<span>11.823 cm
There is a slight ambiguity with this question in that I don't know if the measurements are from the surface of the ball, or the center of the ball. I will take this question literally and as such the point light source will be 124 cm from the wall.
The key thing to remember is that ball won't be showing an effective diameter of 4 cm to the light source. Instead the shadow line is a tangent to the ball's surface. There is a right triangle where the hypotenuse is the distance from the center of the ball to the light source (42 cm), one leg of the triangle is the radius (2cm). That right triangle will define a chord that will be the effective diameter of the disk casting the shadow. The cosine of the half angle of the chord will be 2/42 = 1/21. The sine of the half angle then becomes sqrt(1-(1/21)^2) = sqrt(440/441) = 2sqrt(110) = 0.99886557. Now multiply that sine by 4 (radius of ball multiplied by 2 since it's the half angle and we want the full side of the chord) and we get an effective diameter of 3.995462279 cm.
Now we need to calculate the effective distance that circle is from the wall. It will be slightly larger than 82 cm. The exact value will be 82 + cos(half angle) * radius. So
82 + 1/21 * 2 = 82 + 2/21 = 82.0952381
Now we have the following dimensions with a circle replacing the ball in the original problem.
Distance from wall to effective circle = 82.0952381 cm
Distance from effective circle to point source = 124 - 82.0952381 = 41.9047619 cm
Effective diameter of circle = 3.995462279 cm
And because the geometry makes similar triangles, the following ratio applies.
3.995462279/41.9047619 = X/124
Now solve for X
3.995462279/41.9047619 = X/124
124*3.995462279/41.9047619 = X
495.4373226/41.9047619 = X
11.82293611 = X
The shadow cast on the wall will be a circle with a diameter of 11.823 cm</span>
Answer:
When the ball hits the ground, the velocity will be -34 m/s.
Explanation:
The height and velocity of the ball at any time can be calculated using the following equations:
y = y0 + v0 · t + 1/2 · g · t²
v = v0 + g · t
Where:
y = height of the ball at time "t".
y0 = initial height.
v0 = initial velocity.
t = time.
g = acceleration due to gravity. (-9.8 m/s² considering the upward direction as positive).
v = velocity at time "t".
If we place the origin of the frame of reference on the ground, when the ball hits the ground its height will be 0. Then using the equation of height, we can calculate the time it takes the ball to reach the ground:
y = y0 + v0 · t + 1/2 · g · t²
0 = 60 m + 0 m/s · t - 1/2 · 9.8 m/s² · t²
0 = 60 m - 4.9 m/s² · t²
-60 m / -4.9 m/s² = t²
t = 3.5 s
Now, with this time, we can calculate the velocity of the ball when it reaches the ground:
v = v0 + g · t
v = 0 m/s - 9.8 m/s² · 3.5 s
v = -34 m/s
When the ball hits the ground, the velocity will be -34 m/s.
The angular momentum of the person is given by:
where
is the mass of the person,
is the tangential velocity, and
is the distance of the person from the center of rotation. Using these data, we find
Answer:law of interia(1st law)
Explanation:
