Car at rest:
velocity= 0m/s
Acceleration:
0.2m/s²
Since total time:
3 min = 180s
Formula of acceleration:
acceleration = [final velocity - initial velocity] ÷ [total time]
Velocity at end:
0.2m/s² = [final velocity - 0m/s] ÷ [180s]
0.2m/s² × 180s = [final velocity]
[final velocity] = 36m/s
Distance travelled:
Velocity = displacement(distance) ÷ time
36m/s = displacement(distance) ÷ 180s
displacement(distance) = 36m/s × 180s
displacement(distance) = 6480m
<em><u>Hey I'm sorry but i do not understand why the answer on your worksheet for distance travelled is 3240m... its </u></em><em><u>half</u></em><em><u> of what my answer is...</u></em>
Answer:
The rise in height of combined block/bullet from its original position is 0.45m
Explanation:
Given;
mass of bullet, m₁ = 12 g = 0.012 kg
mass of block of wood, m₂ = 1 kg
initial speed of bullet, u₁ = 250 m/s.
initial speed of block of wood, u₂ = 0
From the principle of conservation of linear momentum, calculate the final speed of the combined block/bullet system.
m₁u₁ + m₂u₂ = v(m₁+m₂)
where;
v is the final speed of the combined block/bullet system.
0.012 x 250 + 0 = v (0.012 + 1)
3 = v (1.012)
v = 3/1.012
v = 2.96 m/s
From the principle of conservation of energy, calculate the rise in height of the block/bullet combined from its original position.
¹/₂mv² = mgh
¹/₂v² = gh
¹/₂ (2.96)² = (9.8)h
4.3808 = 9.8h
h = 4.3808/9.8
h = 0.45 m
Therefore, the rise in height of combined block/bullet from its original position is 0.45m
When light travels from a medium with higher refractive index to a medium with lower refractive index, there is a critical angle after which all the light is reflected (so, there is no refraction).
The value of this critical angle can be derived by Snell's law, and it is equal to
where n2 is the refractive index of the second medium and n1 is the refractive index of the first medium.
In our problem, n1=1.47 and n2=1.33, so the critical angle is
<span>The 2nd truck was overloaded with a load of 16833 kg instead of the permissible load of 8000 kg.
The key here is the conservation of momentum.
For the first truck, the momentum is
0(5100 + 4300)
The second truck has a starting momentum of
60(5100 + x)
And finally, after the collision, the momentum of the whole system is
42(5100 + 4300 + 5100 + x)
So let's set the equations for before and after the collision equal to each other.
0(5100 + 4300) + 60(5100 + x) = 42(5100 + 4300 + 5100 + x)
And solve for x, first by adding the constant terms
0(5100 + 4300) + 60(5100 + x) = 42(14500 + x)
Getting rid of the zero term
60(5100 + x) = 42(14500 + x)
Distribute the 60 and the 42.
60*5100 + 60x = 42*14500 + 42x
306000 + 60x = 609000 + 42x
Subtract 42x from both sides
306000 + 18x = 609000
Subtract 306000 from both sides
18x = 303000
And divide both sides by 18
x = 16833.33
So we have the 2nd truck with a load of 16833.33 kg, which is well over it's maximum permissible load of 8000 kg. Let's verify the results by plugging that mass into the before and after collision momentums.
60(5100 + 16833.33) = 60(21933.33) = 1316000
42(5100 + 4300 + 5100 + 16833.33) = 42(31333.33) = 1316000
They match. The 2nd truck was definitely over loaded.</span>
