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Mariulka [41]
3 years ago
5

When considering that in the past human societies developed in greater isolation from one another than today, each of the follow

ing statements is true EXCEPT:
Physics
1 answer:
galina1969 [7]3 years ago
3 0

Answer:One can easily assume a direct cause and effect relationship between a physical environment and an aspect of culture

Explanation: During the prehistoric and the earlier centuries human societies developed in isolation, there's no interconnection or much of communication between different groups and societies.

Knowledge sharing is not prominent, but in recent times people are more connected to each other,communities and countries interaction takes place through different forums, during the earlier centuries there are no mobile communication equipments,no Television set or the level of sophistication as it is today.

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The answer is wind forces and Earth’s rotation
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Describe the initial horizontal and vertical velocity of a horizontally launched projectile on Earth, as well as what happens to
Jobisdone [24]

Answer:

Explained below

Explanation:

To explain this, let's consider a tennis ball being launched from the top of a very high building.

Now, if the tennis ball is launched horizontally without any upward angle but with an initial velocity of 10 m/s. In this motion, If there is no gravity, the tennis ball would continue in motion at that same speed of 10 m/s in the horizontal direction. However, in reality, gravity causes the tennis ball to accelerate downwards at a rate of 9.8 m/s for every second. This implies that the vertical velocity component is changing at the rate of 9.8 m/s every second.

Thus, after 1 second, horizontal velocity component will remain 10 m/s and vertical component will be 9.8 m/s × 1 = 9.8 m/s downwards.

Also, after 2 seconds, the vertical velocity component will remain 10 m/s, however the vertical component will now be 9.8 × 2 = 19.6 m/s downwards.

Same procedure is repeated as t increases by 1 second.

5 0
3 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.

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.

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