Answer:
250 N
433 N
Explanation:
N = Normal force by the surface of the inclined plane
W = Weight of the block = 500 N
f = static frictional force acting on the block
Parallel to incline, force equation is given as
f = W Sin30
f = (500) Sin30
f = 250 N
Perpendicular to incline force equation is given
N = W Cos30
N = (500) Cos30
N = 433 N
Answer:
0
Explanation:
F1 = G•2.2•4.66/3² (pointed right)
F2 = G•2.2•4.66/3² (pointed left)
subtract the two to get zero
Answer:
At time 10.28 s after A is fired bullet B passes A.
Passing of B occurs at 4108.31 height.
Explanation:
Let h be the height at which this occurs and t be the time after second bullet fires.
Distance traveled by first bullet can be calculated using equation of motion

Here s = h,u = 450m/s a = -g and t = t+3
Substituting

Distance traveled by second bullet
Here s = h,u = 600m/s a = -g and t = t
Substituting

Solving both equations

So at time 10.28 s after A is fired bullet B passes A.
Height at t = 7.28 s

Passing of B occurs at 4108.31 height.
Answer:
The size of the force that pushes the wall is 12,250 N.
Explanation:
Given;
mass of the wrecking ball, m = 1500 kg
speed of the wrecking ball, v = 3.5 m/s
distance the ball moved the wall, d = 75 cm = 0.75 m
Apply the principle of work-energy theorem;
Kinetic energy of the wrecking ball = work done by the ball on the wall
¹/₂mv² = F x d
where;
F is the size of the force that pushes the wall
¹/₂mv² = F x d
¹/₂ x 1500 x 3.5² = F x 0.75
9187.5 = 0.75F
F = 9187.5 / 0.75
F = 12,250 N
Therefore, the size of the force that pushes the wall is 12,250 N.
This question is incomplete, the complete question is;
Two plane mirrors intersect at right angles. A laser beam strikes the first of them at a point 11.5 cm from their point of intersection, as shown in the figure.
For what angle of incidence at the first mirror will this ray strike the midpoint of the second mirror (which is 28.0 cmcm long) after reflecting from the first mirror
Answer: angle of incidence is 39.4°
Explanation:
Given that;
two plain mirrors intersect at right angle (90°)
distance d = 11.5 cm
S = 28.0 cm
Now the angle that the reflection ray males with first the mirror equal theta (∅)
so
tan∅ = (S/2) / d
tan∅ = (28/2) / 11.5
tan∅ = 14 / 11.5
tan∅ = 1.2173
∅ = tan⁻¹ (1.2173)
∅ = 50.6°
so angle of incidence = 90° - ∅
= 90° - 50.6°
= 39.4°
Therefore angle of incidence is 39.4°