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3 years ago

Help please its 30 marks thanks for willing to help me

Physics

1 answer:

andrew11 [14]3 years ago

6 0

Answer:

Independent variable: Height

Dependent variable: time

Explanation:

Not sure what the experiment was, but assuming that this was the amount of time for a ball to reach the ground, the independent variable would be height since you determine and change the initial height. The dependent variable would be time since it changes with height, and you have to measure it.

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A town is considering building a biodiesel power plant that would burn biomass to generate electricity. Which of the following c

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Answer: it reduces dependence on fossil fuels. Disadvantage: it would emit pollution into the air.

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4 years ago

Newton’s law of universal gravitation stars that every object in the universe attracts every other object

LenaWriter [7]

Yes it does ! Uh huh. Right you are. Truer words are seldom written.

You have quoted the law quite accurately but also incompletely.

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3 years ago

Find p1, the gauge pressure at the bottom of tube 1. (Gauge pressure is the pressure in excess of outside atmospheric pressure.)

Taya2010 [7]

Answer:

(a). The gauge pressure at the bottom of tube 1 is $P_{1}=\rho g h_{1}$

(b). The speed of the fluid in the left end of the main pipe $\sqrt{\dfrac{2g(h_{2}-h_{1})}{(1-(\gamma)^2)}}$

Explanation:

Given that,

Gauge pressure at bottom = p₁

Suppose, an arrangement with a horizontal pipe carrying fluid of density p . The fluid rises to heights h1 and h2 in the two open-ended tubes (see figure). The cross-sectional area of the pipe is A1 at the position of tube 1, and A2 at the position of tube 2.

Find the speed of the fluid in the left end of the main pipe.

(a). We need to calculate the gauge pressure at the bottom of tube 1

Using bernoulli equation

$P_{1}=\rho g h_{1}$

(b). We need to calculate the speed of the fluid in the left end of the main pipe

Using bernoulli equation

Pressure for first pipe,

$P_{1}=\rho gh_{1}$ .....(I)

Pressure for second pipe,

$P_{2}=\rho gh_{2}$ .....(II)

From equation (I) and (II)

$P_{2}-P_{1}=\dfrac{1}{2}\rho(v_{1}^2-v_{2}^2)$

Put the value of P₁ and P₂

$\rho g h_{2}-\rho g h_{1}=\dfrac{1}{2}\rho(v_{1}^2-v_{2}^2)$

$gh_{2}-gh_{1}=\dfrac{1}{2}(v_{1}^2-v_{2}^2)$

$2g(h_{2}-h_{1})=v_{1}^2-v_{2}^2$ ....(III)

We know that,

The continuity equation

$v_{1}A_{1}=v_{2}A_{2}$

$v_{2}=v_{1}(\dfrac{A_{1}}{A_{2}})$

Put the value of v₂ in equation (III)

$2g(h_{2}-h_{1})=v_{1}^2-(v_{1}(\dfrac{A_{1}}{A_{2}}))^2$

$2g(h_{2}-h_{1})=v_{1}^2(1-(\dfrac{A_{1}}{A_{2}}))^2$

Here, $\dfrac{A_{1}}{A_{2}}=\gamma$

So, $2g(h_{2}-h_{1})=v_{1}^2(1-(\gamma)^2)$

$v_{1}=\sqrt{\dfrac{2g(h_{2}-h_{1})}{(1-(\gamma)^2)}}$

Hence, (a). The gauge pressure at the bottom of tube 1 is $P_{1}=\rho g h_{1}$

(b). The speed of the fluid in the left end of the main pipe $\sqrt{\dfrac{2g(h_{2}-h_{1})}{(1-(\gamma)^2)}}$

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3 years ago

Which of the following remains constant for a satellite in an elliptical orbit.

Softa [21]

If the satellite doesn't have little rocket engines or other thrusters on it, AND it stays far enough from Earth that it doesn't have to plow through any air molecules, AND no pieces break off of it and drift away, AND there are no hamsters inside it running on treadmills connected to external thrusters, then there's no way for it to gain or lose energy, and its total energy remains constant.

Some of its energy is always changing, either from potential to kinetic or from kinetic to potential, as its distance from Earth changes. But the total stays constant.

5 0

3 years ago

serious [3.7K]

<span>1 pound is equal to .454 kilograms.
115 pounds equals
115 x .454 kilograms = 52.163 kg.

The total dose is equal to the dose per kilograms multiplied by the total mass in kilograms.

6.00mg/kg x 52.163kg = 312.978mg.

The total dose is 313.0 mg</span>

3 0

3 years ago