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maria [59]
2 years ago
10

Two carts are free to slide along the frictionless track shown in figure below. Cart A of mass m1 = 8 kg is released from 12m. A

cart B of mass m2 = 4 kg, initially at rest. The two carts combine together and move as one object. Calculate the height reached
by the two carts after collision.
Physics
1 answer:
Harrizon [31]2 years ago
3 0

The height reached by the two carts after collision is determined as 5.34 m.

<h3>Initial velocity of Cart A</h3>

Apply the principle of conservation of mechanical energy.

K.E = P.E

v = √2gh

v = √(2 x 9.8 x 12)

v = 15.34 m/s

<h3>Final velocity of the two carts after the collision</h3>

Apply the principle of conservation of linear momentum for inelastic collision.

m₁u₁ + m₂u₂ = v(m₁ + m₂)

8(15.34) + 4(0) = v(8 + 4)

122.72 = 12v

v = 10.23 m/s

<h3>Height reached by both carts</h3>

Apply the principle of conservation of mechanical energy.

P.E = K.E

mgh = ¹/₂mv²

h = v²/(2g)

h = (10.23²) / (2 x 9.8)

h = 5.34 m

Learn more about linear momentum here: brainly.com/question/7538238

#SPJ1

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tigry1 [53]

Answer:

A.compared

Explanation:

Fossils help figure out the time that organisms lived. If you know one of the fossils, it can be used as a reference for others around.

5 0
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When you drop a paper clip, why doesn't it fall toward you instead of toward<br> Earth?
svetoff [14.1K]
Because gravity has been known to define as a force of attraction between things that have mass.
5 0
3 years ago
The electron gun in a television tube accelerates electrons (mass = 9.11x10^-31 kg, charge = 1.6 x 10^-19C) from rest to 3 x10^7
stich3 [128]

Answer:

Electric field acting on the electron is  127500 N/C.

Explanation:

It is given that,

Mass of an electron, m=9.11\times 10^{-31}\ kg

Charge on electron, q=1.6\times 10^{-19}\ C

Initial speed of electron, u = 0

Final speed of electron, v=3\times 10^7\ m/s

Distance covered, s = 2 cm = 0.02 m

We need to find the electric field required. Firstly, we will find the acceleration of the electron from third equation of motion as :

a=\dfrac{v^2-u^2}{2s}

a=\dfrac{(3\times 10^7)^2-0}{2\times 0.02}

a=2.25\times 10^{16}\ m/s^2

According to Newton's law, force acting on the electron is given by :

F = ma

F=9.1\times 10^{-31}\times 2.25\times 10^{16}

F=2.04\times 10^{-14}\ N

Electric force is given by :

F = q E, E = electric field

E=\dfrac{F}{q}

E=\dfrac{2.04\times 10^{-14}}{1.6\times 10^{-19}}

E = 127500 N/C

So, the electric field is 127500 N/C. Hence, this is the required solution.

8 0
3 years ago
The resistance between 2 points in an electrical circuit is 1.1 Ω. What additional resistance
valentinak56 [21]

Answer:

the new resister is 11 ohms.

Explanation:

Set it  up like this.

1/x + 1/1.1 = 1                    Subtract 1/1.1 from both sides

1/x = 1 - 1/1.1

1 - 1/1.1 = 1/11

1/x = 1/11                          Cross multiply

11 = x

If 1/11 bothers you, you could do it it another way.

1 - 1/1.1 = (1.1 - 1 ) / 1.1 = 0.1 / 1.1  Multiply top and bottom by 10

0.1*10/(1.1 * 10 ) = 1 / 11

5 0
3 years ago
outward from a wall just above floor level. A 1.5 kg box sliding across a frictionless floor hits the end of the spring and comp
sweet [91]

Answer:

v = 0.489 m/s

Explanation:

It is given that,

Mass of a box, m = 1.5 kg

The compression in the spring, x = 6.5 cm = 0.065 m

Let the spring constant of the spring is 85 N/m

We need to find the velocity of the box (v) when it hit the spring. It is based on the conservation of energy. The kinetic energy of spring before collision is equal to the spring energy after compression i.e.

\dfrac{1}{2}mv^2=\dfrac{1}{2}kx^2

v=\sqrt{\dfrac{kx^2}{m}} \\\\v=\sqrt{\dfrac{85\times (0.065)^2}{1.5}} \\\\v=0.489\ m/s

So, the speed of the box is 0.489 m/s.

3 0
3 years ago
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