Answer:
Here,
Initial velocity(u)=0 m/s
acceleration(a)=1.5m/s
time(t)=0.5s
Now,
distance covered(s)=ut+(1/2)at^2
=0*0.5+(1/2)*1.5*0.5*0.5
=0+(1/2)*0.375
=0+0.1875
=0.1875(nearly 0.19)
Hence,0.19 is correct answer.
Answer:
Pressure
Explanation:
The measure of how much force is applied to an object from gas particle bouncing into it is called pressure.
Pressure is defined as the force per unit area on a body.
Mathematically;
Pressure = 
The pressure of a gas is the combined force with which gas molecules bombard a unit area of the wall of the container. It is the sum of all tiny pushes on the wall of the container.
Various units of pressure are atm, mmHg, torr, pascal e.t.c
Answer:Im guessing Mechanical Energy
Explanation:
We are learning that work and energy work hand in hand so im completely guessing this
Answer:
Explanation:
Moment of inertia of the rod = 1/12 m L²
m is mass of the rod and L is its length
= 1/2 x 2.3 x 2 x 2
= 4.6 kg m²
Moment of inertia of masses attached with the rod
= m₁ d² + m₂ d²
m₁ and m₂ are masses attached , and d is their distance from the axis of rotation
= 5.3 x 1² + 3.5 x 1²
= 8.8 kg m²
Total moment of inertia = 13.4 kg m²
B )
Rotational kinetic energy = 1/2 I ω²
I is total moment of inertia and ω is angular velocity
= .5 x 13.4 x 2²
= 26.8 J .
C )
when mass of rod is negligible , moment of inertia will be due to masses only
Total moment of inertia of masses
= 8.8 kg m²
D )
kinetic energy of the system
= .5 x 8.8 x 2²
= 17.6 J .
Answer: a) C decreases; b) Q stays the same; c) E is the same
d) ΔV increase
Explanation: In order to explain this problem we have to consider the following:
C=εoA/d where A and d are the area and the separation of the plates, respectively.
Increasing d, produces a decrease of C.
Q remain constant becasuse the plates are charges and the wire are isoloted each other.
We also know that ΔV=E*d where E is electric field between the plates.
And E= Q/εo*A ( a constant between the plates)
As we can see from above, ΔV depends directely of the d so if d increase ΔV also increase. To do that we have to do work on the system.
