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3 years ago

FIRST CORRECT ANSWER(A, B, C, or D) GETS BRAINLIEST.

Physics

1 answer:

Alona [7]3 years ago

5 0

Answer:

the last one

Explanation:

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Because the silt (dirt particles) in muddy water eventually settles out, the muddy water is a ______________ .

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The sensor in the torso of a crash test dummy records the magnitude znd direction of the net force acting on the dummy.If the du

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Answer : The force will be 4501.9

We can see that, two forces acting on the dummy in two different direction.

We know that, here two forces are given in perpendicular direction with each other.

We know the force is the vector addition law so, we will use the Pythagoras theorem for the resultant of the vectors

Now, the net force is

$F_{net} = \sqrt{{F_{1}}^{2}+ {F_{2}}^{2}}$

Two forces is given,

$F_{1} = 130.0 N$

$F_{2} = 4500.0 N$

Now, the net force is

$F_{net} = \sqrt{(130)^{2}+ (4500)^{2}}N$

$F_{net} = \sqrt{20266900}N$

$F_{net} = 4501.9 N$

Hence, the force will 4501.9 N

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4 years ago

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Stamina is a measurement of how long you can do an activity? Question 4 options: True False

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3 years ago

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A cylinder is closed by a piston connected to a spring of constant 1.60 103 N/m. With the spring relaxed, the cylinder is filled

Wittaler [7]

Answer:

The height is 0.247 m.

Explanation:

Given that,

Spring constant = 1.60\times10^3\ N/m[/tex]

Pressure = 1.00 atm

Temperature = 20.0°C

Suppose if the piston has a cross section area of 0.0120 m² and negligible mass, how high will it rise when the temperature is raised to 250°C

The equation of the pressure is

$P'=P_{0}+\dfrac{kh}{A}$ ...(I)

The equation of the volume is

$V'=V_{0}+Ah$ ....(II)

We need to calculate the height

Using equation of ideal gas

$\dfrac{P_{0}V}{T}=\dfrac{P'V'}{T'}$

$P'V'=P_{0}V\dfrac{T'}{T}$

Put the value of P' and V' from equation (I) and (II)

$P_{0}+\dfrac{kh}{A}\times V_{0}+Ah=P_{0}V\dfrac{T'}{T}$

Put the value into the formula

$1.013\times10^{5}+\dfrac{1.60\times10^{3}\times h}{0.0120 }\times5.00\times10^{-3}+0.0120 \times h=1.013\times10^{5}\times5.00\times10^{-3}\times\dfrac{523}{293}$

$1.013\times10^{5}\times5.00\times10^{-3}+1.013\times10^{5}\times0.0120\times h+1600h^2=904.09$

$1600h^2+1215.6h=904.09-506.5$

$1600h^2+1215.6h=397.59$

$h=0.247\ m$

Hence, The height is 0.247 m.

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3 years ago