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3 years ago

A 700- kg vehicle is traveling at the speed of 6m/s how much kinetic energy does it have ???

Physics

2 answers:

Sedbober [7]3 years ago

8 0

Kinetic energy = 1/2 x mass x velocity²

Kinetic energy = 1/2 x 700 x 6²

Kinetic energy = 12600 J

lutik1710 [3]3 years ago

6 0

<h2>Answer:</h2>

The kinetic energy will be 12600 J

<h2>Explanation:</h2>

As we know that

Kinetic energy = 1/2 (m) (v)²

By putting the values

Kinetic energy = 1/2 x 700 x 6²

We get

Kinetic energy = 12600 J

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A ballast is dropped from a stationary hot-air balloon that is at an altitude of 576 ft. Find (a) an expression for the altitude

Nadya [2.5K]

Answer:

Explanation:

If a ball dropped from a stationary hot-air balloon that is at an altitude of 576 ft, an expression for the altitude of the ballast after t seconds can be expressed using the equation of motion;

$S = ut + \frac{1}{2}at^{2}$

S is the altitude of the ballest

u is the initial velocity

a is the acceleration of the body

t is the time taken to strike the ground

Since the body is dropped from a stationary air balloon, the initial velocity u will be zero i.e u = 0m/s

Also, since the ballast is dropped from a stationary hot-air balloon, the body is under the influence of gravity, the acceleration will become acceleration due to gravity i.e a = +g

Substituting this values into the equation of the motion;

$S = 0 + \frac{1}{2}gt^2\\ S = \frac{1}{2}gt^2\\$

a) An expression for the altitude of the ballast after t seconds is therefore

$S = \frac{1}{2}gt^2\\$

b) Given S = 576ft and g = 32ft/s², substituting this into the formula in (a);

$576 = \frac{1}{2}(32)t^2\\\\\\576*2 = 32t^2\\1152 = 32t^2\\t^2 = \frac{1152}{32} \\t^2 = 36\\t = \sqrt{36}\\ t = 6.0secs$

This means that the ballast strikes the ground after 6secs

c) To get the velocity when it strikes the ground, we will use the equation of motion v = u + gt.

v = 0 + 32(6)

v = 192ft

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3 years ago

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Answer:

The frequency of a wave is inversely proportional to its wavelength. That means that waves with a high frequency have a short wavelength, while waves with a low frequency have a longer wavelength.

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The greatest height reported for a jump into an airbag is 99.4 m by stuntman Dan Koko. In 1948 he jumped from rest from the top

vaieri [72.5K]

Answer:

v = 44,16 m/s

Explanation:

We will fixate our reference in the starting point from where Dan jumped of, at the top of the Casino. Therefore, the displacement made when dan reached the airbag would be of y= -99,4 m viewed from our reference. We describe the motion of dan with the equation:

$v_y^2 =v_0^2 +2ay$

Dan jumped from the rest, that means that the initial velocity v_0=0, therefore:

$v_y^2 =2ay \rightarrow v_y = \sqrt{2ay}$

Since Dan is moving in the negative axis regarding our reference point, we take the negative root of the equation.

v_y=-√(2*(-9,81 m/s^2 )*(-99,4 m) )=44,1613 m/s $v_y =- \sqrt{2*(-9,81 m/s^2)*(-99,4 m)} = 44,1613 m$

So, if we don’t take the air resistance into account, Dan would have achieved an velocity of 44,16 m/s when he reached the airbag.

I hope everything was clear with my explanation. If you need anything else, let me know. Have a great day :D

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4 years ago

Is it true or false that the displacement always equals the product of the average velocity and the time interval?

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Answer:

True.

Explanation:

Applying the definition of average velocity, we know that we can always write the following expression:

$v_{avg} = \frac{\Delta x}{\Delta t}$ (1)

By definition, Δx is just the displacement, and Δt is the time interval.
So, just rearranging terms in (1), we get:

$\Delta x} = v_{avg}* {\Delta t}$

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3 years ago