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3 years ago

13

Who expressed particles by wave equations?

1 answer:

ehidna [41]3 years ago

7 0

I believe it was famous quantum physicist <span>Erwin Schrödinger
</span>

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If the action force is a player kicking a soccer ball, then what is the reaction force?

LiRa [457]

The force acting on his feet.

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3 years ago

The rate at which energy is transferred is called a. joules. b. power. c. work. d. time.

Marrrta [24]

B. Power
I hope this helps you, have a great day!

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3 years ago

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Time-outs are ineffective because

maria [59]

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4 years ago

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The specific heat of soil is 0.20 kcal/kg*C and the specific heat of water is 1.00 kcal/kg*C. This means that if 1 kg of soil an

stiv31 [10]

Answer: The soil will be $4\°C$ warmer than the water.

Explanation:

The heat (thermal energy) absorbed can be found using the following equation:

$Q=m.C.\Delta T$

Where:

$Q$ is the heat

$m$ is the mass of the element

$C$ is the specific heat capacity of the material.

$\Delta T$ is the variation in temperature

<u>In the case of soil we have:</u>

$Q_{soil}=m_{soil}.C_{soil}.\Delta T_{soil}$ (1)

Where:

$Q_{soil}=1 kcal$

$m_{soil}=1 kg$

$C_{soil}=0.2 kcal/kg \°C$

$\Delta T_{soil}$

<u>In the case of water we have:</u>

$Q_{water}=m_{water}.C_{water}.\Delta T_{water}$ (2)

Where:

$Q_{water}=1 kcal$

$m_{water}=1 kg$

$C_{water}=1 kcal/kg \°C$

$\Delta T_{water}$

Isolating $\Delta T$ from both equations:

$\Delta T_{soil}=\frac{Q_{soil}}{m_{soil}.C_{soil}}$ (3)

$\Delta T_{soil}=\frac{1 kcal}{1 kg(0.2 kcal/kg \°C)}$

$\Delta T_{soil}=5\°C$ (4)

$\Delta T_{water}=\frac{Q_{water}}{m_{water}.C_{water}}$ (5)

$\Delta T_{water}=\frac{1 kcal}{1 kg(1 kcal/kg \°C)}$

$\Delta T_{water}=1\°C$ (6)

Comparing (4) and (6) we can find the soil will be $4\°C$ warmer than the water.

8 0

4 years ago

The closest stars are 4 light years away from us. How far away must you be from a 854 kHz radio station with power 50.0 kW for t

Lubov Fominskaja [6]

Answer:

The distance from the radio station is 0.28 light years away.

Solution:

As per the question:

Distance, d = 4 ly

Frequency of the radio station, f = 854 kHz = $854\times 10^{3}\ Hz$

Power, P = 50 kW = $50\times 10^{3}\ W$

$I_{p} = 1\ photon/s/m^{2}$

Now,

From the relation:

P = nhf

where

n = no. of photons/second

h = Planck's constant

f = frequency

Now,

$n = \frac{P}{hf} = \frac{50\times 10^{3}}{6.626\times 10^{- 34}\times 854\times 10^{3}} = 8.836\times 10^{31}\ photons/s$

Area of the sphere, A = $4\pi r^{2}$

Now,

Suppose the distance from the radio station be 'r' from where the intensity of the photon is $1\ photon/s/m^{2}$

$I_{p} = \frac{n}{A} = \frac{n}{4\pi r^{2}}$

$1 = \frac{8.836\times 10^{31}}{4\pi r^{2}}$

$r = \sqrt{\frac{8.836\times 10^{31}}{4\pi}} = 2.65\times 10^{15}\ m$

Now,

We know that:

1 ly = $9.4607\times 10^{15}\ m$

Thus

$r = \frac{2.65\times 10^{15}}{9.4607\times 10^{15}} = 0.28\ ly$

5 0

3 years ago