Are there conditions under which the incident light ray undergoes reflection but not transmission at the boundary? if so, then w
1 answer:
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Answer:
THE RUBBER BALL
Explanation:
From the question we are told that
The mass of the rubber ball is
The initial speed of the rubber ball is
The final speed at which it bounces bank
The mass of the clay ball is
The initial speed of the clay ball is
The final speed of the clay ball is
Generally Impulse is mathematically represented as
where is the change in the linear momentum so
For the rubber is
=>
For the clay ball
=>
So from the above calculation the ball with the a higher magnitude of impulse is the rubber ball
Answer:
In m/s^2:
a=11.3778 m/s^2
In units of g:
a=1.161 g
Explanation:
Since the racing greyhounds are capable of rounding corners at very high speed so we are going use the following formula of acceleration for circular paths.
where:
v is the speed
r is the radius
Now,
In g units:
Answer:
He should run at least at 1.5 m/s
The diver will enter the water at an angle of 87° below the horizontal.
Explanation:
Hi there!
The position and velocity of the diver are given by the following vectors:
r = (x0 + v0x · t, y0 + v0y · t + 1/2 · g · t²)
v = (v0x, v0y + g · t)
Where:
r = position vector at time t
x0 = initial horizontal position
v0x = initial horizontal velocity
t = time
y0 = initial vertical position
v0y = initial vertical velocity
g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive)
v = velocity vector at time t
Please, see the attached figure for a description of the problem. Notice that the origin of the frame of reference is located at the jumping point so that x0 and y0 = 0.
We know that, to clear the rocks, the position vector r final (see figure) should be:
r final = ( > 5.0 m, -50 m)
So let´s find first at which time the y-component of the vector r final is - 50 m:
y = y0 + v0y · t + 1/2 · g · t²
-50 m = 2.1 m/s · t - 1/2 · 9.8 m/s² · t²
0 = -4.9 m/s² · t² + 2.1 m/s · t + 50 m
Solving the quadratic equation
t = 3.4 s
Now, we can calculate the initial horizontal velocity using the equation of the x-component of the position vector knowing that at t =3.4 the horizontal component should be greater than 5.0 m:
x = x0 + v0x · t (x0 = 0)
5.0 m < v0x · 3.4 s
v0x > 5.0 m / 3.4 s
v0x > 1.5 m/s
The initial horizontal velocity should be greater than 1.5 m/s
To find the angle at which the diver enters the water, we have to find the magnitude of the final velocity (vector vf in the figure). We already know the magnitude of the x-component of the vector vf, since the horizontal velocity is constant. So:
vfx > 1.5 m/s
Now, let´s calculate vfy:
vfy = v0y + g · t
vfy = 2.1 m/s - 9.8 m/s² · 3.4 s
vfy = -31 m/s
Let´s calculate the minimum magnitude that the final velocity will have if the diver safely clears the rocks. Let´s consider the smallest value allowed for vfx: 1.5 m/s. Then:
Then the final velocity of the diver will be greater or equal than 31 m/s.
To find the angle, we have to use trigonometry. Notice in the figure that the vectors vf, vfx and vy form a right triangle in which vf is the hypotenuse, vfx is the adjacent side and vfy is the opposite side to the angle. Then:
cos θ = adjacent / hypotenuse = vfx / vf = 1.5 m/s / 31 m/s
θ = 87°
The diver will enter the water at an angle of 87° below the horizontal.
Answer:
A. 0.199 J
B. 0.0663 C
C = 0.0221 F
D. 12.68 ohms
Explanation:
From the question:
time duration, t = 0.28 seconds
Average power, P = 0.71 W
Average voltage, V = 3 V
A) Energy is given as:
E = P * t
=> E = 0.71 * 0.28 = 0.199 J
B) Electrical energy is also given as:
E = qV
where q = charge
=> q = E / V
∴ q = 0.199 / 3 = 0.0663 C
C) Capacitance is given as charge over voltage:
C = q / V
=> C = 0.0663 / 3 = 0.0221 F
D) Electrical power, P, can also be given as:
P =
where R = resistance
=> R =
R =