A 1:50 scale model of a ship is towed at 4.8 km/hr using a force of 9 N. If we assume the same fluid in the model as in the prot
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Answer:
a) The additional time required for the truck to stop is <u>8.5 seconds</u>
b) The additional distance traveled by the truck is <u>230.05 ft</u>
Explanation:
Since the acceleration is constant, the average speed is:
(final speed - initial speed) / 2 = 0.75 v0
Since travelling at this speed for 8.5 seconds causes the vehicle to travel 690 ft, we can solve for v0:
0.75v0 * 8.5 = 690
v0 = 108.24 ft/s
The speed after 8.5 seconds is: 108.24 / 2 = 54.12 ft/s
We can now use the following equation to solve for acceleration:
a = -6.367 m/s^2
Additional time taken to decelerate: 54.12/6.367 = 8.5 seconds
Total distance traveled:
0 - 108.24^2 = 2 * (-6.367) * s
solving for s we get total distance traveled = 920.05 ft
Additional Distance Traveled: 920.05 - 690 = 230.05 ft
Answer:
Magnitude of the velocity = 16.82 m/s
Angular direction, θ = 52.41°
Explanation:
As given ,
mass of bullet, m₁= 50g = 0.05 kg
speed of bullet , u₁ = 600 m/s
mass of the block , m₂ = 4 kg
speed of the block before collision , u₂ = 12 m/s
direction , θ = 30°
Now,
Assume that the combined velocity of bullet and block after collision = v
and the direction = θ
Now, from the conservation of momentum in x - direction :
m₁ u₁ + m₂ u₂ = ( m₁ + m₂ ) vₓ
where v = final velocity after collision
u₁ = initial velocity of bullet before collision = 0
m₁ = mass of the bullet before collision = 0.05 kg
u₂ = velocity of block before collision = 12 cos(30° )
m₂ = mass of block before collision
m₁ + m₂ = combined mass of bullet and block after collision = 0.05 + 4
∴ we get
0.05 (0) + 4(12 cos(30° ) ) = ( 0.05 + 4 ) vₓ
⇒ 0 + 4(6√3) = 4.05 vₓ
⇒24√3 = 4.05 vₓ
⇒vₓ = 10.26 m/s
Now, from the conservation of momentum in y - direction :
m₁ u₁ + m₂ u₂ = ( m₁ + m₂ )
where v = final velocity after collision
u₁= initial velocity of bullet before collision = 600
m₁ = mass of the bullet before collision = 0.05 kg
u₂ = velocity of block before collision = 12 sin(30° )
m₂= mass of block before collision
m₁+ m₂= combined mass of bullet and block after collision = 0.05 + 4
∴ we get
0.05 (600) + 4(12 sin(30° ) ) = ( 0.05 + 4 )
⇒ 30 + 4(6) = 4.05
⇒30 +24 = 4.05
⇒54 = 4.05
⇒ = 13.33 m/s
Now, the magnitude of the velocity = √vₓ² + ² = √(10.26)² + (13.33)²
= √105.26 + 177.68
= √282.95 = 16.82
The angular direction, θ = (
) =
(
) =
(1.299) = 52.41°
Answer:
(a) 0 kJ
(b) 9.81 kJ
(c) 31.32 m/s
Explanation:
(a)
From the law of conservation of energy, energy can only be transformed from one state to another. At a height of 50 m, all the kinetic energy is converted to potential energy hence KE=0
(b)
Potential energy, PE=mgh where m is the mass, g is acceleration due to gravity and h is the height
Substituting 50 m for h and 20 Kg for m, taking g as 9.81 then
PE=20*9.81*50=9810 J=9.81 kJ
(c)
Relating the equation of potential energy to the equation of kinetic energy, which is
where v is the velocity of the mass
Substituting 50 m for h and taking g as 9.81 then