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3 years ago

The medium at point P will?

Physics

1 answer:

Tanya [424]3 years ago

3 0

<h2>Answer: remain stationary</h2>

Stationary waves (so called because they seem to be immobile) occur when two waves interfere with the <u>same frequency, amplitude but with different direction</u>, along a line with a phase difference of half wavelength.

In this kind of waves there are two types of points:

The nodes, which are points that remain motionless or stationary and do not vibrate. They are due to the destructive interference of both waves when they meet.

The antinodes, which are points that vibrate with a maximum vibration amplitude. They are due to the non-destructive interference of both waves.

According to this explanation and comparing it with the description, when this two waves pass through each other, the point P will become a node, hence<u> it will remain stationary</u>.

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Solve: A car travels 2 km North , 10 km East, then 3 km West. pythagorean theorem

Evgen [1.6K]

________b_____ 7 km east

| 2km north.

pythagorean theorem : ✓a² + b² = c²

c² = a² + b² = 4 + 49 = 53

c = ✓53 km

displacement = c = ✓53 km

distance = 10 + 3 + 2 = 15 km

5 0

3 years ago

The 100-m dash can be run by the best sprinters in 10.0 s. A 66-kg sprinter accelerates uniformly for the first 45 m to reach to

irga5000 [103]

(a) 154.5 N

Let's divide the motion of the sprinter in two parts:

- In the first part, he starts with velocity u = 0 and accelerates with constant acceleration $a_1$ for a total time $t_1$ During this part of the motion, he covers a distance equal to $s_1 = 45 m$ , until he finally reaches a velocity of $v_1 = u + a_1t_1$ . We can use the following suvat equation:

$s_1 = u t_1 + \frac{1}{2}a_1t_1^2$

which reduces to

$s_1 = \frac{1}{2}a_1 t_1^2$ (1)

since u = 0.

- In the second part, he continues with constant speed $v_1 = a_1 t_1$ , covering a distance of $d_2 = 55 m$ in a time $t_2$ . This part of the motion is a uniform motion, so we can use the equation

$s_2 = v_1 t_2 = a_1 t_1 t_2$ (2)

We also know that the total time is 10.0 s, so

$t_1 + t_2 = 10.0 s\\t_2 = (10.0-t_1)$

Therefore substituting into the 2nd equation

$s_2 = a_1 t_1 (10-t_1)$

From eq.(1) we find

$a_1 = \frac{2s_1}{t_1^2}$ (3)

And substituting into (2)

$s_2 = \frac{2s_1}{t_1^2}t_1 (10-t_1)=\frac{2s_1}{t_1}(10-t_1)=\frac{20 s_1}{t_1}-2s_1$

Solving for t,

$s_2+2s_1=\frac{20 s_1}{t_1}\\t_1 = \frac{20s_1}{s_2+2s_1}=\frac{20(45)}{55+2(45)}=6.2 s$

So from (3) we find the acceleration in the first phase:

$a_1 = \frac{2(45)}{(6.2)^2}=2.34 m/s^2$

And so the average force exerted on the sprinter is

$F=ma=(66 kg)(2.34 m/s^2)=154.5 N$

b) 14.5 m/s

The speed of the sprinter remains constant during the last 55 m of motion, so we can just use the suvat equation

$v_1 = u +a_1 t_1$

where we have

u = 0

$a_1 =2.34 m/s^2$ is the acceleration

$t_1 = 6.2 s$ is the time of the first part

Solving the equation,

$v_1 = 0 +(2.34)(6.2)=14.5 m/s$

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4 years ago

How is mass affected by the location of an object in space?

lorasvet [3.4K]

Answer:

We weigh more on Earth because of gravity. Gravity pulls our body towards the Earh and hence our mass is more. But in space there is no gravity and thus the mass will accordingly be less due to absence of centre of gravity. Thus, mass is affected by the location of a body or an object in space.

Explanation:

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3 years ago

I have 17 liters of air to a balloon the is 200 kelvin. If I take the balloon to a place where the temperature is 157 kelvin, wh

amm1812

Answer:

The new volume of the ballon will be 13.345 L

Explanation:

Charles's Law consists of the relationship that exists between the volume and the temperature of a certain quantity of ideal gas, which is maintained at a constant pressure, by means of a constant of proportionality that is applied directly. For a given sum of gas at a constant pressure, as the temperature increases, the volume of the gas increases and as the temperature decreases, the volume of the gas decreases.

Charles's law is a law that says that when the amount of gas and pressure are kept constant, the quotient that exists between the volume and the temperature will always have the same value:

$\frac{V}{T}=k$

It is possible to assume that you have a certain volume of gas V1 that is at a temperature T1 at the beginning of the experiment. If you vary the volume of gas to a new value V2, then the temperature will change to T2, and it will be true:

$\frac{V1}{T1}=\frac{V2}{T2}$