PH of 4 is Acidic and its property is to turn blue litmus red
Answer:
D) momentum of cannon + momentum of projectile= 0
Explanation:
The law of conservation of momentum states that the total momentum of an isolated system is constant.
In this case, the system cannon+projectile can be considered as isolated, because no external forces act on it (in fact, the surface is frictionless, so there is no friction acting on the cannon). Therefore, the total momentum of the two objects (cannon+projectile) must be equal before and after the firing:
But the initial momentum is zero, because at the beginning both the cannon and the projectile are at rest:
So the final momentum, which is sum of the momentum of the cannon and of the projectile, must also be zero:
I’m pretty sure you times them so 1 with A, 2 with e, 3 with C, and 4 with B
We know the formulas for momentum and energy. But they both involve the mass of
the object, and we don't know the mass of the baseball. What can we do ?
It's not a catastrophe. The question only asks which one is bigger. If we're clever,
we can answer that without ever knowing how much the momentum or the energy
actually is. We know that both baseballs have the same mass, so let's just call it
' M ' and not worry about what it really is.
<u>Momentum of anything = (mass) x (speed)</u>
Momentum of the first baseball = (M) x (4 m/s) = 4M
Momentum of the second one = (M) x (16 m/s) = 16M
The second baseball has 4 times as much momentum as the first one has.
<u>Kinetic energy of anything = 1/2 (mass) x (speed squared)</u>
KE of the first baseball = 1/2 (M) x (4 squared) = 8M
KE of the second one = 1/2 (M) x (16 squared) = 128M
The second baseball has 16 times as much kinetic energy as the first one has.
Answer:
Electric Field a the centre
Explanation:
<u>Given:</u>
Total charge on the semicircle =Q
Radius of the semicircle=R
Let consider a elemental charge on the semicircle at an angle 
with the horizontal. Since the The charge distribution is symmetrical about y axis so the there will symmetrical charge on other side. The horizontal component will cancel out each other and vertical component will get added so let dE be the electric Field due to elemental charge
Let 
be the charge per unit length such that
Total Electric Field at the centre
integrating 0 to 
So the Electric field at the centre is calculated.
