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3 years ago

Help please? thank you

Physics

1 answer:

Furkat [3]3 years ago

4 0

Hey!

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$\Large\boxed{\mathsf{D.~The~Ability~To~Do~Work}}$

In physics, energy is classified as doing some form of work. We classify energy as:

Chemical Energy
Thermal Energy
Potential Energy
Thermal Energy
Electrical Energy

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Hope This Helped! Good Luck!

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Which of the following biomes might be found at a latitude of 33° South?

cupoosta [38]

The answer is going be desert.

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3 years ago

A rocket fires two engines simultaneously. One produces a thrust of 725Ndirectly forward while the other gives a 513N thrust at

faust18 [17]

The magnitude of the resultant force, F = 1,190.3 acting at a direction X = 13.35°.

<h3>What is the resultant force the two engines exert on the rocket?</h3>

The resultant force on the rocket is calculated thus:

The 513N thrust is resolved into vertical and horizontal components;

Horizontal component: 513N cos(32.4°) = 433.14 N

Vertical component: 513N sin(32.4°) = 274.88 N

Total forward force on the rocket = 725 N + 433.14 N = 1,158.14 N

Total force at right angles:

0 + 274.88 N = 274.88 N

The resultant force (F) is then given as follows:

F² = a² + b²

F² = (1158.14 N)² + (274.88 N)²

F = √1,416,847.27

F = 1,190.3

To find the direction:

tan X 274.88 N / 1,158.14 N

X = tan⁻¹ 0.237346089419241

X = 13.35°

Therefore, the magnitude of the resultant force, F = 1,190.3 acting at a direction X = 13.35°.

In conclusion, the resultant force is obtained by resolving the forces into vertical and horizontal components.

Learn more about resultant force at: brainly.com/question/17434363

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1 year ago

Two types of energy can be either kinetic or potential energy. Identify those two types. For each type, explain why it can be ei

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Answer:

All forms of energy are either kinetic or potential. The energy associated with motion is called kinetic energy . The energy associated with position is called potential energy . Potential energy is not "stored energy".

Explanation:

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2 years ago

Read 2 more answers

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 $=0.33\ m$ .