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Nataly_w [17]
3 years ago
11

Which ocean borders the west coast of this United States

Physics
1 answer:
Gennadij [26K]3 years ago
3 0
The Pacific.
Hope it helps!
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A child is riding in a wagon. What reference frame might have been used if an observer said the child was not moving.
GarryVolchara [31]

Answer:

the wagon should be used as frame of reference if an observer said the child was not moving.

Explanation:

The state of motion of a body depends upon the frame of reference. It is the set of co-ordinates according to which the motion is analyzed. If a child is riding in a wagon, then he will be considered in motion to a person standing outside the wagon. Hence, if we take a frame of reference outside the wagon then the child must be in motion with respect to the observer. On the other hand if the observer is inside the wagon, then the child must be in rest with respect to the observer. Hence, if we take the wagon to be the frame of reference, then the child will be at rest with respect to the observer.

<u>Therefore, the wagon should be used as frame of reference if an observer said the child was not moving.</u>

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What is matter explain verifly​
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Answer:

<u>Matter is a substance that has inertia and occupies physical space. According to modern physics, matter consists of various types of particles, each with mass and size.Matter can exist in several states, also called phases. The three most common states are known as solid, liquid and gas.</u><u>Matter is the Stuff Around You</u><u> </u><u>o</u><u>r</u><u> </u><u>Atoms and compounds are all made of very small parts of matter. Those atoms go on to build the things you see and touch every day. Matter is defined as anything that has mass and takes up space (it has volume).</u><u>Solid ice, water and steam are few examples of matter touched in everyday life. Subatomic particles are also considered as matter.</u>

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Leyes de newton primera ley
Nadusha1986 [10]

Answer:

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5 0
3 years ago
A frictionless piston–cylinder device contains 7 kg of nitrogen at 100 kPa and 250 K. Nitrogen is now compressed slowly accordin
lora16 [44]

Answer: -1038.8 kJ

Explanation:

From the question, we can see that PV^n = constant. And as such, we can deduce that it is a polytropic process. Thus, we can use the polytropic work equation to calculate the needed work input.

from the question we were given

Mass of nitrogen, m = 7kg

initial temperature, T1 = 250k

Final temperature, T2 = 450k

Polytropic index, n = 1.4

Specific gas constant, R = 0.2968kJ/kgK

W = [p2 * v2 - p1 * v1] / 1 - n

W = [m * R * T2 - T1] / 1 - n

W = 7*0.2968*(450 - 250)] / 1 - 1.4

W = [7*0.2968*200] / -0.4

W = 415.52 / -0.4

W = -1038.8 kJ

5 0
3 years ago
Water is leaking out of an inverted conical tank at a rate of 10,500 cm3/min at the same time that water is being pumped into th
satela [25.4K]

The tank has a volume of \dfrac\pi3R^2H, where H=6\,\rm m is its height and R=\dfrac d2=2\,\rm m is its radius.

At any point, the water filling the tank and the tank itself form a pair of similar triangles (see the attached picture) from which we obtain the following relationship:

\dfrac26=\dfrac rh\implies r=\dfrac h3

The volume of water in the tank at any given time is

V=\dfrac\pi3r^2h

and can be expressed as a function of the water level alone:

V=\dfrac\pi3\left(\frac h3\right)^2h=\dfrac\pi{27}h^3

Implicity differentiating both sides with respect to time t gives

\dfrac{\mathrm dV}{\mathrm dt}=\dfrac\pi9h^2\,\dfrac{\mathrm dh}{\mathrm dt}

We're told the water level rises at a rate of \dfrac{\mathrm dh}{\mathrm dt}=20\,\frac{\rm cm}{\rm min} at the time when the water level is h=2\,\mathrm m=200\,\mathrm{cm}, so the net change in the volume of water \dfrac{\mathrm dV}{\mathrm dt} can be computed:

\dfrac{\mathrm dV}{\mathrm dt}=\dfrac\pi9(200\,\mathrm{cm})^2\left(20\,\dfrac{\rm cm}{\rm min}\right)=\dfrac{800,000\pi}9\,\dfrac{\mathrm{cm}^3}{\rm min}

The net rate of change in volume is the difference between the rate at which water is pumped into the tank and the rate at which it is leaking out:

\dfrac{\mathrm dV}{\mathrm dt}=(\text{rate in})-(\text{rate out})

We're told the water is leaking out at a rate of 10,500\,\frac{\mathrm{cm}^3}{\rm min}, so we find the rate at which it's being pumped in to be

\dfrac{800,000\pi}9\,\dfrac{\mathrm{cm}^3}{\rm min}=(\text{rate in})-10,500\,\dfrac{\mathrm{cm}^3}{\rm min}

\implies\text{rate in}\approx289,753\,\dfrac{\mathrm{cm}^3}{\rm min}

4 0
3 years ago
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