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victus00 [196]
3 years ago
5

It is found experimentally that the electric field in a certain region of Earth's atmosphere is directed vertically down. At an

altitude of 390 m the field has magnitude 60.0 N/C. At an altitude of 240 m, the magnitude is 100 N/C. Find the net amount of charge contained in a cube 150 m on edge, with horizontal faces at altitudes of 240 and 390 m. Neglect the curvature of Earth.
Physics
1 answer:
Maru [420]3 years ago
5 0

Answer:

q=7.965*10^-^6C

Explanation:

From the question we are told that

Altitude of  d_1 390m

MagnitudeM_1=60.0 N/C

Altitude of d_2=240 m

Magnitude is M_2= 100 N/C

Distance of cube d_c=150 m

Generally the flux \phi is mathematical given as

 \phi=60(150)^2cos180+100(150)^2*cos0

 \phi=-9*10^5

Generally Quantity of charge  q is mathematically given as

 q=\varepsilon _0 *\phi

 q=8.85*10^-^1^2 *9*10^5

 q=7.965*10^-^6C

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Answer:

Your answer would be C <u><em>Hope this helps</em></u>

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Which of these statements are true about hydroelectric power? Check all that are true. It is the most widely used renewable ener
IRINA_888 [86]

Answer:

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7 0
3 years ago
On Earth, a spring stretches by 5.0 cm when a mass of 3.0 kg is suspended from one end.
Neko [114]

Answer:

Mass = 18.0 kg

Explanation:

From Hooke's law,

F = ke

where: F is the force, k is the spring constant and e is the extension.

But, F = mg

So that,

mg = ke

On the Earth, let the gravitational force be 10 m/s^{2}.

3.0 x 10 = k x 5.0

30 = 5k

⇒ k = \frac{30}{5} ................ 1

On the Moon, the gravitational force is \frac{1}{6} of that on the Earth.

m x \frac{10}{6} = k x 5.0

\frac{10m}{6} = 5k

⇒ k = \frac{10m}{30} ............. 2

Equating 1 and 2, we have;

\frac{30}{5}  = \frac{10m}{30}

m = \frac{900}{50}

    = 18.0

m = 18.0 kg

The mass required to produce the same extension on the Moon is 18 kg.

8 0
3 years ago
Read 2 more answers
A heat engine performs (245 + A) J of work in each cycle while also delivering (142 + B) J of heat to the cold reservoir. Find t
Ganezh [65]

Answer:

The value is \eta  =  54.4 \%

Explanation:

From the question we are told that

    The work input is  W = ( 245 + A ) \  J

     The heat delivered is Q =  (142 + B) \  J

      The value of A is  A =  14

        The value of B  is  B  = 72

Generally the efficiency of the heat engine is mathematically represented as

          \eta  =  \frac{W}{Q_t}

Here  Q_t is the total out energy produce by the heat engine and this is mathematically represented as

           Q_t= Q + W

=>         Q_t=  245 + A + 142 + B

=>         Q_t=  390 + A+B

So

               \eta  =  \frac{245 + A }{390 + A+ B}

=>          \eta  =  0.544

=>          \eta  =  0.544 *100

=>          \eta  =  54.4 \%

5 0
3 years ago
Three masses are located in the x-y plane as follows: a mass of 6 kg is located at (0 m, 0 m), a mass of 4 kg is located at (3 m
Natali [406]
<h2>Answer:</h2>

D. (1m, 0.5m)

<h2>Explanation:</h2>

The center of mass (or center of gravity) of a system of particles is the point where the weight acts when the individual particles are replaced by a single particle of equivalent mass. For the three masses, the coordinates of the center of mass C(x, y) is given by;

x = (m₁x₁ + m₂x₂ + m₃x₃) / M       ----------------(i)

y = (m₁y₁ + m₂y₂ + m₃y₃) / M       ----------------(ii)

Where;

M = sum of the masses

m₁ and x₁ = mass and position of first mass in the x direction.

m₂ and x₂ = mass and position of second mass in the x direction.

m₃ and x₃ = mass and position of third mass in the x direction.

y₁ , y₂ and y₃ = positions of the first, second and third masses respectively in the y direction.

From the question;

m₁ = 6kg

m₂ = 4kg

m₃ = 2kg

x₁ = 0m

x₂ = 3m

x₃ = 0m

y₁ = 0m

y₂ = 0m

y₃ = 3m

M = m₁ + m₂ + m₃ = 6 + 4 + 2 = 12kg

Substitute these values into equations (i) and (ii) as follows;

x = ((6x0) + (4x3) + (2x0)) / 12

x = 12 / 12

x = 1 m  

y = (6x0) + (4x0) + (2x3)) / 12

y = 6 / 12

y = 0.5m

Therefore, the center of mass of the system is at (1m, 0.5m)

7 0
3 years ago
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