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3 years ago

9

Suppose the Pathfinder moves at a rate of 0.2m/s for 20 seconds and then turns around and travels at the same speed for 3 second

s, what is the Pathfinder's current position?

1 answer:

kotykmax [81]3 years ago

7 0

Answer:

PLEASE BE MORE SPECIFIC

Explanation:

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A certain ideal gas has molar heat capacity at constant volume CV. A sample of this gas initially occupies a volume V0 at pressu

ANTONII [103]

Answer:

Explanation:

The processes are described on the image attached below. The isobaric process consists of an horizontal line, the adiabatic expansion is described by a polytropic curve:

$P_{2} \cdot V_{2}^{\gamma} = P_{3} \cdot V_{3}^{\gamma}$

Where:

$\gamma = \frac{c_{p}}{c_{v}}$

$\gamma = 1 + \frac{R}{c_{v}}$

Final pressure is:

$P_{3} = P_{2}\cdot \left(\frac{V_{2}}{V_{3}} \right)^{\gamma}$

$P_{3} = P_{o}\cdot \left(\frac{1}{2}\right)^{\gamma}$

$P_{3} = \frac{P_{o}}{2^{\gamma}}$

8 0

3 years ago

Which of the following are electromagnetic waves?

olchik [2.2K]

Answer:

Microwaves

Radiowaves.

5 0

3 years ago

Read 2 more answers

Please view image attached!

Marat540 [252]

Answer:

you would expect a change in the graph

Explanation:

8 0

3 years ago

What fraction of all the electrons in a 25 mg water

mihalych1998 [28]

Answer:

$9.11\times 10^{-15}.$

Explanation:

The water droplet is initially neutral, it will obtain a 40 nC of charge when a charge of -40 nC is removed from the water droplet.

The charge on one electron, $\rm e=-1.6\times 10^{-19}\ C.$

Let the N number of electrons have charge -40 nC, such that,

$\rm Ne=-40\ nC\\\Rightarrow N=\dfrac{-40\ nC}{e}=\dfrac{-40\times 10^{-9}\ C}{-1.6\times 10^{-19}\ C}=2.5\times 10^{11}.$

Now, mass of one electron = $\rm 9.11\times 10^{-31}\ kg.$

Therefore, mass of N electrons = $\rm N\times 9.11\times 10^{-31}=2.5\times 10^{11}\times 9.11\times 10^{-31}=2.2775\times 10^{-19}\ kg.$

It is the mass of the of the water droplet that must be removed in order to obtain a charge of 40 nC.

Let it is m times the total mass of the droplet which is $25\ \rm mg = 25\times 10^{-6}\ kg.$

Then,

$\rm m\times (25\times 10^{-3}\ kg) = 2.2775\times 10^{-19}\ kg.\\m=\dfrac{2.2775\times 10^{-19}\ kg}{25\times 10^{-3}\ kg}=9.11\times 10^{-15}.$

It is the required fraction of mass of the droplet.

3 0

3 years ago

A tennis ball is thrown from a 25 m tall building with a zero initial velocity. At the same moment, another ball is thrown from

Nataly [62]

Answer:

The two balls meet in 1.47 sec.

Explanation:

Given that,

Height = 25 m

Initial velocity of ball= 0

Initial velocity of another ball = 17 m/s

We need to calculate the ball

Using equation of motion

$s=ut+\dfrac{1}{2}gt^2+h$

Where, u = initial velocity

h = height

g = acceleration due to gravity

Put the value in the equation

For first ball

$s_{1}=0-\dfrac{1}{2}gt^2+25$ ....(I)

For second ball

$s_{2}=17t-\dfrac{1}{2}gt^2+0$ ....(II)

From equation (I) and (II)

$-\dfrac{1}{2}gt^2+25=17t-\dfrac{1}{2}gt^2+0$

$t=\dfrac{25}{17}$

$t=1.47\ sec$

Hence, The two balls meet in 1.47 sec.

6 0

4 years ago