Answer:
(1) passed through the foil
Explanation:
Ernest Rutherford conducted an experiment using an alpha particle emitter projected towards a gold foil and the gold foil was surrounded by a fluorescent screen which glows upon being struck by an alpha particle.
Answer:
Q = 913.9 gpm
Explanation:
The Hazen Williams equation can be written as follows:
where,
P = Friction Loss per foot of pipe = = 4 x 10⁻⁴
Q = Flow Rate in gallon/min (gpm) = ?
d = pipe diameter in inches = (400 mm)(0.0393701 in/1 mm) = 15.75 in
C = roughness coefficient = 100
Therefore,
<u>Q = 913.9 gpm</u>
Answer:
1.045 m from 120 kg
Explanation:
m1 = 120 kg
m2 = 420 kg
m = 51 kg
d = 3 m
Let m is placed at a distance y from 120 kg so that the net force on 51 kg is zero.
By use of the gravitational force
Force on m due to m1 is equal to the force on m due to m2.
3 - y = 1.87 y
3 = 2.87 y
y = 1.045 m
Thus, the net force on 51 kg is zero if it is placed at a distance of 1.045 m from 120 kg.
Explanation:
Let me know if you have questions
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.
