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3 years ago

The equation for momentum is p = mv, where p is momentum, m is mass, and v

Physics

1 answer:

Tasya [4]3 years ago

3 0

Answer:

p = mv, where p is momentum, m is mass, and v is velocity.

a. ) m = 12kg v = 14m/s

Momentum (p) = mv

= 12kg × 14m/s

= 168kg•m/s

b.) momentum (p) = 35 kg•m/s

velocity = 3m/s

p = mv

make m the subject

divide both sides by v

we get

m = p/v

Therefore m is

m = 35 kg•m/s / 3m/s

m = 11.67kg

Therefore the mass of the object is 11.67kg

Hope this helps

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A car battery seems to be malfunctioning (not providing the proper voltage and current to start your car). While the car is off

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No. You cannot rule out the battery even after the open circuit voltage measurement. The open-circuit voltage may not have changed but the battery's internal resistance may have greatly increased.

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2 years ago

On the Apollo 14 mission to the moon, astronaut Alan Shepard hit a golf ball with a golf club improvised from a tool. The free-f

aliya0001 [1]

Answer:

15.3 s and 332 m

Explanation:

With the launch of projectiles expressions we can solve this problem, with the acceleration of the moon

gm = 1/6 ge

gm = 1/6 9.8 m/s² = 1.63 m/s²

We calculate the range

R = Vo² sin 2θ / g

R = 25² sin (2 30) / 1.63

R= 332 m

We will calculate the time of flight,

Y = Voy t – ½ g t2

Voy = Vo sin θ

When the ball reaches the end point has the same initial height Y=0

0 = Vo sin  t – ½ g t2

0 = 25 sin (30) t – ½ 1.63 t2

0= 12.5 t – 0.815 t2

We solve the equation

0= t ( 12.5 -0.815 t)

t=0 s

t= 15.3 s

The value of zero corresponds to the departure point and the flight time is 15.3 s

Let's calculate the reach on earth

R2 = 25² sin (2 30) / 9.8

R2 = 55.2 m

R/R2 = 332/55.2

R/R2 = 6

Therefore the ball travels a distance six times greater on the moon than on Earth

5 0

2 years ago

A package of mass m is released from rest at a warehouse loading dock and slides down a 3.0-m-high frictionless chute to a waiti

LuckyWell [14K]

Answer:

The speed of the package of mass m right before the collision $= 7.668\ ms^-1$

Their common speed after the collision $= 2.56\ ms^-1$

Height achieved by the package of mass m when it rebounds $= 0.33\ m$

Explanation:

Have a look to the diagrams attached below.

a.To find the speed of the package of mass m right before collision we have to use law of conservation of energy.

$K_{initial} + U_{initial} = K_{final}+U_{final}$

where $K$ is Kinetic energy and $U$ is Potential energy.

$K= \frac{mv^2}{2}$ and $U= mgh$

Considering the fact $K_{initial} = 0\ and U_{final} =0$ we will plug out he values of the given terms.

So $V_{1}{(initial)} =\sqrt{2gh} = \sqrt{2\times9.8\times3} = 7.668\ ms^-1$

Keypoints:

Sum of energies and momentum are conserved in all collisions.
Sum of KE and PE is also known as Mechanical energy.
Only KE is conserved for elastic collision.
for elastic collison we have $e=1$ that is co-efficient of restitution.

<u>KE = Kinetic Energy and PE = Potential Energy</u>

b.Now when the package stick together there momentum is conserved.

Using law of conservation of momentum.

$m_1V_1(i) = (m_1+m_2)V_f$ where $V_1{i} =7.668\ ms^-1$ .

Plugging the values we have

$m\times 7.668 = (3m)\times V_{f}$

Cancelling m from both sides and dividing 3 on both sides.

$V_f = 2.56\ ms^-1$

Law of conservation of energy will be followed over here.

c.Now the collision is perfectly elastic $e=1$

We have to find the value of $V_{f}$ for m mass.

As here $V_{f}=-2.56\ ms^-1$ we can use that if both are moving in right ward with $2.56$ then there is a $-2.56$ velocity when they have to move leftward.

The best option is to use the formulas given in third slide to calculate final velocity of object $1$ .

$V_{1f} = \frac{m_1-m_2}{m_1+m_2} \times V_{1i}= \frac{m-2m}{3m} \times7.668=\frac{-7.668}{3} = -2.56\ ms^-1$

Now using law of conservation of energy.

$K_{initial} + U_{initial} = K_{final}+U_{final}$

$\frac{m\times V(f1)^2}{2} + 0 = 0 +mgh$

$\frac{v(f1)^2}{2g} = h$

$h= \frac{(-2.56)^2}{9.8\times 3} =0.33\ m$

The linear momentum is conserved before and after this perfectly elastic collision.

So for part a we have the speed $=7.668\ ms^-1$ for part b we have their common speed $=2.56\ ms^-1$ and for part c we have the rebound height $=0.33\ m$ .

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2 years ago

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otez555 [7]

Answer:

180

60 minutes in a hour

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2 years ago

Using Hooke's law, F spring=k delta x, find the distance a spring with an elastic constant of 4 N/cm will stretch if a 2 newton

pishuonlain [190]

Hello!

Using Hooke's law, F spring=k delta x, find the distance a spring with an elastic constant of 4 N/cm will stretch if a 2 newton force is applied to it.

Data:

Hooke represented mathematically his theory with the equation:

F = K * Δx

On what:

F (elastic force) = 2 N

K (elastic constant) = 4 N/cm

Δx (deformation or elongation of the elastic medium or distance from a spring) = ?

Solving:

$F = K * \Delta{x}$