required force, F₃ = 277.142857 MPa × (70 mm × 50 mm) = 970 kN
