Answer:
Explanation:
Let the charge on proton be q .
energy gain by proton in a field having potential difference of V₀
= V₀ q
Due to gain of energy , its kinetic energy becomes 1/2 m v₀²
where m is mass and v₀ is velocity of proton
V₀ q = 1/2 m v₀²
In the second case , gain of energy in electrical field
= 2 V₀q , if v be the velocity gained in the second case
2 V₀q = 1/2 m v²
1/2 m v² = 2 V₀q = 2 x 1/2 m v₀²
mv² = 2 m v₀²
v = √2 v₀
Answer:
F = - 3.53 10⁵ N
Explanation:
This problem must be solved using the relationship between momentum and the amount of movement.
I = F t = Δp
To find the time we use that the average speed in the contact is constant (v = 600m / s), let's use the uniform movement ratio
v = d / t
t = d / v
Reduce SI system
m = 26 g ( 1 kg/1000g) = 26 10⁻³ kg
d = 50 mm ( 1m/ 1000 mm) = 50 10⁻³ m
Let's calculate
t = 50 10⁻³ / 600
t = 8.33 10⁻⁵ s
With this value we use the momentum and momentum relationship
F t = m v - m v₀
As the bullet bounces the speed sign after the crash is negative
F = m (v-vo) / t
F = 26 10⁻³ (-500 - 630) / 8.33 10⁻⁵
F = - 3.53 10⁵ N
The negative sign indicates that the force is exerted against the bullet
Answer:
Mass and velocity.
Explanation:
Kinetic energy <u>is the energy that an object has due to its movement</u>, mathematically it is represented as follows:
where 
is the mass of the object, and 
is its velocity at a given point in time.
So we can see that to find the kinetic energy just before the ball hits the gound, we need the quantities:
- mass of the ball
- velocity of the ball before it hits the ground
With the knowledge of these two quantities the kinetic energy of the ball before touching the gound can be determined.
