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3 years ago

14

I need help with the second one asap

2 answers:

dangina [55]3 years ago

6 0

Answer:

what I don't know show a question mark me as brainleast

katrin [286]3 years ago

3 0

I dont know hwo to help sorry :(

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Car A has a mass of 1,200 kg and is traveling at a rate of 22 km/hr. It collides with car B. Car B has a mass of 1,900 kg and is

anastassius [24]

The car A has a mass of 1200 kg.

The car B has the mass of 1900 kg.

It is given that velocity of car A is given as 22 Km/hr

The car B has the velocity of 25 Km/hr.

Let the mass of two bodies are denoted as $m_{1} \ and\ m_{2}$

Let the velocity of cars A and B are denoted as $v_{1} \ and\ v_{2}$

The momentum before collision is-

$p_{i} =m_{1} v_{1} +m_{2} v_{2}$

[Here p stand for momentum.]

We are asked to calculate the final momentum of the system after collision.

The answer of the question is based law of conservation of linear momentum.

As per law of conservation of linear momentum the sum total linear momentum for an isolated system is always constant.Hence irrespective of the type of collision[elastic and inelastic],the momentum of the system is always constant which is a universal truth.

Let after the collision the velocity of A and B are $v'_{1} \ and\ v'_{2}$

Hence the final momentum of the system is-

$p_{f} = m_{1} v'_{1} +m_{2} v'_{2}$

As per the law of conservation of linear momentum, the initial and final momentum must be equal i.e

$p_{i} =p_{f}$

$m_{1} v_{1} +m_{2}v_{2} =m_{1} v'_{1} +m_{2} v'_{2}$

Hence the option A is right.

7 0

3 years ago

Read 2 more answers

A hot-air balloon is rising upward with a constant speed of 2.85 m/s. When the balloon is 2.50 m above the ground, the balloonis

NISA [10]

Answer:

Explanation:

You have to declare which way is plus -- up or down. I will use down.

vi = - 2.85 The balloon is going up. That is the minus direction.

a = 9.81

d = 2.50 meters distance in this case is from the object to the ground.

d = vi*t + 1/2 a t^2

-2.50 = -2.85*t + 1/2 * 9.81 * t^2

-2.50 = -2.85*t + 4.905 * t^2 transfer the left to the right.

-4.905 t^2 + 2.85*t + 2.50 = 0

Use the quadratic formula to solve for t.

It turns out that t = 1.06

5 0

3 years ago

Quiero conoce amiga que tenga correo

34kurt

Hola , soy daniela .

4 0

3 years ago

A block of mass 0.510 kg is pushed against a horizontal spring of negligible mass until the spring is compressed a distance x. T

maria [59]

Answer:

x=0.46m, speed=7.9m/s

Explanation:

Using the concept of conservation of energy:

1. kinetic energy of mass m and velocity v: $E_k=\frac{1}{2}mv^2$

2. gravitational potential energy of mass m, grav. acc. g and height h: $E_g=mgh$

3. potential energy in a spring with spring constant k and displacement from equilibrium x: $E_s=\frac{1}{2}kx^2$

Calculating x:

$\frac{1}{2}mv_a^2=\frac{1}{2}kx^2$

$x=\sqrt{\frac{m}{k}}v_a$

Calculating the speed:

$\frac{1}{2}mv_a^2 +mgh_a=\frac{1}{2}mv_b^2+mgh_b + W_{friction}$

$h_a=0, h_b=2R,W_{friction}=F_{friction}\times distance=7\pi R$

$\frac{1}{2}mv_a^2=\frac{1}{2}mv_b^2+2mgR+7\pi R$

Solving for $v_b$ :

$v_b=\sqrt{v_a^2-4gR-14\pi\frac{R}{m}}$

7 0

3 years ago

How many questions are on cumulative tests? on edgenutiy

Evgesh-ka [11]

Roughly 50 for me i dont know about anyone else

8 0

2 years ago