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Step2247 [10]
3 years ago
5

What other forces affect the soccer ball after the student's foot stops touching it?

Physics
2 answers:
hodyreva [135]3 years ago
8 0
Gravity, acceleration, velocity
svet-max [94.6K]3 years ago
4 0
Gravity I hope this helps you I'm not very good at this kind of stuff
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The Earth's place in space can be described as _________.
Olin [163]
The answer is B because if you use process of elimination, you find that A is invalid because Venus is the second planet. C is out because Mars is the 4th planet. D is out because we are nowhere  near the Andromeda Galaxy. We are millions of light years away.
4 0
3 years ago
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The radioactive 60co isotope is used in nuclear medicine to treat certain types of cancer. Calculate the wavelength and frequenc
Ivanshal [37]

1. Frequency: 3.23\cdot 10^{20} Hz

The energy given is the energy per mole of particles:

E=1.29\cdot 10^{11} J/mol

1 mole contains a number of Avogadro of particles, N_A, equal to

N_A=6.022\cdot 10^{23} particles

So, by setting the following proportion, we can calculate the energy of a single photon:

1.29 \cdot 10^{11} J/mol : 6.022 \cdot 10^{23} ph/mol = E_1 : 1 ph\\E_1 = \frac{(1.29\cdot 10^{11} J/mol)(1 ph)}{6.022\cdot 10^{23} ph/mol}=2.14\cdot 10^{-13} J

This is the energy of a single photon; now we can calculate its frequency by using the formula:

E_1 = hf

where

h=6.63\cdot 10^{-34} Js is the Planck's constant

f is the photon frequency

Solving for f, we find

f=\frac{E_1}{h}=\frac{2.14\cdot 10^{-13} J}{6.63\cdot 10^{-34} Js}=3.23\cdot 10^{20} Hz

2. Wavelength: 9.29\cdot 10^{-13} m

The wavelength of the photon is given by the equation:

\lambda=\frac{c}{f}

where

c=3\cdot 10^8 m/s

is the speed of the photon (the speed of light). Substituting,

\lambda=\frac{3 \cdot 10^8 m/s}{3.23\cdot 10^{20} Hz}=9.29\cdot 10^{-13} m

6 0
3 years ago
How does light from a headlamp use a lens and a mirror to produce a narrow beam?
Kryger [21]
The headlamp's concave mirror is open on one end, and the light bulb's filament is placed at or near the focus. (Sorry if this is Wrong)
4 0
2 years ago
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A backyard swimming pool with a circular base of diameter 6.00m is filled to depth 1.50m. (b) Two persons with combined mass 150
Mashutka [201]

The pressure increase at the bottom of the pool after they enter the pool and float is 106.103 Pa.

<h3>What is absolute pressure?</h3>

Absolute pressure is the force that exists in a space when there is no matter present, or when there is a perfect vacuum. This absolute zero serves as the baseline for measurements in absolute pressure. The measurement of barometric pressure is the greatest illustration of an absolute referenced pressure. In order to determine absolute pressure, a complete vacuum is used. In contrast, gauge pressure is the amount of pressure that is measured in relation to atmospheric pressure, also referred to as barometric pressure.

given,

diameter = 6 m

depth = h = 1.5 m

Atmospheric pressure = P₀ = 10⁵ Pa

a) absolute pressure

P = P₀ + ρ g h

P = 10⁵ + 1000 x 10 x 1.5

P = 1.15 x 10⁵ Pa

b) When two person enters into the pool,

mass of the two person = 150 Kg

weight of water level displaced exists equal to the weight of person.

\rho \mathrm{Vg}=2 \mathrm{mg} \\

V=\frac{2 m}{\rho} \\

V=\frac{2 \times 150}{1000} \\

\mathrm{~V}=0.3 \mathrm{~m}^3

Area of pool $=\frac{\pi}{4} d^2$

&=\frac{\pi}{4} \times 6^2 \\

&=28.27 \mathrm{~m}^2

Height of the water rise

h &=\frac{V}{A} \\

h &=\frac{0.3}{28.27} \\

& \mathrm{~h}=0.0106 \mathrm{~m}

  • Pressure increased

P = ρ g h

P = 1000 x 10 x 0.0106

P = 106.103 Pa

To learn more about absolute pressure refer to:

brainly.com/question/17200230

#SPJ4

4 0
2 years ago
Light with an intensity of 1 kW/m2 falls normally on a surface and is completely absorbed. The radiation pressure is
kobusy [5.1K]

Answer:

The radiation pressure of the light is 3.33 x 10⁻⁶ Pa.

Explanation:

Given;

intensity of light, I = 1 kW/m²

The radiation pressure of light is given as;

Radiation \ Pressure = \frac{Flux \ density}{Speed \ of \ light}

I kW = 1000 J/s

The energy flux density = 1000 J/m².s

The speed of light = 3 x 10⁸ m/s

Thus, the radiation pressure of the light is calculated as;

Radiation \ pressure = \frac{1000}{3*10^{8}} \\\\Radiation \ pressure =3.33*10^{-6} \ Pa

Therefore, the radiation pressure of the light is 3.33 x 10⁻⁶ Pa.

6 0
3 years ago
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