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3 years ago

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A commonly used unit of mass in the English system is the pound-mass, abbreviated lbm, where 1 lbm = 0.454 kg. What is the densi

ty of glycerine in pound-mass per cubic foot? (Assume the density of glycerine is 1.26 ✕ 103 kg/m3.)

1 answer:

Gemiola [76]3 years ago

3 0

Answer: The density of glycerine will be $0.028lbm/foot^3$

Explanation:

Density is defined as the mass contained per unit volume.

$Density=\frac{mass}{Volume}$

Given:

Density of glycerine= $1.26\times 10^3kg/m^3$

1 lbm = 0.454 kg

1 kg = $\frac{1}{0.454}=2.2lbm$

Thus $0.454kg=\frac{2.2}{1}\times 0.454=0.99$

Also $1m^3=35.3147foot^3$

Putting in the values we get:

$Density=\frac{0.99lbm}{35.3147}=0.028lbm/foot^3$

Thus density of glycerine will be $0.028lbm/foot^3$

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Answer:

change in momentum, $\Delta p=7.65 \,kg.m.s^{-1}$

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Explanation:

Given:

angle of kicking from the horizon, $\theta= 30^{\circ}$

velocity of the ball after being kicked, $v=18 m.s^{-1}$

mass of the ball, $m=0.425\, kg$

time of application of force, $t=5.3\times 10^{-2}\,s$

We know, since body is starting from the rest

$\Delta p=m.v$ .....................(1)

$\Delta p=0.425\times 18$

$\Delta p=7.65 \,kg.m.s^{-1}$

Now the components:

$\Delta p_x= 7.65\times cos 30^{\circ}$

$\Delta p_x= 6.6251 \,kg.m.s^{-1}$

similarly

$\Delta p_y= 7.65\times sin 30^{\circ}$

$\Delta p_y= 3.825 \,kg.m.s^{-1}$

also, impulse

$I=F\times t$ .........................(2)

where F is the force applied for t time.

Then from eq. (1) & (2)

$F\times t=m.v$

$F\times 5.3\times 10^{-2}= 7.65$

$F=144.3396\,N$

Now, the components

$F_x=144.3396\times cos 30^{\circ}$

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$F_y=144.3396\times sin 30^{\circ}$

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3 years ago

What is the wavelength associated with 0.113kg ball traveling with velocity of 43 m/s?

lesya [120]

Answer:

2.73×10¯³⁴ m.

Explanation:

The following data were obtained from the question:

Mass (m) = 0.113 Kg

Velocity (v) = 43 m/s

Wavelength (λ) =?

Next, we shall determine the energy of the ball. This can be obtained as follow:

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Velocity (v) = 43 m/s

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E = ½ × 0.113 × 43²

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Next, we shall determine the frequency. This can be obtained as follow:

Energy (E) = 104.4685 J

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Divide both side by 6.63×10¯³⁴

f = 104.4685 / 6.63×10¯³⁴

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Velocity (v) = 43 m/s

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Wavelength (λ) =?

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Divide both side by 15.76×10³⁴

λ = 43 / 15.76×10³⁴

λ = 2.73×10¯³⁴ m

Therefore, the wavelength of the ball is 2.73×10¯³⁴ m.

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