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3 years ago

In a isovolumetric process the is constant.

Physics

1 answer:

V125BC [204]3 years ago

3 0

Answer:

An isothermal process is a change of a system, in which the temperature remains constant: ΔT = 0.

Explanation:

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Two particles are traveling through space. At time t the first particle is at the point (−1 + t, 4 − t, −1 + 2t) and the second

Pie

Answer:

Yes, the paths of the two particles cross.

Location of path intersection = ( 1 , 2 , 3)

Explanation:

In order to find the point of intersection, we need to set both locations equal to one another. It should be noted however, that the time for each particle can vary as we are finding the point where the <u>paths</u> meet, not the point where the particles meet themselves.

So, we can name the time of the first particle $T_F$ , and the time of the second particle $T_S$ .

Setting the locations equal, we get the following equations to solve for $T_F$ and $T_S$ :

$(-1 + T_F) = (-7 + 2T_S)$ Equation 1

$(4 - T_F) = (-6 + 2T_S)$ Equation 2

$(-1 + 2T_F) = (-1 + T_S)$ Equation 3

Solving these three equations simultaneously we get:

$T_F =$ 2 seconds

$T_S =$ 4 seconds

Since, we have an answer for when the trajectories cross, we know for a fact that they indeed do cross.

The point of crossing can be found by using the value of $T_F$ or $T_S$ in the location matrices. Doing this for the first particle we get:

Location of path intersection = ( -1 + 2 , 4 - 2 , -1 + 2(2) )

Location of path intersection = ( 1 , 2 , 3)

5 0

3 years ago

A marble on a frictionless track, starting from point A in the drawing, is projected down the curved runway. (This means that th

EleoNora [17]

Answer:

v = 4.4 m / s

Explanation:

Unfortunately, the exercise scheme does not appear. Let's analyze the problem the marble leaves point A with an initial velocity, goes down and then rises to a given height where its velocity is zero, in the whole trajectory they tell us that the resistance is zero, so we can use the conservation relations of the enegy.

Starting point. Point A

Em₀ = K + U = ½ m v2 + mg y_a

point B.

Em_f = U = m g y

the energy is conserved

Em₀ = Em_f

½ m v² + mg y_a = m g y

½ m v² = m g (y -y_a)

v = $\sqrt {2g ( y - y_a)}$

In the exercise the diagram is not seen, but the height of point A must be known, suppose that y_a = 4 m

v = $\sqrt{ 2 \ 9.8 ( 5 -4)}$

v = 4.4 m / s

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3 years ago

The day VFR visibility and cloud clearance requirements to operate over the town of Cooperstown, after departing and climbing ou

pshichka [43]

Answer:

The correct option is;

C. 1 mile clear of clouds

Explanation:

Given that the indicated airspace location is at or below 700 feet AGL therefore, it is taken as being in the region of a class G airspace which covers the airspace regions from the base up to and equal to 1,200 feet beneath the class E airspace and the requirement for VFR flight for class G are 1 mile and clear of clouds.

4 0

3 years ago

How much heat does 56 L of water need to absorb to raise its temperature from 350C to 740C?

LekaFEV [45]

Answer:

300 L of water

Explanation:

7 0

3 years ago

Describe in your own words, What is insulation?

serg [7]

Insulators are the materials that are bad conductors of heat/electricity.insulation is tge pricess of covering any thing with such materials in order to prevent heat or electricity.

6 0

3 years ago