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2 years ago

An object is dropped from a 15 m ledge. How fast it is moving just before it hits the ground?

Physics

2 answers:

loris [4]2 years ago

7 0

Answer:

$v_f = 17.15 m/s$

Explanation:

Since the ball is dropped from the height

h = 15 m

so here initial speed of the ball is

$u = 0$

acceleration of the ball due to gravity is given as

$a = 9.8 m/s^2$

now the final speed of the ball is given as

$v_f^2 - v_i^2 = 2 a d$

$v_f^2 - 0 = 2(9.8)(15)$

$v_f = 17.15 m/s$

Sergeeva-Olga [200]2 years ago

4 0

By v^2 = u^2 + 2gh
v^2 = 0 + 2 x 9.8 x 15
v = √294
v = 17.15 m/s

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Scorpion4ik [409]

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A 13-cm-diameter cd has a mass of 25 g . part a what is the cd's moment of inertia for rotation about a perpendicular axis throu

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2 years ago

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A circuit node has current ia entering and currents ib and ic exiting, where it is known that ia=5 ma and ib=9 ma. what is curre

Rudik [331]

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2 years ago

The energy from 0.015 moles of octane was used to heat 250 grams of water. The temperature of the water rose from 293.0 K to 371

arsen [322]

Answer : The correct option is, (B) -5448 kJ/mol

Explanation :

First we have to calculate the heat required by water.

$q=m\times c\times (T_2-T_1)$

where,

q = heat required by water = ?

m = mass of water = 250 g

c = specific heat capacity of water = $4.18J/g.K$

$T_1$ = initial temperature of water = 293.0 K

$T_2$ = final temperature of water = 371.2 K

Now put all the given values in the above formula, we get:

$q=250g\times 4.18J/g.K\times (371.2-293.0)K$

$q=81719J$

Now we have to calculate the enthalpy of combustion of octane.

$\Delta H=\frac{q}{n}$

where,

$\Delta H$ = enthalpy of combustion of octane = ?

q = heat released = -81719 J

n = moles of octane = 0.015 moles

Now put all the given values in the above formula, we get:

$\Delta H=\frac{-81719J}{0.015mole}$

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Therefore, the enthalpy of combustion of octane is -5448 kJ/mol.

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2 years ago

A disk is spinning about its center with a constant angular speed at first. Let the turntable spin faster and faster, with const

hoa [83]

Answer:

4 (please see the attached file)

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This acceleration is related with the tangential acceleration, by this expression:

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