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Fynjy0 [20]
3 years ago
12

A 3000-kg truck moving with a velocity of 25 m/s hits a 1000-kg parked car. The impact causes the 1000-kg car to be set in motio

n at 7 m/s. What is the velocity of the truck after the collision?
Physics
1 answer:
Alborosie3 years ago
4 0

Answer:

v₃ = 22.67 [m/s]

Explanation:

In order to solve this problem, we must use the principle of conservation of the quantity of linear momentum. Where momentum is conserved before and after the collision, i.e. remains the same.

The terms on the left of the equation represent the amount of linear momentum before the collision and the members on the right represent the momentum after the collision.

(m_{1}*v_{1})+(m_{2}*v_{2})=(m_{1}*v_{3})+(m_{2}*v_{4})

where:

m₁ = mass of the truck = 3000 [kg]

m₂ = mass of the car = 1000 [kg]

v₁ = velocity of the truck before the coliision = 25 [m/s]

v₂ = velocity of the car parked = 0 (without movement)

v₃ = velocity of the truck after the collision [m/s]

v₄ = velocity of the car after the collision = 7 [m/s]

Now replacing:

(3000*25)+(1000*0)=(3000*v_{3})+(1000*7)\\75000-7000 = 3000*v_{3}\\v_{3}=22.67 [m/s]

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julia-pushkina [17]

4.) Quantitative: notice the root word 'quantity', which is a certain percentage of something.

a.) Quantitative, since it represents a certain percentage. The weight before the camp could also be written as an improper fraction.

b.) This is also quantitative because it can be written as a ratio(2.99:500), which can also be written as a fraction(\frac{2.99}{500} or percentage.

c.) This is qualitative. Joan has a higher percentage than Emile.

d.) Qualitative. It judges the quality of something instead of amount. To make it quantitative requires a fraction of percentage. In this case, maybe you could add how he can run y miles every x seconds.

5.)

a.)This depends on whether or not the air is moving. If the classroom's AC is on, or the window is open, then the correct unit to use is CFM (cubic feet per minute). If not, the common unit is milliliters, but must also include temperature and air pressure as well as the volume of the room.

b.)  Simply calculate the volume of the can using inches cubed, since volume is the measure of how much space an object takes up.

c.) The unit would be \frac{kg}{ft^3}. If \frac{mass}{volume} = density, then simply multiply your density by your mass.

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e.) Use the same formula for c.), which means kilograms per cubic foot.

Hope this helped.

8 0
3 years ago
The mass of an atom of carbon-12 (Z = 6) is less than the mass of 6 protons and 6 neutrons because of
Orlov [11]
Because 2 carbon attach to one's of each of the others
4 0
3 years ago
Water flows at 10 m/s through a pipe with radius 0.025 m. The pipe goes up to the second floor of the building, 2.5 m higher, an
Lera25 [3.4K]

Answer: from the information given, the velocity of the water will decrease but the pipe size will remain the same.

This can be proved with bernoulli's equation.

Explanation: careful analysis of the system using bernoulli's equation of flow is shown in the image attached

5 0
3 years ago
Runner A is initially 5.7 km west of a flagpole and is running with a constant velocity of 8.9 km/h due east. Runner B is initia
Oksanka [162]

Answer:

B meet A 0.01 km east of flagpole

Explanation:

given data

distance A = 5.7 km  west

velocity V1 = 8.9 km/h

distance B = 4.5 km  east

velocity V2 = 7 km/h

to find out

How far runners from the flagpole, when paths cross

solution

we know A and B are 5.7 + 4.5 = 10.2 km apart

and we consider here B will run distance x km for meet

so time will be for B is

time B = distance / velocity

time B = x / 7     ...................1

and

for A distance for meet = ( 10.2 - x ) km

so time A = distance / velocity

time A = ( 10.2 - x )  / 8.9      .............2

now equating equation 1 and 2

time A = time B

x / 7  =   ( 10.2 - x )  / 8.9

x = 4.490

so distance of B run for meet is 4.490 km

so distance from the flagpole when their paths cross is 4.5 - 4.490 = 0.01 km

so B meet A 0.01 km east of flagpole

4 0
3 years ago
With a velocity of 15 m/s, how long would it take the dog to run 100 meters?
Zinaida [17]

Answer

6.66 seconds imma seem like an idiot if something like this is werong

Explanation:

dogg is speed

100/15

3 0
3 years ago
Read 2 more answers
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