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emmasim [6.3K]
4 years ago
11

Two air track carts move along an air track towards each other. Cart A has a mass of 450 g and moves toward the right with a spe

ed of 0.850 m/s and air track cart B has a mass of 300 g and moves toward the left with a speed of 1.12 m/s. What is the total momentum of the system
Physics
1 answer:
ra1l [238]4 years ago
6 0

Answer:

0.465 kgm/s

Explanation:

Given that

Mass of the cart A, m1 = 450 g

Speed of the cart A, v1 = 0.85 m/s

Mass of the cart B, m2 = 300 g

Speed of the cart B, v2 = 1.12 m/s

Now, using the law of conservation of momentum.

It is worthy of note that our cart B is moving in opposite directions to A

m1v1 + m2v2 =

(450 * 0.85) - (300 * 1.12) =

382.5 - 336 =

46.5 gm/s

If we convert to kg, we have

46.5 / 100 = 0.465 kgm/s

Thus, the total momentum of the system is 0.465 kgm/s

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mel-nik [20]

a = F/(m1+m2)
1.2 = F/48,000
F = 57,600N

a = F/(m1+m3)
a = 57,600/24,000
a = 2.4 m/s^2
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4 years ago
A container has a large cylindrical lower part with a long thin cylindrical neck open at the top. The lower part of the containe
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Answer:

8.6*10^5N

Explanation:

Pressure of the water = density of water * height of the water * acceleration due to gravity

Pressure of water = 1000*9.81*(2.5 + 8.50)

Pressure of the water = 107910 = 1.08*10^5

And Pressure = force / A

Force = Pressure * A(surface area of the container)

Force = 1.08*10^5 * 8 = 863280 = 8.6*10^5N

6 0
3 years ago
Jana is observing a young autistic boy in a preschool classroom. Her job is to note each time the boy gets up from his chair wit
svp [43]
The answer is tallying
3 0
4 years ago
The sides of a square increase in length at a rate of 3 ​m/sec. a. At what rate is the area of the square changing when the side
maks197457 [2]

The area of a square is given by:

A = s²

A is the square's area

s is the length of one of the square's sides

Let us take the derivative of both sides of the equation with respect to time t in order to determine a formula for finding the rate of change of the square's area over time:

d[A]/dt = d[s²]/dt

The chain rule says to take the derivative of s² with respect to s then multiply the result by ds/dt

dA/dt = 2s(ds/dt)

A) Given values:

s = 14m

ds/dt = 3m/s

Plug in these values and solve for dA/dt:

dA/dt = 2(14)(3)

dA/dt = 84m²/s

B) Given values:

s = 25m

ds/dt = 3m/s

Plug in these values and solve for dA/dt:

dA/dt = 2(25)(3)

dA/dt = 150m²/s

6 0
3 years ago
A cylinder fitted with a piston exists in a high-pressure chamber (3 atm) with an initial volume of 1 L. If a sufficient quantit
Bess [88]

Answer:

C. 85%

Explanation:

A cylinder fitted with a piston exists in a high-pressure chamber (3 atm) with an initial volume of 1 L. If a sufficient quantity of a hydrocarbon material is combusted inside the cylinder to produce 1 kJ of energy, and if the volume of the chamber then increases to 1.5 L, what percent of the fuel's energy was lost to friction and heat?

A. 15%

B. 30%

C. 85%

D. 100%

work done by the system will be

W=PdV

p=pressure

dV=change in volume

3tam will be changed to N/m^2

3*1.01*10^5

W=3.03*10^5*(1.5-1)

convert 0.5L to m^3

5*10^-4

W=3.03*10^5*5*10^-4

W=152J

therefore

to find the percentage used

152/1000*100

15%

100%-15%

85% uf the fuel's energy was lost to friction and heat

6 0
3 years ago
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