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Allisa [31]
3 years ago
12

At a certain altitude above the Earth's surface, the electric field has a magnitude of 135 V/m. How much energy is stored in 1.0

0 m3 of air due to this field?
Physics
1 answer:
Paul [167]3 years ago
4 0

Answer:

Explanation:

Given that,

Electric field E=135V/m

Energy stored in 1m³of air=?

The energy stored in an electric field is given as

u = ½ εo E²

Where

U is the energy stored

εo is permissivity and it value is 8.85×10^-12C²/N..m²

And E is the electric field

Then,

U=½×8.85×10^-12×135²

U=8.06×10^-8J/m³

Then, the energy stored in 1m³ of air is 8.06×10^-8 J/m³

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Answer:

Earth's atmosphere is a mixture of nitrogen (79%), oxygen (20%), and a small fraction of carbon dioxide, water vapours and other gases. However, the atmospheres on Venus and Mars mainly consist of carbon dioxide.

Explanation:

3 0
3 years ago
A student solving a physics problem for the range of a projectile has obtained the expression
Alika [10]

Answer:

The range of the projectile is 66.7 meters.

Explanation:

The range of a projectile is given by the following expression as :

R=\dfrac{v_o^2\ sin2\theta}{g}..............(1)

v_o=37.2\ m/s

\theta=14.1^{\circ}

g=9.8\ m/s^2

The range can be calculated using equation (1). Putting the values of all parameters we get :

R=\dfrac{(37.2)^2\ sin2(14.1)}{9.8}

R = 66.7 meters

So, the range of the projectile is 66.7 meters. Hence, this is the required solution.

6 0
3 years ago
The hindenburg zeppelin contained 200,000 m^3 of hydrogen, with a density of 0.0899 kg/m^3. What is the mass of the hydrogen?
oksian1 [2.3K]

d=m/v

0.0899=m/200000

or,0.0899*200000=m

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6 0
3 years ago
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A force F = (2.75 N)i + (7.50 N)j + (6.75 N)k acts on a 2.00 kg mobile object that moves from an initial position of d = (2.75 m
Natasha2012 [34]

Answer:

The work done on the object by the force in the 5.60 s interval is 40.93 J.

Explanation:

Given that,

Force F=(2.75\ N)i+(7.50\ N)j+(6.75\ N)k

Mass of object = 2.00 kg

Initial position d=(2.75)i-(2.00)j+(5.00)k

Final position d=-(5.00)i+(4.50)j+(7.00)k

Time = 4.00 sec

We need to calculate the work done on the object by the force in the 5.60 s interval.

Using formula of work done

W=F\cdot d

W=F\cdot(d_{f}-d_{i})

Put the value into the formula

W=(2.75)i+(7.50)j+(6.75)k\cdot (-(5.00)i+(4.50)j+(7.00)k-(2.75)i+(2.00)j-(5.00)k)

W=(2.75)i+(7.50)j+(6.75)k\cdot((-7.75)i+6.50j+2.00k)

W=2.75\times-7.75+7.50\times6.50+6.75\times2.00

W=40.93\ J

Hence, The work done on the object by the force in the 5.60 s interval is 40.93 J.

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