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3 years ago

Give me 3 events of matter and energy that you can't explain yourself ( need some ideas)

Physics

1 answer:

nalin [4]3 years ago

8 0

Answer:

Plasma, Supersolid, Superfluid

Explanation:

I don't know if this is what you are looking for, but I have no idea what these even are.

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Three thermometers are in the same water bath. After thermal equilibrium is established, it is found that the Celsius thermomete

Nikolay [14]

Answer:

OPTION A, Kelvin Thermometer is Incorrect

Explanation:

Now, if you consider best two out of three results, then celsius and Fahrenheit thermometers read the same value, meaning both are right.

1) K = °C + 273

K = 100°C + 273

k = 373°C

Kelvin Thermometer is Incorrect

2) $C = \frac{F - 32}{1.8}$

when we have 212°F

$C = \frac{212 - 32}{1.8} \\\\= 100^\circ C$

which is correct

3 0

3 years ago

A sound wave has a frequency of 300 Hz. If the wavelength is .50 m, then what is the speed of

kherson [118]

Answer:

15000 m/s

Explanation:

You just need to multiply the wavelength with the frequency.

7 0

3 years ago

Barack is playing basketball in his back yard. He takes a shot 7.0 m from the basket (measured along the ground), shooting at an

VMariaS [17]

Answer:

The time of flight of the ball is 1.06 seconds.

Explanation:

Given $\Delta x=7\ m$

$\theta=45 \°$

Also, $\Delta y=(3.5-2)=1.5\ m$

$a_x=0\ and\ a_y=-9.81\ m/s^2$

Let us say the velocity in the x-direction is $v_x$ and in the y-direction is $v_y$ . And acceleration in the x-direction is $a_x$ and in the y-direction is $a_y$ .

Also, $\Delta x\ and\ \Delta y$ is distance covered in x and y direction respectively. And $t$ is the time taken by the ball to hit the backboard.

We can write $v_x=v_0cos(45)\ and\ v_y=v_0sin(45)$ . Where $v_0$ is velocity of ball.

Now,

$\Delta x=v_x\times t+\frac{1}{2}\times a_x\times t^2\\ \Delta x=v_x\times t+\frac{1}{2}\times 0\times t^2\\\Delta x=v_xt$

$\Delta x=v_0cos(45)\times t\\7=v_0cos(45)\times t\\\\t=\frac{7}{v_0cos(45)}$

Also,

$\Delta y=v_y\times t+\frac{1}{2}\times a_y\times t^2\\ 1.5=v_0sin(45)\times \frac{7}{v_0cos(45)}+\frac{1}{2}\times (-9.81)\times(\frac{7}{v_0cos(45)} )^2\\\\1.5=7-\frac{481}{(v_0)^2}\\ \\\frac{481}{(v_0)^2}=5.5\\\\(v_0)^2=\frac{481}{5.5}\\ \\(v_0)^2=87.45\\\\v_0=\sqrt{87.45}=9.35\ m/s$ .