What is another unit for momentum besides kg-m/s? a. N b. N-s c. N-s2 d. N/s

Football player 1 has a mass of 80 kg and a velocity of 2 m/s east while player 2 has a mass of 70 kg and a velocity of 3 m/s we

sukhopar [10]

The total momentum of the system is equal to 50 Kgm/s.

Given the following data:

Mass 1 = 80 kg

Velocity 1 = 2 m/s east.

Mass 2 = 70 kg

Velocity 2 = 3 m/s west.

To determine the total momentum of the system:

Mathematically, momentum is given by the formula;

$Momentum = mass \times velocity$

For Football player 1:

$Momentum = 80 \times 2$

Momentum 1 = 160 Kgm/s.

For Football player 2:

$Momentum = 70 \times 3$

Momentum 1 = 210 Kgm/s.

Now, we can calculate the total momentum of the system:

$Total\;momentum = momentum \;2 - momentum \;1\\\\Total\;momentum = 210-160$

Total momentum = 50 Kgm/s.

Note: We subtracted because the football players were moving in opposite directions.

6 0

2 years ago

Explain why a moving object cannot come to a stop instantaneously (in zero seconds). Hint: Think about the acceleration that wou

gizmo_the_mogwai [7]

To stop instantly, you would need infinite deceleration. This in turn, requires infinite force, as demonstrable with this equation:F=maSo when you hit a wall, you do not instantly stop (e.g. the trunk of the car will still move because the car is getting crushed). In a case of a change in momentum, mv⃗ mv→, we can use the following equation to calculate force:F=p/hHowever, because the force is nowhere close to infinity, time will never tend to zero either, which means that you cannot come to an instantaneous stop.

7 0

3 years ago

An ideal spring hangs from the ceiling. A 1.25-kg mass is hung from the spring. After all vibrations have died away, the spring

ch4aika [34]

The kinetic energy of the mass at the instant it passes back through its equilibrium position is about 1.20 J

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<h3>Further explanation</h3>

Let's recall Elastic Potential Energy formula as follows:

$\boxed{E_p = \frac{1}{2}k x^2}$

where:

Ep = elastic potential energy ( J )

k = spring constant ( N/m )

x = spring extension ( compression ) ( m )

Let us now tackle the problem!

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Given:

mass of object = m = 1.25 kg

initial extension = x = 0.0275 m

final extension = x' = 0.0735 - 0.0275 = 0.0460 m

Asked:

kinetic energy = Ek = ?

Solution:

Firstly , we will calculate the spring constant by using Hooke's Law as follows:

$F = k x$

$mg = k x$

$k = mg \div x$

$k = 1.25(9.8) \div 0.0275$

$k = 445 \frac{5}{11} \texttt{ N/m}$

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Next , we will use Conservation of Energy formula to solve this problem:

$Ep_1 + Ek_1 = Ep_2 + Ek_2$

$\frac{1}{2}k (x')^2 + mgh + 0 = \frac{1}{2}k x^2 + Ek$

$Ek = \frac{1}{2}k (x')^2 + mgh - \frac{1}{2}k x^2$

$Ek = \frac{1}{2}k ( (x')^2 - x^2 ) + mgh$

$Ek = \frac{1}{2}(445 \frac{5}{11}) ( 0.0460^2 - 0.0275^2 ) + 1.25(9.8)(0.0735)$

$\boxed {Ek \approx 1.20 \texttt{ J}}$

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<h3>Learn more</h3>

Kinetic Energy : brainly.com/question/692781

Acceleration : brainly.com/question/2283922

The Speed of Car : brainly.com/question/568302
Young Modulus : brainly.com/question/9202964
Simple Harmonic Motion : brainly.com/question/12069840

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<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Elasticity

8 0

3 years ago

Read 2 more answers

how many times does the kinectic energy of a car increase when traveling 60 mph as opposed to traveling 30 mph

Bumek [7]

The car at 60 kph has 9 times more kinetic energy than the car traveling at 20 kph. This assumes that both cars have the same mass. Kinetic energy depends on the square of thee speed so if one car is going 3 times faster, its kinetic energy will be 3^2 ( = 9 ) greater. The car going at 60 kph will have 4 times the KE of the car going at 30 kph ( again assuming that the cars have the same mass.)

3 0

3 years ago

An automobile to be transported by ship is raised 8.2m above the dock. If it's GPE is 63,405 J what is the car's mass?

Pepsi [2]

Answer:

m = 788.2[kg]

Explanation:

The potential energy of a body is defined as the product of mass by gravitational acceleration by height. And it can be calculated by means of the following equation.

$E_{pot}=m*g*h$

where:

Epot = potential energy = 63405 [J]

m = mass [kg]

g = gravity acceleration = 9.81[m/s²]

h = elevation = 8.2[m]

Now replacing:

$63405=m*9.81*8.2\\m=788.2[kg]$

5 0

3 years ago