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1 year ago

The kinetic energy of a 10-kilogram object is 2000 joules. What is the speed of the object, in m/s?

Physics

1 answer:

Svetllana [295]1 year ago

6 0

Answer:

20 m/s.

Explanation:

Mass of body m: 10kg

The kinetic energy of the body is given as; $K.E= \frac{1}{2} * m * v^{2}$

Value of Kinetic energy K.E = 2000 Joules

V is the body's velocity, and M is the body's mass.

Calculating the value of velocity; $V=\frac{\sqrt{2K*E} }{m}$ $V= \frac{\sqrt{2*2000} }{10}$ = $V=20m/s$

Velocity maintained by the body having a kinetic energy of 2000J is

20 m/s.

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Savatey [412]

Answer:

The answer is "Option B".

Explanation:

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3 years ago

What is the speed of a car that travels 14 meters in 2 seconds?

mars1129 [50]

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Total time taken =2sec

Speed =distance / time

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3 years ago

In class we calculated the range of a projectile launched on flat ground. Consider instead, a projectile is launched down-slope

zysi [14]

Answer:

With an initial speed of 10m/s at an angle 30° below the horizontal, and a height of 8m, the projectile travels 7.49m horizontally before it lands.

Explanation:

Since the horizontal motion is independent from the vertical motion, we can consider them separated. The horizonal motion has a constant speed, because there is no external forces in the horizontal axis. On the other hand, the vertical motion actually is affected by the gravitational force, so the projectile will be accelerated down with a magnitude $g$ .

If we have the initial velocity $v_o$ and its angle $\theta$ , we can obtain the vertical component of the velocity $v_{oy}$ using trigonometry:

$v_{oy}=v_osin\theta$

Therefore, if we know the height at which the projectile was launched, we can obtain the final velocity using the formula:

$v_{fy}^{2} =v_{oy}^{2}+2gy\\\\ v_{fy}=\sqrt{v_{oy}^{2}+2gy }$

Next, we compute the time the projectile lasts to reach the ground using the definition of acceleration:

$g=\frac{v_{fy}-v_{oy}}{\Delta t} \\\\\Delta t= \frac{v_{fy}-v_{oy}}{g}=\frac{\sqrt{v_{oy}^{2}+2gy} -v_{oy}}{g}$

Finally, from the equation of horizontal motion with constant speed, we have that:

$x=v_{ox}\Delta t= v_{ox}\frac{\sqrt{v_{oy}^{2}+2gy} -v_{oy}}{g}$

For example, if the projectile is launched at an angle 30° below the horizontal with an initial speed of 10m/s and a height 8m, we compute:

$v_{ox}=10\frac{m}{s} cos30=8.66\frac{m}{s}\\v_{oy}=10\frac{m}{s} sin30=5\frac{m}{s}\\\\x=8.66\frac{m}{s} \frac{\sqrt{(5\frac{m}{s}) ^{2}+2(9.8\frac{m}{s^{2}})8m}-5\frac{m}{s} }{9.8\frac{m}{s^{2} } } =7.49m$

In words, the projectile travels 7.49m horizontally before it lands.

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4 years ago

Heather and Matthew take 45 s to walk eastward along a straight road to a store 72 m away. What is their average velocity?

vladimir1956 [14]

Answer:

v = 1.6 m/s

Explanation:

Given that,

Distance, d = 72 m

Time taken, t = 45 s

We need to find their average velocity. Average velocity of an object is given by total distance divided by total time taken.

$v=\dfrac{d}{t}\\\\v=\dfrac{72\ m}{45\ s}\\\\v=1.6\ m/s$

So, their average velocity is 1.6 m/s.

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4 years ago

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3 years ago