A car moving at 95 km/h passes a 1.00-km-long train traveling in the same direction on a track that isparallel to the road. If t
1 answer:
Answer:
Time to pass the train=0.05 h
How far the car traveled in this time=4.75 Km
Explanation:
We have that the train and the car are moving in the same direction, the difference between the speed of the vehicles is:
We will use this difference in the speed of the car an train to calculate how much time take the car to pass the train. For this we have that the train is 1km long and the car is moving with a speed of 20km/h (we use this value because is the speed that the car have in advantage of the train) then for a movement with a constant speed we have:
Where x is the distance, t is the time and v is the speed. using the data that we have:
This is the time that the car take to pass the train. Now to calculate how far the car have traveled in this time we have to considered the speed of 95Km/h of the car, then:
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Answer:
Explanation:
The form of Newton's 2nd Law that we use for this is:
F - f = ma where F is the Force pulling the mass down the ramp forward, f is the friction trying to keep it from moving forward, m is the mass and a is the acceleration (and our unknown).
We know mass and we can find f, but we don't have F. But we can solve for that by rewriting our main equation to reflect F:
That's everything we need.
w is weight: 6.0(9.8). Filling in:
6.0(9.8)sin20 - .15(6.0)(9.8) = 6.0a and
2.0 × 10¹ - 8.8 = 6.0a and
11 = 6.0a so
a = 1.8 m/s/s
Answer:
y = 17 m
Explanation:
For this projectile launch exercise, let's write the equation of position
x = v₀ₓ t
y = t - ½ g t²
let's substitute
45 = v₀ cos θ t
10 = v₀ sin θ t - ½ 9.8 t²
the maximum height the ball can reach where the vertical velocity is zero
v_{y} = v_{oy} - gt
0 = v₀ sin θ - gt
0 = v₀ sin θ - 9.8 t
Let's write our system of equations
45 = v₀ cos θ t
10 = v₀ sin θ t - ½ 9.8 t²
0 = v₀ sin θ - 9.8 t
We have a system of three equations with three unknowns for which it can be solved.
Let's use the last two
v₀ sin θ = 9.8 t
we substitute
10 = (9.8 t) t - ½ 9.8 t2
10 = ½ 9.8 t2
10 = 4.9 t2
t = √ (10 / 4.9)
t = 1,429 s
Now let's use the first equation and the last one
45 = v₀ cos θ t
0 = v₀ sin θ - 9.8 t
9.8 t = v₀ sin θ
45 / t = v₀ cos θ
we divide
9.8t / (45 / t) = tan θ
tan θ = 9.8 t² / 45
θ = tan⁻¹ ( 9.8 t² / 45 )
θ = tan⁻¹ (0.4447)
θ = 24º
Now we can calculate the maximum height
v_y² = - 2 g y
vy = 0
y = v_{oy}^2 / 2g
y = (20 sin 24)²/2 9.8
y = 3,376 m
the other angle that gives the same result is
θ‘= 90 - θ
θ' = 90 -24
θ'= 66'
for this angle the maximum height is
y = v_{oy}^2 / 2g
y = (20 sin 66)²/2 9.8
y = 17 m
thisis the correct