They can't hear an echo in small room because in it the sound can't be reflected back. For an echo of a sound to be heard,the minimum distance between the source of sound and the walls of the roomshould be 17.2 m. Obviously,in asmall room echoes cannot beheard.
Answer:
<h2><u>Constant</u></h2>
Explanation:
Please don't comment in this question's comment box
<h2>Thanks</h2>
Answer:
4 m/s² down
Explanation:
We'll begin by calculating the net force acting on the object.
The net force acting on the object from the left and right side is zero because the same force is applied on both sides.
Next, we shall determine the net force acting on the object from the up and down side. This can be obtained as follow:
Force up (Fᵤ) = 15 N
Force down (Fₔ) = 25 N
Net force (Fₙ) =?
Fₙ = Fₔ – Fᵤ
Fₙ = 25 – 15
Fₙ = 10 N down
Finally, we shall determine the acceleration of the object. This can be obtained as follow:
Mass (ml= 2.5 Kg
Net force (Fₙ) = 10 N down
Acceleration (a) =?
Fₙ = ma
10 = 2.5 × a
Divide both side by 2.5
a = 10 / 2.5
a = 4 m/s² down
Therefore, the acceleration of the object is 4 m/s² down
Answer:
a)15 N
b)12.6 N
Explanation:
Given that
Weight of block (wt)= 21 N
μs = 0.80 and μk = 0.60
We know that
Maximum value of static friction given as
Frs = μs m g = μs .wt
by putting the values
Frs= 0.8 x 21 = 16.8 N
Value of kinetic friction
Frk= μk m g = μk .wt
By putting the values
Frk= 0.6 x 21 = 12.6 N
a)
When T = 15 N
Static friction Frs= 16.8 N
Here the value of static friction is more than tension T .It means that block will not move and the value of friction force will be equal to the tension force.
Friction force = 15 N
b)
When T= 35 N
Here value of tension force is more than maximum value of static friction that is why block will move .We know that when body is in motion then kinetic friction will act on the body.so the value of friction force in this case will be 12.6 N
Friction force = 12.6 N
I am pretty sure it’s B (carbon dioxide)
