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Amiraneli [1.4K]

1 year ago

7

A major-league pitcher throws a baseball toward home plate. The ball rotates at 1560 rpm, and it travels the 18.5 meters to the

plate at an average translational speed of 40.2 m/s. How many revolutions does the ball make during this trip?

1 answer:

DedPeter [7]1 year ago

4 0

Answer:

no of revolutions = 56.62 rev

Explanation:

given data

rotates = 1560 rpm

travels distance = 18.5 meters

speed = 40.2 m/s

solution

first we convert here rev/min to rev/sec

rotates = 1560 rpm = 1560 × 0.0167 = 26.052 rev/sec

we get here no of revolutions that is

no of revolutions = spin rotate ÷ time ..........1

here time = distance ÷ speed

time = 18.5 ÷ 40.2

time = 0.4601 s

put value in equation 1 we get

no of revolutions = $\frac{26.052}{0.4601}$

no of revolutions = 56.62 rev

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2 years ago

An object initially at rest experiences a constant horizontal acceleration due to the action of a resultant force applied for 10

Marianna [84]

Answer:

a = 18.28 ft/s²

Explanation:

given,

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mass of the object = 15 lb

acceleration, a = ? ft/s²

1 btu = 778.15 ft.lbf

10 btu = 7781.5 ft.lbf

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now,

work done is equal to change in kinetic energy

$W = \dfrac{1}{2} m (v_f^2-v_i^2)$

$7781.5 = \dfrac{1}{2}\times 0.466\times v_f^2$

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now, acceleration of object

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1 year ago

A 0.468 g sample of pentane, C 5H 12, was burned in a bomb calorimeter. The temperature of the calorimeter and 1.00 kg of water

Fittoniya [83]

Answer:

$Q_{lost}$ per mole of pentane = 3157.53 kJ/mol

Explanation:

Given:

Mass of pentane, m = 0.468 gram

Molar mass of pentane, M = 72.15

Now, mol of pentane, n = mass/M = 0.468/72.15 = 0.00648 mol of C5H12

Now,

ΔT = 23.65 - 20.45 = 3.2°C

Heat capacity of the calorimeter, C = 2.21 kJ/°C

Specific heat capacity of the water, Cp = 4.184 J/g.°C

Now,

the heat gained = the heat lost

$Q_{gained} = -Q_{lost}$

also,

$Q_{gained} = Q_{water} + Q_{calorimeter}$

$Q_{water} = m\times C\times(T_f-T_i)$

or

$Q_{water} = 1000\times4.184\times(23.65-20.45) = 13388.8\ J$

and

$Q_{calorimeter} = C\times\Delta T = 2.21\times1000\times3.2 = 7072\ J$

Now,

$Q_{total} = 13388.8 +7072 = 20460.8\ J$

we have,

$Q_{lost} = -Q_{gain} = - 20460.8\ J$ (Here negative sign depicts the release of the heat)

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2 years ago

Three balls with the same radius 21 cm are in water. Ball 1 floats, with

Masteriza [31]

Answer:

Explanation:

A )

The ball floats with half of it exposed above the water level . So it must have density half that of water . In other words its density must have been 500 kg / m³

B )

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2 years ago

A 4 kg box is on a frictionless 35° slope and is connected via a massless string over a massless, frictionless pulley to a hangi

Anarel [89]

Answer:

(a) 19.62 N

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(c) 24.43 N

Explanation:

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2 Kg box causes tension

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(c)

Acceleration a= $\frac {forward force-backward force}{Total mass}= \frac {mg sin \theta -mg}{m1 + m2}$

$a= \frac {22.51-19.62}{2+4}=0.48$

When moving, the box will exert force T"= $mgsin \theta + ma$

T"= 4*9.81 sin 35 +(4*0.48)= 24.43 N

7 0

2 years ago

Read 2 more answers