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Genrish500 [490]
3 years ago
14

4. Heat is added to an ideal gas and the gas expands. In such a process the temperature

Physics
1 answer:
Bumek [7]3 years ago
4 0

Answer:

is high as 100 degrees c

Explanation:

due to high heat gas expands fast than normal

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At what speed (in m/s) will a proton move in a circular path of the same radius as an electron that travels at 8.00 ✕ 106 m/s pe
KonstantinChe [14]

Answer:

4347.8 m/s  

Explanation:

It is given that the radius of the circular path traversed by proton and electron is same. Also, we know that magnitude of charge on an electron and proton is same. Magnetic field strength is same for both.

\frac{m_ev_e^2}{r}=qv_eB\\\frac{m_pv_p^2}{r}=qv_pB\\

Take the ratio:

m_ev_e=m_pv_p\\\Rightarrow v_p=\frac{m_e}{m_p}v_e\\\Rightarrow v_p=\frac{1}{1840}\times 8.0\times 10^6 m/s\\\Rightarrow v_p=4347.8m/s

3 0
3 years ago
Do all of them please and explain
kondaur [170]

At 305 and 309 it was negative because it is going down


at 300 and 301 and 305 it was positive because its going up


at 302/303/304/306/309 she wasn't accelerating at all because it is staying the same

5 0
3 years ago
Read 2 more answers
Calculate the electric field intensity at a point 3 cm away from point charge of 3 x 10^-9 C.
lukranit [14]

Answer:

The electric field intensity is <u>30000 N/C.</u>

Explanation:

Given:

Magnitude of the point charge is, q=3\times 10^{-9}\ C

Distance of the given point from the point charge is, d=3\ cm=0.03\ m

Electric field intensity is directly proportional to the magnitude of point charge and inversely proportional to the square of the distance of the point and the given charge.

Therefore, electric field intensity 'E' at a distance of 'd' from a point charge 'q' is given as:

E=\frac{kq}{d^2}

Plug in k=9\times 10^9\ N\cdot m^2/C^2, q=3\times 10^{-9}\ C, d=0.03\ m. Solve for 'E'.

E=\frac{(9\times 10^9\ N\cdot m^2/C^2)(3\times 10^{-9}\ C)}{(0.03\ m)^2}\\\\E=\frac{27}{0.0009}\ N/C\\\\E=30000\ N/C

Therefore, the electric field intensity at a point 3 cm from the point charge is 30000 N/C.

3 0
3 years ago
How do your results from ray tracing compare to your results from using the thin-lens equation?
EastWind [94]

Answer:

20cm

Explanation:

A convex lens has a positive focal length and the object placed in front of it produce both virtual and real image <em>(image distance can be negative or positive depending on the nature of the image</em>).

According to the lens equation

\frac{1}{f} = \frac{1}{u} + \frac{1}{v} where;

f is the focal length  of the lens

u is the object distance

v is the image distance

If the magnification is - 0.6

mag = v/u = -0.5

v = -0.5u

since v = 10cm

10 = -0.5u

u = -10/0.5

u =-20 cm

Substitute u = -20cm ( due to negative magnification)and v = 10cm into the lens formula to get the focal length f

\frac{1}{f} = \frac{1}{u} + \frac{1}{v}\\\frac{1}{f} = \frac{1}{-20} + \frac{1}{10}\\\frac{1}{f} = \frac{1}{-20} + \frac{1}{10}\\\frac{1}{f} = \frac{-1+2}{20} \\\frac{1}{f} = \frac{1}{20} \\cross \ multiply\\f = 20\\f = 20 cm

Hence the focal length of the convex lens is 20cm

7 0
3 years ago
At what temperature do the fahrenheit and celsius scales give the same reading?
anygoal [31]
At -40.

-40 gives the same reading for Fahrenheit and Celsius scale.
7 0
3 years ago
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