a ray of light incident on a mirror, at an angle of 45°. Another mirror is placed at an angle of 45° to the first ones as shown.
1 answer:
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1. A-20 km south east
The car's displacement consists of two components into two different directions. Using a system of coordinates in which x represents the east direction and y represents the south direction, the two displacements are:
east
south
Since the two components are orthogonal to each other, we can find the resultant displacement by using Pythagorean's theorem:
and the direction is between the two original directions, so south-east.
2. D. 10 m/s
First of all, we need to calculate the total time the stone took to hit the ground. Since the vertical distance covered is S = 78.4 m, and since the motion is an accelerated motion with constant acceleration g=9.8 m/s^2, we have
From which we find the total time of the fall, t:
Now we can consider the horizontal motion of the stone: we know that the stone travels for d = 40 m in a time of t = 4 s, therefore the horizontal velocity of the stone is
3. B=32.32 m
As in the previous problem, we have to calculate the total time it takes for the stone to reach the river first. Since the vertical distance covered is S = 20 m, we have
And since the stone is traveling horizontally at v = 16 m/s, the horizontal distance covered is
So, the closest answer is B.
Answer:
T = 0.017s
Explanation:
period is the time it takes a particle to make one oscillation
An electric current is periodic in nature
The current reaches 3.8A ten times.
So there must have been 10 cycles (10 periods) in 0.17s. let 'T' be the period:
t is the total time interval
n is the number of oscillations
10T = 0.17
T = 0.17/10 = 0.017s
Answer:
a = 1 m/s² and
Explanation:
The first two parts can be seen in attachment
We use Newton's second law on each axis
Y axis
Ty - W = 0
Ty = w
X axis
Tx = m a
With trigonometry we find the components of tension
Sin θ = Ty / T
Ty = T sin θ
Cos θ = Tx / T
Tx = T cos θ
We calculate the acceleration with kinematics
Vf = Vo + a t
a = (Vf -Vo) / t
a = (20 -10) / 10
a = 1 m/s²
We substitute in Newton's equations
T Sin θ = mg
T cos θ = ma
We divide the two equations
Tan θ = g / a
θ = tan⁻¹ (g / a)
θ = tan⁻¹ (9.8 / 1)
θ = 84º
We see that in the expression of the angle the mass does not appear therefore you should not change the angle
