Train cars are coupled together by being bumped into one another. Suppose two loaded train cars are moving toward one another, t
2 answers:
Answer:
final velocity = 0.08585m/s
Explanation:
We are taking train cars as our system. In this system no external force is acting. So we can apply the law of conservation of linear momentum.
The law of conservation of linear momentum states that the total linear momentum of a system remains constant if there is no external force acting on the system. That is total linear momentum before = total linear momentum after
total linear momentum before = linear momentum of first train car + linear momentum of second train car
We know that linear momentum = mv
where,
m = mass
v = velocity
thus,
total linear momentum before = m₁v₁ + m₂v₂
m₁ = mass of first train car = 135,000kg
v₁ = velocity of first train car = 0.305m/s
m₂ = mass of first second car = 100,000kg
v₂ = velocity of second train car = −0.210m/s
Note: Momentum is a vector. So while adding momentum we should take account of its direction too. Here since second train car is moving in a direction opposite to that of the first one, we have taken its velocity as negative.
total linear momentum before = m₁v₁ + m₂v₂
= 135,000x0.305 + 100,000x(−0.210)
= 135,000x0.305 - 100,000x0.210
= 20,175 kgm/s
Now we have to find total linear momentum after bumping. After the bumping both the train cars will be moving together with a common velocity(say v).
Therefore, total linear momentum after = mv
m = m₁ + m₂ = 135,000 + 100,000 = 235,000
total linear momentum before = total linear momentum after
235,000v = 20,175
v =
= 0.08585m/s
You might be interested in
Answer:
The dilation of time.
The falling of objects.
The changing of paths of light.
Explanation:
I have explained in the image attached below.
From the explanation, the correct ones are;
The dilation of time.
The falling of objects.
The changing of paths of light.
Answer:
1) E = 2.25 i^+ 0.809j^) 10⁹ N / C , 2) E = 2.39 10⁹ N / C , 3) θ = 19.8º , 4) F = 19.12 10⁹ N , 5) E = (1.32 i^+ 3.56 j^) 109 N/C
Explanation:
1) The equation for the electric field is
E = k q / r²
Where K is the Coulomb constant that is worth 8.99 10⁹ N m² /C², q is the load and r is the distance of the load to the test point
Since the electric field is a vector magnitude, we can find its component
X axis
Ex = k q / x²
where the distance on the axis is
x = √ (X₂-x₁)²
x = √ (-15 + 9)² = 6 m
Eₓ = 8.99 10⁹ 9/6²
Eₓ = 2.25 10⁹ N /C
Y Axis
y = √ (y₂-y₁)² = √ (15-5)² = 10 m
= 8.99 10⁹ 9/10²
= 0.809 10⁹ N / C
E = Eₓ i^+ j^
E = 2.25 i^+ 0.809j^) 10⁹ N / C
2) the magnitude can be found using the Pythagorean triangle
E = √ (Eₓ² + ²)
E = √ (2.25² + 0.809²) 10⁹
E = 2.39 10⁹ N / C
3) to find the angle let's use trigonometry
tan θ = / Eₓ
θ = tan⁻¹ / Eₓ
θ = tan⁻¹ (0.809 / 2.25)
θ = 19.8º
Regarding the positive side of the x axis
4) a charge q2 = 8C is placed, let's calculate the force
F = q E
F = 8 2.39 10⁹
F = 19.12 10⁹ N
5) The total electric field at the origin, let's look for its components
q₁ = 9C
r₁ = -9 i ^ + 5 j ^
q₂ = 8 C
r₂ = -15 i ^ + 15 j ^
X axis
Eₓ = E₁ₓ + E₂ₓ
Eₓ = k q₁ / Dx₁² + k q₂ / Dx₂²
Eₓ = 8.99 10⁹ (9 / (9-0)² + 8 / (15-0)²)
Eₓ = 1.32 109 N / A
Y Axis
=
+
= k q₁ / Δy₁² + k q₂ / Δy₂²
= 8.99 109 (9/5² + 8/15²)
= 3.56 109 N / A
E = (1.32 i^+ 3.56 j^) 109 N/C
Answer :
(a) The acceleration of the car is,
(b) The distance covered by the car is, 150 m
Explanation :
By the 1st equation of motion,
...........(1)
where,
v = final velocity = 5 m/s
u = initial velocity = 25 m/s
t = time = 10 s
a = acceleration of the car = ?
Now put all the given values in the above equation 1, we get:
The acceleration of the car is,
By the 2nd equation of motion,
...........(2)
where,
s = distance covered by the car = ?
u = initial velocity = 25 m/s
t = time = 10 s
a = acceleration of the car =
Now put all the given values in the above equation 2, we get:
By solving the term, we get:
The distance covered by the car is, 150 m