Answer:
u₂ = 3.7 m/s
Explanation:
Here, we use the law of conservation of momentum, as follows:
where,
m₁ = mass of the car = 1250 kg
m₂ = mass of the truck = 2020 kg
u₁ = initial speed of the car before collision = 17.4 m/s
u₂ = initial speed of the tuck before collision = ?
v₁ = final speed of the car after collision = 6.7 m/s
v₂ = final speed of the truck after collision = 10.3 m/s
Therefore,
<u>u₂ = 3.7 m/s</u>
Answer:
Explanation:
Range of projectile R = 20 m
formula of range
R = u² sin2θ / g
u is initial velocity , θ is angle of projectile
putting the values
20 = u² sin2x 40 / 9.8
u² = 199
u = 14.10 m /s
At the initial point
vertical component of u
= u sin40 = 14.1 x sin 40
= 9.06 m/s
Horizontal component
= u cos 30
At the final point where the ball strikes the ground after falling , its speed remains the same as that in the beginning .
Horizontal component of velocity
u cos 30
Vertical component
= - u sin 30
= - 9.06 m /s
So its horizontal component remains unchanged .
change in vertical component = 9.06 - ( - 9.06 )
= 18.12 m /s
change in momentum
mass x change in velocity
= .050 x 18.12
= .906 N.s
Impulse = change in momentum
= .906 N.s .
To solve this problem it is necessary to apply the equations related to the conservation of momentum. Mathematically this can be expressed as
Where,
= Mass of each object
= Initial velocity of each object
= Final Velocity
Since the receiver's body is static for the initial velocity we have that the equation would become
Therefore the velocity right after catching the ball is 0.0975m/s
S= 343m/s
F=256Hz
WL= 343ms/256-1
WL=V/F
= 1.339844m
Answer:
20m/second
Explanation:
The reason the answer is 20m/second is because to find the speed of the ball in this question you have to divide the distance over the time giving you the result of 20m/second
