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NemiM [27]
3 years ago
12

Two charges of 15 pC and −40 pC are inside a cube with sides that are of 0.40-m length. Determine the net electric flux through

the surface of the cube.
Physics
1 answer:
Kobotan [32]3 years ago
4 0

To solve this problem, we will apply the concepts related to Gaus' law that establishes that the electric flow through a surface is given by the charge of the object on the Vacuum permittivity. Mathematically this would be

\phi = \frac{Q_{net}}{\epsilon_0}

We must bear in mind that the net charge with the difference of the two charges given, therefore, replacing,

\phi = \frac{(15-40)*10^{-12}C}{8.85*10^{-12}C^2/Nm^2}

\phi = 2.82WB

Negative sign indicate flux is inward the surface

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A 5 kg pineapple is hanging completely still in mid air on a string and suddenly explodes
11111nata11111 [884]

Answer:

Explanation:

Conservation of momentum

Initial momentum is zero

3(15) + 2(v) = 0

v = - 22.5 m/s

v = 22.5 m/s downward

3 0
3 years ago
An object is projected from the ground with an upward speed of ų m/s has a speed of 23m/s when it is at a height of 5m above the
vovikov84 [41]

Answer:

25.08m/s

Explanation:

mgh1 + 0.5mv1² = mgh2 + 0.5mv2²

h1 = 0m

v1 = u

h2 = 5m

v2 = 23m/s

putting the values into the formula above;

m(10)(0) + 0.5m(u²) = m(10)(5) + 0.5m(23²)

0 + 0.5mu² = 50m + 264.5m

0.5mu² = 314.5m

dividing through by m

0.5u² = 314.5

u² = 629

u = <u>2</u><u>5</u><u>.</u><u>0</u><u>8</u><u>m</u><u>/</u><u>s</u>

<u>Theref</u><u>ore</u><u>,</u><u> </u><u>the</u><u> </u><u>init</u><u>ial</u><u> </u><u>speed</u><u> </u><u>"</u><u>u</u><u>"</u><u> </u><u>=</u><u> </u><u>2</u><u>5</u><u>m</u><u>/</u><u>s</u>

6 0
3 years ago
Where does the cart have the most velocity(speed)?
Varvara68 [4.7K]

Answer:

point c

Explanation:

the cart has accelerated and is at the lowest point on the path .

consider the acceleration due to gravity converting potential to kinetic energy

6 0
3 years ago
Read 2 more answers
If the average pitcher is releasing the ball from a height of 1.8 m above the ground, and the pitcher's mound is 0.2 m higher th
mina [271]

The catcher can catch the ball at a height of 0.96 m from the ground.

The distance between the pitcher's mound and the catcher's box is about 60'6", which translates to 18.44 m. An average pitcher can pitch with speeds ranging from 88 mph to 97 mph, which is from 39.3 m/s to 43.4 m/s.

Assume the pitcher pitches a ball horizontally with a speed of 40 m/s. If the catcher catches the ball in a time t, then the ball travels a horizontal distance x of 18.44 m and at the same time falls through a height y.

The horizontal motion of the ball is uniform motion since no force acts on the ball ( assuming no air resistance) and hence the acceleration of the ball along the horizontal direction is zero.

Therefore,

x=ut

Calculate the time t by substituting 18.44 m for x and 40 m/s for u.

t=\frac{x}{u} \\ =\frac{18.44 m}{40 m/s} \\ =0.461s

The ball is acted upon by the earth's gravitational attraction and hence it accelerates downwards with an acceleration equal to the acceleration due to gravity g.

Since a horizontal projection is assumed, the ball has no component of velocity in the downward direction.

Therefore, for vertical motion, which is an accelerated motion, the distance y, the ball falls in the time t taken by it to reach the catcher's box is given by the equation,

y=\frac{1}{2} gt^2

Substitute 9.8 m/s² for g and 0.461 s for t.

y=\frac{1}{2} gt^2\\ y=\frac{1}{2}(9.8 m/s^2)(0.461s)^2=1.04 m

The pitcher releases the ball at a height of 1.8 m from a mound which is at a height of 0.2 m. Thus, the ball is released at a height of 2.0 m from the ground. It falls through a distance of 1.04 m in the time it takes to reach the catcher.

Therefore, the height at which the catcher needs to keep his glove so as to catch the ball is given by,(2.0 m)-(1.04 m)=0.96 m

The catcher needs to hold his glove at a height of <u>0,96 m from the ground.</u>

8 0
3 years ago
What is the slope of the line if the rise of a line on a distance versus-time graph is 900 meters and the run is 3 minutes?
Alexandra [31]
600/3 = 200
the slope is 200m/min
 
OR

600/ (3/60) =
600 x 60/3 =
600 x 20 = 12,000 meters per hour 

6 0
3 years ago
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