Answer:
The horizontal distance traveled by the projectile is 15.23 m.
Explanation:
Given;
angle of projection, θ = 25⁰
initial velocity of the projectile, u = 15 m/s
time of flight, t = 1.12 s
The the travelling path of the object is calculated as the range of the projectile
Therefore, the horizontal distance traveled by the projectile is 15.23 m.
Answer:
Explanation:
We shall apply the formula for velocity in case of elastic collision which is given below
v₁ = (m₁ - m₂)u₁ / (m₁ + m₂) + 2m₂u₂ / (m₁ + m₂)
m₁ and u₁ is mass and velocity of first object , m₂ and u₂ is mass and velocity of second object before collision and v₁ is velocity of first velocity after collision.
Here u₁ = 22 cm /s , u₂ = - 14 cm /s . m₁ = 7.7 gm , m₂ = 18 gm
v₁ = ( 7.7 - 18 ) x 22 / ( 7.7 + 18 ) + 2 x 18 x - 14 / ( 7.7 + 18 )
= - 8.817 - 19.6
= - 28.4 cm / s
Answer:
100J
Explanation:
Kinetic energy=1/2mv^2
Kinetic energy=(1/2 x 8)x5^2
Kinetic energy=4x25
Kinetic energy=100
100J
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