Net Force = (mass) x (acceleration) (Newton #2)
Net Force = (50 kg) x (6 m/s² down)
Net Force = (50 * 6) (kg-m/s² down)
<em>Net Force = 300 Newtons down</em>
Answer:
The final pressure will be 6.99 N/m^ 2
Explanation:
Initial volume of tire = 2.00 L
Initial pressure of the tire = 7.00 × 10 5 N/m 2
Initial temperature of tire 31 °C
P1V1 = P2V2
7.00 × 10 5 N/m 2 × V1 = 101325×100 → V1 = 1.45×10^(-5) m^3 = 1.45×10^(-2) L
Therefore new pressure = initial pressure - (pressure of released gas filling the entire 2 L)
pressure of released gas filling the entire 2 L = 2 L/(7.00 × 10^5 N/m^ 2 × 1.45×10^(-5) m^3) = 1.197 ×10^(-4) N/m^2
New pressure = 7.00 × 10^5 N/m^2 - 1.197 ×10^(-4) N/m^2
= 6.99 N/m^ 2
Before we start thinking about the snowball, we need to remind
each other that energy is "conserved". That means that if you
ever see energy decrease in one place, then the missing amount
must have gone somewhere, and if you ever see energy increase
in one place, then the energy that appeared must have come from
somewhere. Energy does not magically appear or disappear.
So you toss a snowball out of your hand. As you let it go, you give
it some kinetic energy, and it starts rolling along the ground.
Once you let go of it, it can't get any more energy (unless it has
some kind of little tiny engine inside it).
Kinetic Energy = (1/2) (mass) (speed)² .
and that amount can't change.
So if extra snow sticks to it as it merrily rolls along, and its mass
increases, then it must slow down.
Answer:
Its not B. So I think its A.
Explanation:
Reporting The Results.
To solve this problem it is necessary to apply the concepts related to the conservation of angular momentum. This can be expressed mathematically as a function of inertia and angular velocity, that is:
Where,
I = Moment of Inertia
= Angular Velocity
For the given object the moment of inertia is equivalent to
Considering that the moment of inertia varies according to distance, and that there are two of these without altering the mass we will finally have to
Our values are given as,
Replacing we have,
Therefore the angular speed after the catch slips is 0.2rad/s
