Answer:
t = 0.28 seconds
Explanation:
Given that,
Force acting on a firework, F = 700 N
The momentum of the firework, p =200 kg-m/s
We need to find the time before it explodes. Ket the time be t. We know that, the rate of change of momentum is equal to external frce. So,
So, the required time is equal to 0.28 seconds.
The force required to push the box upward is 145.3N and the force to pus the box downward is -109.3N
Data given;
To move the plane upward, the box will move against gravity.
Let's solve for F
When the force pushes the box down the inclined plane, it moves towards gravity.
The force required to push the box upward is 145.3N and the force required to push the box downward is -109.3N
Learn more on force across an inclined plane here;
Answer:
a) E = 1.06 10⁻¹⁹ J, b) v = 9.78 10⁵ m / s
Explanation:
The physical magnitudes can be given in several units, but in general all must be reduced to the same system in a given exercise, the most used system is the international (SI)
1 eV =q V= 1.6 10⁻¹⁹ J
let's reduce the quantities requested
a) E = 1.0 eV to Joule
E = 1.0 eV (1.6 10⁻¹⁹ J / 1 eV)
E = 1.06 10⁻¹⁹ J
b) the kinetic energy is given by
K = ½ m v²
v =
the mass of the proton is
m = 1,673 10⁻²⁷ kg
let's reduce the energy to the SI system
E = 5000 ev (1.6 10-19 J / 1 eV) = 8000 10⁻¹⁹ J
let's calculate
v =
v =
v = 9.78 10⁵ m / s