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tresset_1 [31]
3 years ago
10

A 30,000-kg freight car is coasting at 0.850 m/s with negligible friction under a hopper that dumps 110,000 kg of scrap metal in

to it. (a) What is the final velocity of the loaded freight car? (b) How much kinetic energy is lost?(Check 2 choices for the answer) *
Physics
1 answer:
erica [24]3 years ago
6 0

Explanation:

Given that,

Mass of a freight car, m_1=30,000-kg

Speed of a freight car, u_1=0.85\ m/s

Mass of a scrap metal, m_2=110,000\ kg

(a) Let us assume that the final velocity of the loaded freight car is V. The momentum of the system will remain conserved as follows :

30000\times 0.85=(30000+110000)V\\\\V=\dfrac{30000\times 0.85}{30000+110000}\\\\=0.182\ m/s

So, the final velocity of the loaded freight car is 0.182 m/s.

(b) Lost on kinetic energy = final kinetic energy - initial kinetic energy

\Delta K=\dfrac{1}{2}[(m_1+m_2)V^2-m_1u_1^2)]\\\\=\dfrac{1}{2}\times [(30,000+110,000 )0.182^2-30000(0.85)^2]\\\\=-8518.82\ J

Lost in kinetic energy is 8518.82. Negative sign shows loss.

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Answer:

1)4.7334J

2)225.4m/s

Explanation:

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H= Height of the bullet along circular arc= 10cm=0.1m

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R= Radius of the circular arc= 18cm= 0.18m

m= Mass of the bullet= 30g= 0.03kg

M= Mass of the block = 4.8 kg

Using the law of conservation of energy

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1/2 mv^2= mgh..............eqn(1)

But we have two mass m and M

We can write eqn(1) as

0.5(m+M)v^2= (m+M)gh ...........eqn(2)

If we make "v" subject of the formula we have

v = √2gh

Then substitute the values we have

= √2 x 9.81 x 0.1 = 1.40m/s

1) We can now calculate the total energy of the system after collision as

KE = 1/2(m+M)v^2

= 1/2 x (0.03+4.8) x (1.40)^2

KE = 4.7334J

Hence, the total energy of the composite system at any time after the collision is 4.7334J

2)to determine the initial velocity of the bullet.

From law of momentum conservation, which can be expressed as

m1u1+m2u2=(m1+m2)v

Where the initial Velocity of the bullet u1= ?

Final velocity of the bullet = 0

the Velocity of both the bullet and the block after collision=v= 1.40m/s

(0.03×u1) +(u×0)= (4.8+0.03)1.4

0.03u1=6.762

U1=225.4m/s

Hence, the initial velocity of the bullet is 225.4m/s

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If a rock has a weight of 30 N on Earth, would its weight be more or less if it was on Jupiter (gravity on Jupiter = 25 m/s2)?
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The electrons involved in the formation of a chemical bond are called
tatiyna

Answer: Valence electrons

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The radius of the aorta is about 1 cm and the blood flowing through it has a speed of about 30 cm/s. Calculate the average speed
puteri [66]

Answer:

The average speed of the blood in the capillaries is 0.047 cm/s.

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Given;

radius of the aorta, r₁ = 1 cm

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Area of the capillaries, A₂ = 2000 cm²

let the average speed of the blood in the capillaries = v₂

Apply continuity equation to determine the average speed of the blood in the capillaries.

A₁v₁ = A₂v₂

v₂ = (A₁v₁) / (A₂)

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Therefore, the average speed of the blood in the capillaries is 0.047 cm/s.

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