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Nikolay [14]
3 years ago
10

True or False: The variables in the equation 4x-(5y)2=64x-(5y)2=6 are 4, 5, and 6

Physics
2 answers:
iogann1982 [59]3 years ago
5 0

Answer:

true

Explanation:

Luba_88 [7]3 years ago
3 0

Answer:

true

Explanation:

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A vertical cylinder with a heavy piston contains air at 300 K. The initial pressure is 2.0 x 105 Pa and the initial volume is 0.
aleksklad [387]

Find answers and explanations in the attachments

8 0
3 years ago
Problem 6. A negatively charged particle is placed in a uniform electric field directed
Montano1993 [528]

<u>C</u><u>)</u><u> </u><u>South</u>

As we know that, north is considered as negative, and south as positive. Now, the charge on particle is negative; so, when we will release it, it will move towards the south.

8 0
2 years ago
What is the acceleration of a 50 kg object pushed with a net force of 500 newtons?
vodomira [7]
Using the formula F=ma
500N=50kg (a)
a= 10 m/s^2
3 0
3 years ago
A spherical, conducting shell of inner radius r1= 10 cm and outer radius r2 = 15 cm carries a total charge Q = 15 μC . What is t
lutik1710 [3]

a) E = 0

b) 3.38\cdot 10^6 N/C

Explanation:

a)

We can solve this problem using Gauss theorem: the electric flux through a Gaussian surface of radius r must be equal to the charge contained by the sphere divided by the vacuum permittivity:

\int EdS=\frac{q}{\epsilon_0}

where

E is the electric field

q is the charge contained by the Gaussian surface

\epsilon_0 is the vacuum permittivity

Here we want to find the electric field at a distance of

r = 12 cm = 0.12 m

Here we are between the inner radius and the outer radius of the shell:

r_1 = 10 cm\\r_2 = 15 cm

However, we notice that the shell is conducting: this means that the charge inside the conductor will distribute over its outer surface.

This means that a Gaussian surface of radius r = 12 cm, which is smaller than the outer radius of the shell, will contain zero net charge:

q = 0

Therefore, the magnitude of the electric field is also zero:

E = 0

b)

Here we want to find the magnitude of the electric field at a distance of

r = 20 cm = 0.20 m

from the centre of the shell.

Outside the outer surface of the shell, the electric field is equivalent to that produced by a single-point charge of same magnitude Q concentrated at the centre of the shell.

Therefore, it is given by:

E=\frac{Q}{4\pi \epsilon_0 r^2}

where in this problem:

Q=15 \mu C = 15\cdot 10^{-6} C is the charge on the shell

r=20 cm = 0.20 m is the distance from the centre of the shell

Substituting, we find:

E=\frac{15\cdot 10^{-6}}{4\pi (8.85\cdot 10^{-12})(0.20)^2}=3.38\cdot 10^6 N/C

4 0
3 years ago
A ball is thrown into the air with an upward velocity of 20 feet per second. Its height, h, in feet after t seconds is given by
Margarita [4]
Thank you for posting your question here at brainly. Feel free to ask more questions.  
<span>The best and most correct answer among the choices provided by the question is B. Reaches a max height of 8.25 feet after 0.63 seconds</span> .     <span><span>

</span><span>Hope my answer would be a great help for you. </span> </span>  

<span> </span>

3 0
3 years ago
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