Can you please help me with these 7 questions. (THEY ARE DUE TOMORROW)

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 $1$ .

So

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

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 $=0.33\ m$ .

3 0

3 years ago

How are metal bridges built to cope with changes of temperature?

nirvana33 [79]

As the metal expands as does the road bed so neither really effevts those foing over the bridge. as it is hot the metal will expand and so will most tarmac on roads.

8 0

3 years ago

A spring is used as part of a lift system and follows Hooke's law. If the spring is

salantis [7]

Answer:

1.05m or 105cm

Explanation:

Using the hooke's law equation as follows;

F = –k.x

Where;

F = force (N)

x = extension length (m)

k = constant of proportionality (N/m)

According to the information given in this question;

Displacement (x) = 85cm = 85/100 = 0.85m

Force = 12500N

Using F = kx, we find the proportionality constant

k = F/x

K = 12500/0.85

K = 14705.8N/m.

Also, since K = 14705.8N/m, the displacement (x), when the force increases to 15500N is;

F = kx

x = F/k

x = 15500/14705.8

x = 1.05m or 105cm

6 0

3 years ago

Consider the situation||: A child pulls a sled by a rope across the lawn at a constant speed. Of the forces listed, identify whi

Mice21 [21]

Answer:

Gravitational

Tension

Normal

Friction.

Explanation:

The forces acting on the sled are:

Tension: the tension from the rope, this is the force that "moves" the sled.

Friction: kinetic friction between the sled and the ground as the sled moves.

There are another two forces that also act on the sled, but that "has no effect"

Gravitational force: This force pulls the sled down, against the floor.

Normal force: This force "opposes" to the gravitational one, so they cancel each other.

These two forces cancel each other, so they have no direct impact on the movement of the sled. BUT, the friction force depends on the weight of the moving object, and the weight of the moving object depends on the gravitational force, so we need gravitational force in order to have friction force.

Then we can conclude that the forces acting on the sled are:

Gravitational

Tension

Normal

Friction.

6 0

3 years ago

___ charges attract each other

klio [65]

Answer:

Opposite charges attract each other.

6 0

3 years ago

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