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A car of mass 1000 kg is moving at 25 m/s. It collides with a car of mass 1200 kg moving at 30 m/s. When the cars collide, they

Alinara [238K]

Answer:

The total momentum of the cars before the collision is 61,000 kg.m/s

The total momentum of the cars after the collision is 61,000 kg.m/s

The velocity of the cars after the collision is 27.727 m/s

Explanation:

Given;

mass of the first car, m₁ = 1000 kg

initial velocity of the car, u₁ = 25 m/s

mass of the second car, m₂ = 1200 kg

initial velocity of the second car, u₂ = 30 m/s

The common velocity of the cars after collision = v

The total momentum of the cars before collision is calculated as;

P₁ = m₁u₁ + m₂u₂

P₁ = (1000 x 25) + (1200 x 30)

P₁ = 61,000 kg.m/s

The total momentum of the cars after collision is calculated as;

P₂ = m₁v + m₂v

where;

v is the common velocities of the cars after collision since they stick together.

P₂ = v(m₁ + m₂)

To determine "v" apply the principle of conservation of linear momentum for inelastic collision.

m₁u₁ + m₂u₂ = v(m₁ + m₂)

(1000 x 25) + (1200 x 30) = v(1000 + 1200)

61,000 = 2,200v

v = 61,000/2,200

v = 27.727 m/s

The total momentum after collsion = v(m₁ + m₂)

= 27.727(1000 + 1200)

= 61,000 kg.m/s

Thus, momentum before and after collsion are equal.

8 0

2 years ago

Consider a large tank holding 1000 L of pure water into which a brine solution of salt begins to owat a constant rate of 6L/min.

lbvjy [14]

Answer:

T = 693.147 minutes

Explanation:

The tank is being continuously stirred. So let the salt concentration of the tank at some time t be x in units of kg/L.

Therefore, the total salt in the tank at time t = 1000x kg

Brine water flows into the tank at a rate of 6 L/min which has a concentration of 0.1 kg/L

Hence, the amount of salt that is added to the tank per minute = $(6\times0.1)kg/min=0.6kg/min$

Also, there is a continuous outflow from the tank at a rate of 6 L/min.

Hence, amount of salt subtracted from the tank per minute = 6x kg/min

Now, the rate of change of salt concentration in the tank = $\frac{dx}{dt}$

So, the rate of change of salt in the tank can be given by the following equation,

$1000\frac{dx}{dt} =0.6-6x$

or, $\int\limits^{0.05}_0 {\frac{1000}{0.6-6x} } \, dx =\int\limits^T_0 {} \, dt$

or, T = 693.147 min (time taken for the tank to reach a salt concentration

of 0.05 kg/L)

3 0

2 years ago

Water is the only pure substance that exists as a liquid at normal Earth temperatures.

gladu [14]

Here are many things that exist as liquid at "normal" Earth temperatures. I would bet that you've seen a demonstration or someone use something with liquid mercury.

FALSE 

Hope this helps.

4 0

2 years ago

1. A pumpkin with a mass of 2 kg accelerates 2 m/s/s when an unknown force is applied to it. What is the amount of the force?

hram777 [196]

Answer:

4 Newton

Explanation:

F=ma therefore 2kg*2m/s/s=4 Newton

6 0

2 years ago

On the ride "spindletop" at the amusement park six flags over texas, people stood against the inner wall of a hollow vertical cy

Anastasy [175]

a) In this case the forces are the centrifugal force Fcp, which is directed horizontally toward the wall; the force of static friction Ff with the wall, directed upward; the normal force Fn by the wall, which is directed away the wall; the force of gravity Fg, directed downwards. Then we have that the horizontal forces are all equal in magnitude; similarly the vertical forces are also all equal in magnitude.

b) The minimum coefficient s occurs when force of gravity is equals the max friction force, that is

Fg = Ff,max

m g = s Fn

Also, the normal force has equal magnitude to the centrifugal force:

m g = s Fcp

m g = s m w^2 r

g = s w^2 r

s = g / (r w^2)

With values: g = 9.81 m/s^2; r = 2.5 m; and w = 2pi * 0.60 = 3.77 rad/s; we find

s = 9.81 / (2.5 * 3.77^2) = 0.276

3 0

2 years ago

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