OK. So you're pushing on the small box, and on the other side of it, the small
box is pushing on the big box. So you're actually pushing both of them.
-- The total mass that you're pushing is (5.2 + 7.4) = 12.6 kg.
-- You're pushing it with 5.0N of force.
-- Acceleration of the whole thing = (force)/(mass) = 5/12.6 = <em>0.397 m/s²</em> (rounded)
-- Both boxes accelerate at the same rate. So the box farther away from you ...
the big one, with 7.4 kg of mass, accelerates at the same rate.
The force on it to make it accelerate is (mass) x (acceleration) =
(7.4 kg) x (5/12.6 m/s²) = <em>2.936 N.</em>
The only force on the big box comes from the small box, pushing it from behind.
So that same <em>2.936N</em> must be the contact force between the boxes.
Answer:
(a) the angular velocity at θ1 is 11.64 rad/s
(b) the angular acceleration is 0.12 rad/
(c) the angular position was the disk initially at rest is - 428.27 rad
Explanation:
Given information :
θ1 = 16 rad
θ2 = 76 rad
ω2 = 11 rad/s
t = 5.3 s
(a) The angular velocity at θ1
First, we use the angular motion equation for constant acceleration
Δθ = (ω1+ω2)t/2
θ2 - θ1 = (ω1+ω2)t/2
ω1 + ω2 = 2 (θ2 - θ1) / t
ω1 = (2 (θ2 - θ1) / t ) - ω2
= (2 (76-16) / 5.3) - 11
= 11.64 rad/s
(b) the angular acceleration
ω2 = ω1 + α t
α t = ω2 - ω1
α = (ω2 - ω1)/t
= (11.64 - 11) / 5.3
= 0.12 rad/
(c) the angular position was the disk initially at rest, θ0
at rest ω0 = 0
ω2^2 = ω01 t + 2 α Δθ
2 α Δθ = ω2^2
θ2 - θ0 = ω2^2 / 2 α
θ0 = θ2 - (ω2^2) / 2 α
= 76 - (
/ 2 x 0.12
= 76 - 504.16
= - 428.27 rad
Answer:
If the ray of light is deflected by 45 degrees by the first mirror its total deflection by mirror (I) is 90 deg. (incident = 45 and exit ray equals 45 deg)
The second mirror will cause a net deflection of 90 degrees and the total deflection will be 180 deg or in opposite direction to the incident ray.
Answer:
utrasonic
Explanation:
these are sounds beyond our hearing capacity range of 20-20kHz
Answer : The partial pressure of
is, 67.009 atm
Solution : Given,
Partial pressure of
at equilibrium = 30.6 atm
Partial pressure of
at equilibrium = 13.9 atm
Equilibrium constant = 
The given balanced equilibrium reaction is,

The expression of
will be,

Now put all the values of partial pressure, we get


Therefore, the partial pressure of
is, 67.009 atm