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2 years ago

an object is moving with a speed of 35 m/s and has a kinetic energy of 1500 J, what is the mass of the object?

1 answer:

Elena-2011 [213]2 years ago

5 0

You'd use the equation kinetic energy=mass*0.5*speed^2
So you'd rearrange this to get mass =kinetic energy /0.5 *speed^2
Which is mass= 1500J/0.5*35^2
=2.44897959183673469........kg

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If your friend drops a chocolate bar to you from a height of 5.0 m above your hands,

Sladkaya [172]

Answer:

Explanation:

Using the equation of motion S = ut+1/2gt² to solve the problem where;

u is the initial velocity of the chocolate = 0m/s

t is the time taken

g is the acceleration due to gravity = 9.81m/s²

S is the height of fall = 5.0m

Substituting the given parameter into the formula to get the time t we have;

5 = 0(t)+1/2(9.81)t²

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t² = 5/4.905

t² = 1.019

t = √1.019

t = 1.009 secs

<em>Hence it will take 1.01 secs for me to catch the chocolate bar</em>

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3 years ago

A 2 kg ball is thrown upward with a velocity of 15 m/s. What is the kinetic energy of the ball as it is being thrown?

marusya05 [52]

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7 0

3 years ago

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To analyze the motion of a body that is traveling along a curved path, to determine the body's acceleration, velocity, and posit

DiKsa [7]

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$R = 350ft$

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PART A )

$a_c = \frac{V^2}{R}$

$a_c = \frac{V^2}{350}$

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$a = \sqrt{a_t^2+a_r^2}$

$5.25 = \sqrt{(1.1)^2+(\frac{v^2}{350})^2}$

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$v = r\omega$

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$\alpha = 3.1428*10^{-3}rad/s^2$

Calculate the time needed by the motorcycle to reach an acceleration of

$5.25ft/s^2$

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$t = 38.53s$

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$6.75 = \sqrt{(1.1)^2+(\frac{v^2}{350})^2}$

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PART C)

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$6.75 = \sqrt{(1.1)^2+(\frac{v^2}{350})^2}$

$45.5625 = 1.21 + \frac{v^4}{122500}$

$v=48.2796ft/s$

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3 years ago

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1123123213213242141241

schepotkina [342]

Answer:

thanks for the points

Explanation:

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