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9966 [12]
3 years ago
6

Burning oil and coal adds to the atmosphere.

Physics
1 answer:
True [87]3 years ago
8 0

Answer:

carbon dioxide

Explanation:

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What system will break the chewed food down to a form your cell can use
horsena [70]
If I'm not mistaken it should be the digestive system due to the fact that our mouths and stomachs break down food and our intestines absorb any water and nutrients
4 0
3 years ago
An alpha particle (the nucleus of a helium atom) consists of two protons and two neutrons, and has a mass of 6.64 * 10-27 kg. A
melamori03 [73]

Answer:

t = 4.21x10⁻⁷ s

Explanation:

The time (t) can be found using the angular velocity (ω):

\omega = \frac{\theta}{t}

<em>Where θ: is the angular displacement = π (since it moves halfway through a complete circle)</em>

We have:

t = \frac{\theta}{\omega} = \frac{\theta}{v/r}  

<u>Where</u>:      

<em>v: is the tangential speed </em>

<em>r: is the radius</em>

The radius can be found equaling the magnetic force with the centripetal force:

qvB = \frac{mv^{2}}{r} \rightarrow r = \frac{mv}{qB}

Where:

m: is the mass of the alpha particle = 6.64x10⁻²⁷ kg

q: is the charge of the alpha particle = 2*p (proton) = 2*1.6x10⁻¹⁹C

B: is the magnetic field = 0.155 T

Hence, the time is:

t = \frac{\theta*r}{v} = \frac{\theta}{v}*\frac{mv}{qB} = \frac{\theta m}{qB} = \frac{\pi * 6.64 \cdot 10^{-27} kg}{2*1.6 \cdot 10^{-19} C*0.155 T} = 4.21 \cdot 10^{-7} s

Therefore, the time that takes for an alpha particle to move halfway through a complete circle is 4.21x10⁻⁷ s.

I hope it helps you!    

4 0
3 years ago
nan moved 18m to the right and then another 22m to the right if the motion takes 20 seconds what is nans velocity
IrinaVladis [17]
Step 1: list known info

distance(change in position (Δx))= 18m+22m= 40m
time= 20 seconds

Step 2 :solve for velocity

velocity= Δx÷time
v= 40/20= 2m/s

Answer: the velocity is 2 meters per a second (m/s)


7 0
3 years ago
A wire 2.80 m in length carries a current of 5.60 A in a region where a uniform magnetic field has a magnitude of 0.300 T. Calcu
Alekssandra [29.7K]

Complete question:

A wire 2.80 m in length carries a current of 5.60 A in a region where a uniform magnetic field has a magnitude of 0.300 T. Calculate the magnitude of the magnetic force on the wire assuming the following angles between the magnetic field and the current.

a) 60 ⁰

b) 90 ⁰

c) 120 ⁰

Answer:

(a) When the angle, θ = 60 ⁰,  force = 4.07 N

(b) When the angle, θ = 90 ⁰,  force = 4.7 N

(c) When the angle, θ = 120 ⁰,  force = 4.07 N

Explanation:

Given;

length of the wire, L = 2.8 m

current carried by the wire, I = 5.6 A

magnitude of the magnetic force, F = 0.3 T

The magnitude of the magnetic force is calculated as follows;

F = BIl \ sin(\theta)

(a) When the angle, θ = 60 ⁰

F = BIl \ sin(\theta)\\\\F = 0.3 \times 5.6 \times 2.8 \times sin(60)\\\\F = 4.07 \ N

(b) When the angle, θ = 90 ⁰

F = BIl \ sin(\theta)\\\\F = 0.3 \times 5.6 \times 2.8 \times sin(90)\\\\F = 4.7 \ N

(c) When the angle, θ = 120 ⁰

F = BIl \ sin(\theta)\\\\F = 0.3 \times 5.6 \times 2.8 \times sin(120)\\\\F = 4.07 \ N

6 0
3 years ago
The photon energies used in different types of medical x-ray imaging vary widely, depending upon the application. Single dental
pav-90 [236]

A) 5.0\cdot 10^{-11} m

The energy of an x-ray photon used for single dental x-rays is

E=25 keV = 25,000 eV \cdot (1.6\cdot 10^{-19} J/eV)=4\cdot 10^{-15} J

The energy of a photon is related to its wavelength by the equation

E=\frac{hc}{\lambda}

where

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

c=3\cdot 10^8 m/s is the speed of light

\lambda is the wavelength

Re-arranging the equation for the wavelength, we find

\lambda=\frac{hc}{E}=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{4\cdot 10^{-15}J}=5.0\cdot 10^{-11} m

B) 2.0\cdot 10^{-11} m

The energy of an x-ray photon used in microtomography is 2.5 times greater than the energy of the photon used in part A), so its energy is

E=2.5 \cdot (4\cdot 10^{-15}J)=1\cdot 10^{-14} J

And so, by using the same formula we used in part A), we can calculate the corresponding wavelength:

\lambda=\frac{hc}{E}=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{1\cdot 10^{-14}J}=2.0\cdot 10^{-11} m

4 0
3 years ago
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