Answer:
592000 J
Explanation:
We'll begin by converting 3.7×10⁵ Pa to Kg/ms². This can be obtained as follow:
1 Pa = 1 Kg/ms²
Therefore,
3.7×10⁵ Pa = 3.7×10⁵ Kg/ms²
Next, we shall determine the workdone.
Workdone is given by the following equation:
Workdone (Wd) = pressure (P) × change in volume (ΔV)
Wd = PΔV
With the above formula, the work done can be obtained as follow:
Pressure (P) = 3.7×10⁵ Kg/ms²
Change in volume (ΔV) = 1.6 m³
Workdone (Wd) =?
Wd = PΔV
Wd = 3.7×10⁵ × 1.6
Wd = 592000 Kgm²/s²
Finally, we shall convert 592000 Kgm²/s² to Joule (J). This can be obtained as follow:
1 Kgm²/s² = 1 J
Therefore,
592000 Kgm²/s² = 592000 J
Therefore, the Workdone is 592000 J.
<u>Answer:</u>
Option (a)
<u>Explanation :</u>
A stage hand starts sliding a large piece of stage scenery originally at rest by pulling it horizontally with a force of 176 N.
Hence Force applied 
Force on piece of scenery 
µk = 
= 
=
=0.36
coefficient of static friction is 0.36
Average acceleration is the ratio of the change in velocity to the time it takes for that change to occur:
We want to find 
:
Answer:
We know that loudness is directly proportional to amplitude and SI unit of amplitude is Decibel So unit of loudness is decibel
Answer:
u/2 √(1 + 3 cos² θ)
Explanation:
The object is thrown at an angle θ, so the velocity has two components, vertical and horizontal.
Initially, the vertical component is u sin θ and the horizontal component is u cos θ.
At the maximum height, the vertical component is 0 and the horizontal component is u cos θ.
The mean vertical velocity is:
(u sin θ + 0) / 2 = u/2 sin θ
The mean horizontal velocity is:
(u cos θ + u cos θ) / 2 = u cos θ
The net mean velocity can be found with Pythagorean theorem:
v² = (u/2 sin θ)² + (u cos θ)²
v² = u²/4 sin² θ + u² cos² θ
v² = u²/4 (1 − cos² θ) + u² cos² θ
v² = u²/4 (1 − cos² θ) + u²/4 (4 cos² θ)
v² = u²/4 (1 − cos² θ + 4 cos² θ)
v² = u²/4 (1 + 3 cos² θ)
v = u/2 √(1 + 3 cos² θ)
