2e level 1
8e level 2
18e level 3
32e level 4
Answer:
a) 4.2m/s
b) 5.0m/s
Explanation:
This problem is solved using the principle of conservation of linear momentum which states that in a closed system of colliding bodies, the sum of the total momenta before collision is equal to the sum of the total momenta after collision.
The problem is also an illustration of elastic collision where there is no loss in kinetic energy.
Equation (1) is a mathematical representation of the the principle of conservation of linear momentum for two colliding bodies of masses
and
whose respective velocities before collision are
and
;

where
and
are their respective velocities after collision.
Given;

Note that
=0 because the second mass
was at rest before the collision.
Also, since the two masses are equal, we can say that
so that equation (1) is reduced as follows;

m cancels out of both sides of equation (2), and we obtain the following;

a) When
, we obtain the following by equation(3)

b) As
stops moving
, therefore,

Answer:
V = 331.59m/s
Explanation:
First we need to calculate the time taken for the shell fire to hit the ground using the equation of motion.
S = ut + 1/2at²
Given height of the cliff S = 80m
initial velocity u = 0m/s²
a = g = 9.81m/s²
Substitute
80 = 0+1/2(9.81)t²
80 = 4.905t²
t² = 80/4.905
t² = 16.31
t = √16.31
t = 4.04s
Next is to get the vertical velocity
Vy = u + gt
Vy = 0+(9.81)(4.04)
Vy = 39.6324
Also calculate the horizontal velocity
Vx = 1330/4.04
Vx = 329.21m/s
Find the magnitude of the velocity to calculate speed of the shell as it hits the ground.
V² = Vx²+Vy²
V² = 329.21²+39.63²
V² = 329.21²+39.63²
V² = 108,379.2241+1,570.5369
V² = 109,949.761
V = √ 109,949.761
V = 331.59m/s
Hence the speed of the shell as it hits the ground is 331.59m/s
Velocidad inicial = 20 m/s
velocidad final = 0 m/s
aceleracion = -2 m/s^2
aceleracion = (cambio de velocidad)/(cambio de tiempo)
(cambio de tiempo)= (cambio de velocidad)/aceleracion
tiempo = (-20 m/s)/(-2 m/s^2)
= 10 segundos
x = (x(inicial)) + (v(inicial))(tiempo) + 1/2(aceleracion)(tiempo)^2
x(inicial) = 0
x = (20 m/s)(10 s) + 1/2 (-2m/s^2)(10 s)^2
x = 200 m - 100 m
x = 100 m (el espacio recorrido en los dos segundos)
espero que esto te ayude! buena suerte!
Answer:
80km/hr
Explanation:
Total distance: 400km
Total time 5 hours
Average speed= distance/time
400/5= 80km/hr