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3 years ago

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A force of 5 n produces an acceleration of 8m/s2 in mass m1 and an acceleration of 24 m/s2 in mass m2 .what acceleration would i

t give if both the masses are tied together ?

1 answer:

Inessa [10]3 years ago

4 0

Acceleration of the both masses tied together= 6m/s²

Explanation:

The force is given by F= ma

so 5= m1 (8)

m1=0.625 Kg

for m2

5=m2 (24)

m2=0.208 kg

now total mass= m1+m2=0.625+0.208

Total mass=M=0.833 Kg

now F= ma

5= 0.833 (a)

a= 5/0.833

a=6m/s²

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Answer:

a) $(dP)_{v} = 9.692 kPa$

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c) dP = 0 Pa

Explanation:

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Note that:

$dP = \frac{R}{v} dT - \frac{RT}{v^{2} } dV$

1% increase in temperature at specific volume:

$dT = \frac{0.01}{1} *350\\dT = 3.5 K$

a) Change in pressure of helium at constant volume:

$(dP)_{v} = \frac{R}{v} dT$

R = 2.0769 kJ/kg-K

dT = 3.5 K

v = 0.75 m³/kg

$(dP)_{v} = \frac{2.0769}{0.75} * 3.5\\(dP)_{v} = 9.692 kPa$

b)

dv = (1%/100%) *0.75

dv = 0.0075 m³/kg

Change in pressure of helium at constant temperature:

$(dP)_{T} = \frac{-RT}{v^{2} } dv$

R = 2.0769 kJ/kg-K

T = 350 K

v = 0.75 m³/kg

dv = 0.0075 m³/kg

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c) The change in pressure of helium :

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dP = 0

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4 years ago

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Answer:

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Answer:

the ball bounces during 2.633s.

Explanation:

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equaling Y to zero to find the time we have:

$0=h-\frac{g*t^{2} }{2}$

and then:

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and for the next one:

[ $t=\sqrt{\frac{2(h/4)}{g} }$

and so on, so the total time will be an infinty serie, which will look like this:

$t=\sqrt{\frac{2}{g}*h }+2\sqrt{\frac{2}{g}*\frac{h}{2}}+2\sqrt{\frac{2}{g}*\frac{h}{4}}+2\sqrt{\frac{2}{g}*\frac{h}{8}}...$

factorizing, and replacing the value of h we have:

$t=\sqrt{\frac{2}{g} }(1+2*\sqrt{\frac{1}{2} }+2*\sqrt{\frac{1}{2} } ^{2}+2*\sqrt{\frac{1}{2} } ^{3}+... )$

here we have to write 1=2-1 to complete the serie:

$t=\sqrt{\frac{2}{g} }(-1+2*\sqrt{\frac{1}{2} } ^{0}+2**\sqrt{\frac{1}{2} } ^{1}+2*\sqrt{\frac{1}{2} } ^{2}+2*\sqrt{\frac{1}{2} } ^{3}+... )$

and this is a geometric serie, of the form sum from n=0 to n=infinity of $a*r^{n}$ where a=2 and $r=\sqrt{\frac{1}{2} }$ .

The solution for this serie is well known and it is $\frac{a}{1-r}$

which gives: $\frac{a}{1-r}=\frac{2}{1-\sqrt{\frac{1}{2} } }=6.828$

and replacing this we have:

$t=\sqrt{\frac{2}{g} }(-1+6.828 )$

replacing g and solving:

$t=\sqrt{\frac{2}{9.8} }(5.828 )=2.633s$

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3 years ago

Is defined as a state of balance<br> A weight<br> B mass<br> C equilibrium

jonny [76]

Answer:

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Mass is the amount of particles of an object.

But equilibrium describes the state of balance.

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3 years ago

You and your friend Peter are putting new shingles on a roof pitched at 20degrees . You're sitting on the very top of the roof w

Anit [1.1K]

Answer:

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Explanation:

Newton's second law of the box:

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∑F : algebraic sum of the forces in Newton (N)

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a : acceleration in meters over second square (m/s²)

Known data

m=2.1 kg mass of the box

d= 5.4m length of the roof

θ = 20° angle θ of the roof with respect to the horizontal direction

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g = 9.8 m/s² : acceleration due to gravity

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We define the x-axis in the direction parallel to the movement of the box on the roof and the y-axis in the direction perpendicular to it.

W: Weight of the box : In vertical direction

N : Normal force : perpendicular to the direction the roof

fk : Friction force: parallel to the direction to the roof

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x-y weight components

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N-Wy= 0

N=Wy =19.34 N

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We apply the formula (1) to calculated acceleration of the block:

∑Fx = m*ax , ax= a : acceleration of the block

Wx-f = ( 2.1)*a

7.039 - 9.86 = ( 2.1)*a

-2.821 = ( 2.1)*a

a=(-2.821) /( 2.1)

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Kinematics of the box

Because the box moves with uniformly accelerated movement we apply the following formula to calculate the final speed of the block :

vf²=v₀²+2*a*d Formula (2)

Where:

d:displacement = 5.4 m

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vf: final speed = 0

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We replace data in the formula (2)

0²=v₀²+2*(-1.34)*(5.4)

2*(1.34)*(5.4)= v₀²

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3 years ago