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3 years ago

Explain ALL THREE of Newton’s Laws in your own words.

Physics

1 answer:

Svet_ta [14]3 years ago

5 0

Explanation:

(1) Every object moves in a straight line unless acted upon by a force. (2) The acceleration of an object is directly proportional to the net force exerted and inversely proportional to the object's mass. (3) For every action, there is an equal and opposite reaction.

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What is the energy of a rock with a mass of 10.2 kg on a cliff that is 300 m height?

Anuta_ua [19.1K]

The potential energy of the rock is 30,000 J

Explanation:

The mechanical energy of an object is equal to the sum of its gravitational potential energy (PE) and its kinetic energy (KE):

$E=PE+KE$

where

PE is the gravitational potential energy, which is the energy possessed by the object due to its position in the gravitational field

KE is the kinetic energy, which is the energy possessed by the object due to its motion

In this problem, the rock is at rest, so its kinetic energy is zero:

KE = 0

Therefore, the energy of the rock is just equal to its potential energy, which is:

$E=PE=mgh$

where

m = 10.2 kg is the mass of the rock

$g=9.8 m/s^2$ is the acceleration of gravity

h = 300 m is the height of the rock above the ground

Substituting and solving, we find

$PE=(10.2)(9.8)(300)=30,000 J$

Learn more about potential energy:

brainly.com/question/1198647

brainly.com/question/10770261

#LearnwithBrainly

4 0

3 years ago

You launch a cannonball at an angle of 35° and an initial velocity of 36 m/s (assume y = y₁=

velikii [3]

Answer:

Approximately $4.2\; {\rm s}$ (assuming that the projectile was launched at angle of $35^{\circ}$ above the horizon.)

Explanation:

Initial vertical component of velocity:

$\begin{aligned}v_{y} &= v\, \sin(35^{\circ}) \\ &= (36\; {\rm m\cdot s^{-1}})\, (\sin(35^{\circ})) \\ &\approx 20.6\; {\rm m\cdot s^{-1}}\end{aligned}$ .

The question assumed that there is no drag on this projectile. Additionally, the altitude of this projectile just before landing $y_{1}$ is the same as the altitude $y_{0}$ at which this projectile was launched: $y_{0} = y_{1}$ .

Hence, the initial vertical velocity of this projectile would be the exact opposite of the vertical velocity of this projectile right before landing. Since the initial vertical velocity is $20.6\; {\rm m\cdot s^{-1}}$ (upwards,) the vertical velocity right before landing would be $(-20.6\; {\rm m\cdot s^{-1}})$ (downwards.) The change in vertical velocity is:

$\begin{aligned}\Delta v_{y} &= (-20.6\; {\rm m\cdot s^{-1}}) - (20.6\; {\rm m\cdot s^{-1}}) \\ &= -41.2\; {\rm m\cdot s^{-1}}\end{aligned}$ .

Since there is no drag on this projectile, the vertical acceleration of this projectile would be $g$ . In other words, $a = g = -9.81\; {\rm m\cdot s^{-2}}$ .

Hence, the time it takes to achieve a (vertical) velocity change of $\Delta v_{y}$ would be:

$\begin{aligned} t &= \frac{\Delta v_{y}}{a_{y}} \\ &= \frac{-41.2\; {\rm m\cdot s^{-1}}}{-9.81\; {\rm m\cdot s^{-2}}} \\ &\approx 4.2\; {\rm s} \end{aligned}$ .

Hence, this projectile would be in the air for approximately $4.2\; {\rm s}$ .

8 0

1 year ago

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A roller coaster's velocity at the top of the hill is 10 m/s. Two seconds later it reaches the bottom of the hill with a velocit

svet-max [94.6K]

your answer is 8 m / sec²

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2 years ago

Amount of pressure of liquid increases with ?

Pachacha [2.7K]

Answer: Pressure increases as the depth increases.

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2 years ago

Read 2 more answers

1) mechanical

Dafna1 [17]

I am pretty sure about these answers.Thermal goes in the 4th blank.
Mechanical goes in the 2nd blank.
Electrical goes in the 3rd blank.
I think chemical goes in the 1st blank and light goes in the 5th blank
Hope this helps

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3 years ago