Answer:
i have no clue i just need brailnly points
Explanation:
Answer:
A) d = 11.8m
B) d = 4.293 m
Explanation:
A) We are told that the angle of incidence;θ_i = 70°.
Now, if refraction doesn't occur, the angle of the light continues to be 70° in the water relative to the normal. Thus;
tan 70° = d/4.3m
Where d is the distance from point B at which the laser beam would strike the lakebottom.
So,d = 4.3*tan70
d = 11.8m
B) Since the light is moving from air (n1=1.00) to water (n2=1.33), we can use Snell's law to find the angle of refraction(θ_r)
So,
n1*sinθ_i = n2*sinθ_r
Thus; sinθ_r = (n1*sinθ_i)/n2
sinθ_r = (1 * sin70)/1.33
sinθ_r = 0.7065
θ_r = sin^(-1)0.7065
θ_r = 44.95°
Thus; xonsidering refraction, distance from point B at which the laser beam strikes the lake-bottom is calculated from;
d = 4.3 tan44.95
d = 4.293 m
Answer:
A car engine has more power than a horse because a car engine does the same amount of work in time. Yasmin and Raj each had 10 boxes of equal weight to stack next to each other on the same shelf, at the same height and in the same arrangement. Yasmin completed the task in 2 minutes, while Raj took 3 minutes to stack his boxes. Raj applied less power than Yasmin because his stacking took more time to do the same amount of work.
Explanation:
m = mass of burrito thrown by the student = 0.5 kg
a = acceleration of the burrito thrown by the student = 3 m/s²
F = force applied by the student on the burrito = ?
According to newton's second law , the net force on an object is the product of its mass and acceleration. it is given as
F = ma
inserting the values
F = (0.5) (3)
F = 1.5 N
hence the net force on the burrito comes out to be 1.5 N
Answer:
a) The initial momentum of the ball is 22.5 kg·m/s
b) The magnitude of the momentum imparted by the ball is 30 kg·m/s
Explanation:
The question is based on change in momentum
The mass of the ball, m = 0.75 kg
The velocity with which the ball is thrown against the wall, u = 30 m/s
The time duration it takes while the ball is in contact with the wall, Δt = 0.05 sec
The velocity of the ball as it bounce back, v = -10 m/s (The ball moves in the opposite direction)
a) The initial momentum of the ball, 
= m × u = 0.75 kg × 30 m/s = 22.5 kg·m/s
b) The final momentum of the ball, 
= m × v = 0.75 kg × (-10 m/s) = -7.5 kg·m/s
The momentum imparted by the ball, ΔP = The final momentum - The initial momentum
∴ ΔP = - = -7.5 kg·m/s - 22.5 kg·m/s = -30 kg·m/s
The magnitude of the momentum imparted by the ball, 
= 30 kg·m/s
