Answer:
I know I am a very good answerable teacher but I can't answer this question I don't know what
Answer:Force on -7 uC charge due to charge placed at x = - 10cm
now we will have
towards left
similarly force due to -5 uC charge placed at x = 6 cm
now we will have
towards left
Now net force on 7 uC charge is given as
towards left
Explanation:
<u>Complete Question:</u>
A hockey player swings her hockey stick and strikes a puck. According to Newton’s third law of motion, which of the following is a reaction to the stick pushing on the puck?
A. the puck pushing on the stick
.
B. the stick pushing on the player
.
C. the player pushing on the stick
.
D. the puck pushing on the player.
<u>Correct Option:</u>
According to Newton’s third law of motion the puck pushing on the stick is a reaction to the stick pushing on the puck.
<u>Option: A</u>
<u>Explanation:</u>
As when the hockey exert force on the puck (which is a flat ball basically used in ice hockey) then this action by hockey will receive equal and opposite reaction by puck. Thus when the stick is pushing on the this flat ball, then puck also push the stick. This is understood by newton's third law pf motion, where action and reaction forces are subject of discussion, displaying their is pair of forces applied among the interacting objects. This form is observed more practically in life and very frequent.
Answer:
cross out the false piece in blue and write the true piece in red
Answer: 0m/s²
Explanation:
Since the forces acting along the plane are frictional force(Ff) and moving force(Fm), we will take the sum of the forces along the plane
According newton's law of motion
Summation of forces along the plane = mass × acceleration
Frictional force is always acting upwards the plane since the body will always tends to slide downwards on an inclined plane and the moving acts down the plane
Ff = nR where
n is coefficient of friction = tan(theta)
R is normal reaction = Wcos(theta)
Fm = Wsin(theta)
Substituting in the formula of newton's first law we have;
Fm-Ff = ma
Wsin(theta) - nR = ma
Wsin(theta) - n(Wcos(theta)) = ma... 1
Given
W = 562N, theta = 30°, n = tan30°, m = 56.2kg
Substituting in eqn 1,
562sin30° - tan30°(562cos30°) = 56.2a
281 - 281 = 56.2a
0 = 56.2a
a = 0m/s²
This shows that the trunk is not accelerating
