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Answer:
The force exerted by the wood on the bullet is 399.01 N
Explanation:
Given;
mass of bullet, m = 0.0021 kg
initial velocity of the bullet, u = 497 m/s
final velocity of the bullet, v = 0
distance traveled by the bullet, S = 0.65 m
Determine the acceleration of the bullet which is the deceleration.
Apply kinematic equation;
V² = U² + 2aS
0 = 497² - (2 x 0.65)a
0 = 247009 - 1.3a
1.3a = 247009
a = 247009 / 1.3
a = 190006.92 m/s²
Finally, apply Newton's second law of motion to determine the force exerted by the wood on the bullet;
F = ma
F = 0.0021 x 190006.92
F = 399.01 N
Therefore, the force exerted by the wood on the bullet is 399.01 N
Answer:
E = {(Charge Density/2e0)*(1 - [z/(sqrt(z^2 - R^2))]}
R is radius = Diameter/2 = 0.210m.
At z = 0.2m,
Put z = 0.2m, and charge density = 2.92 x 10^-2C/m2, and constant value e0 in the equation,
E can be calculated at distance 0.2m away from the centre of the disk.
Put z = 0.3m and all other values in the equation,
E can be calculated at distance 0.3m away from the centre of the disk
6. 0N. This questions requires understanding of how friction functions. Friction is a resistive force, meaning it opposes the direction of any applied or unbalanced forces. The box in the question experiences no horizontal force, so there is no resistive force in response to it, making it 0N.
7. This question tests your understanding of static friction. Static friction only applies when an applied or unbalanced force is applied to an object which does not move. The static friction always equals the magnitude of the applied or unbalanced force component parallel to the surface which the object rests on. The question states that the crate starts to move only when the applied force exceeds 313N, so we use this value to determine the force of static friction. The additional info in the question pertaining to when the crate is moving is irrelevant when determining static friction (only relevant if determining kinetic friction). Knowing this we solve for the weight of the crate:
F = mg
F = (45)(9.8)
F = 441N = Normal Force
The weight of the crate is also equal to the Normal Force since the object rests on a horizontal surface and the applied force is horizontal as well. In this question, since the object is not moving at 313N of applied force, the magnitude of static friction equals the applied force:
Ff = μs * Fn
(313) = μs (441)
0.71 (rounded) = μs
The portion of the flux leaves the curved surface of the cylinder is 60%.
<h3 /><h3>What are electrons?</h3>
The electrons are the spinning objects around the nucleus of the atom of the element in an orbit.
If a point charge is located at the center of a cylinder and the electric flux leaving one end of the cylinder is 20% of the total flux leaving the cylinder.
If 20% of the flux leave from one end, then another 20% will leave from another end.
So, the net flux through curved surface is
100 -20 -20 = 60%
Thus, the total flux leaves the curved surface of the cylinder is 60%
Learn more about electrons.
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