Answer:
θ = 22.2
Explanation:
This is a diffraction exercise
a sin θ = m λ
The extension of the third zero is requested (m = 3)
They indicate the wavelength λ = 630 nm = 630 10⁻⁹ m and the width of the slit a = 5 10⁻⁶ m
sin θ = m λ / a
sin θ = 3 630 10⁻⁹ / 5 10⁻⁶
sin θ = 3.78 10⁻¹ = 0.378
θ = sin⁻¹ 0.378
to better see the result let's find the angle in radians
θ = 0.3876 rad
let's reduce to degrees
θ = 0.3876 rad (180º /π rad)
θ = 22.2º
Here is the rule for see-saws here on Earth, and there is no reason
to expect that it doesn't work exactly the same anywhere else:
(weight) x (distance from the pivot) <u>on one side</u>
is equal to
(weight) x (distance from the pivot) <u>on the other side</u>.
That's why, when Dad and Tiny Tommy get on the see-saw, Dad sits
closer to the pivot and Tiny Tommy sits farther away from it.
(Dad's weight) x (short length) = (Tiny Tommy's weight) x (longer length).
So now we come to the strange beings on the alien planet.
There are three choices right away that both work:
<u>#1).</u>
(400 N) in the middle-seat, facing (200 N) in the end-seat.
(400) x (1) = (200) x (2)
<u>#2).</u>
(200 N) in the middle-seat, facing (100 N) in the end-seat.
(200) x (1) = (100) x (2)
<u>#3).</u>
On one side: (300 N) in the end-seat (300) x (2) = <u>600</u>
On the other side:
(400 N) in the middle-seat (400) x (1) = 400
and (100 N) in the end-seat (100) x (2) = 200
Total . . . . . . . . . . . . <u>600</u>
These are the only ones to be identified at Harvard . . . . . . .
There may be many others but they haven't been discarvard.
Answer:
t = 1.41 sec.
Explanation:
If we assume that the acceleration of the blocks is constant, we can apply any of the kinematic equations to get the time since the block 2 was released till it reached the floor.
First, we need to find the value of acceleration, which is the same for both blocks.
If we take as our system both blocks, and think about the pulley as redirecting the force simply (as tension in the strings behave like internal forces) , we can apply Newton's 2nd Law, as they were moving along the same axis, aiming at opposite directions, as follows:
F = m₂*g - m₁*g = (m₁+m₂)*a (we choose as positive the direction of the acceleration, will be the one defined by the larger mass, in this case m₂)
⇒ a = (
= g/5 m/s²
Once we got the value of a, we can use for instance this kinematic equation, and solve for t:
Δx = 1/2*a*t² ⇒ t² = (2* 1.96m *5)/g = 2 sec² ⇒ t = √2 = 1.41 sec.
