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RoseWind [281]
3 years ago
7

24 A uniform electric field of magnitude 1.1×104 N/C is perpendicular to a square sheet with sides 2.0 m long. What is the elect

ric flux through the sheet?
Physics
1 answer:
Tatiana [17]3 years ago
4 0

Answer:

44,000 Nm^2/C

Explanation:

The electric flux through a certain surface is given by (for a uniform field):

\Phi = EA cos \theta

where:

E is the magnitude of the electric field

A is the area of the surface

\theta is the angle between the direction of the field and of the normal to the surface

In this problem, we have:

E=1.1\cdot 10^4 N/C is the electric field

L = 2.0 m is the side of the sheet, so the area is

A=L^2=(2.0)^2=4.0 m^2

\theta=0^{\circ}, since the electric field is perpendicular to the surface

Therefore, the electric flux is

\Phi =(1.1\cdot 10^4)(4.0)(cos 0^{\circ})=44,000 Nm^2/C

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Shtirlitz [24]

Answer:

d = 90 ft

Explanation:

As we know that after each bounce it reaches to 4/5 times of initial height

so we can say

h_2 = \frac{4}{5}h

so the distance covered is given as

d = h + 2(\frac{4}{5}h) + 2(\frac{4}{5})^2h + 2(\frac{4}{5})^3h........

here we know that

h = 10 feet

d = h + 2(\frac{4}{5}h)(1 + \frac{4}{5} + (\frac{4}{5})^2 + ...........)

d = 10 + 2(\frac{4}{5}(10))(\frac{1}{1 - \frac{4}{5}})

d = 90 ft

8 0
3 years ago
What is the specific heat of the masses in this experiment? Infer the substance the masses are made of and explain your inferenc
Liono4ka [1.6K]

The metal whose specific heat capacity is close to the obtained value is aluminum.

<h3>What is specific heat capacity</h3>

The specific heat capacity of an object is the heat required to raise a unit mass of the substance by 1 kelvin.

Q= mc\Delta \theta

where;

  • c is the specific heat capacity
  • Δθ is change in temperature

Let the mass of the water = 50 g

mass of the metal for this first trial = 50 g

The heat gained by the water is calculated as follows

Q = 50 \times 4.184 \times 8.4\\\\Q = 1757.28 \ J

Specific heat capacity of the metal for the first trial is calculated as follows;

Heat gained by water = Heat lost by metal

C = \frac{Q}{m\Delta T} = \frac{1757.28}{50\times 8.4} = 4.184 \ J/g^oC

Specific heat capacity of the metal for the second trial;

mass of metal = 200 - mass of water = 150 g

C_2 = \frac{1757.28}{150 \times 15.2} = 0.77 \ J/g^oC

Specific heat capacity of the metal for the third trial;

C_3 = \frac{1757.28}{250 \times 20.8} = 0.34\ J/g^oC

Specific heat capacity of the metal for the fourth trial;

C_4 = \frac{1757.28}{350 \times 25.4} = 0.19\ J/g^oC

Specific heat capacity of the metal for the fifth trial;

C_5 = \frac{1757.28}{450 \times 29.6} = 0.13\ J/g^oC

Average specific heat capacity

C = \frac{4.184 + 0.77 + 0.34+ 0.19 + 0.13 }{5} = 1.12 \ J/g^oC = 1120 J/kg^oC

The metal whose specific heat capacity is close to the above value is aluminum.

Learn more about specific heat capacity here: brainly.com/question/16559442

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2 years ago
Divergent boundaries are also called
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Continental drifts i'm pretty sure
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3 years ago
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Answer:

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 },1000);

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} else {

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3 years ago
A 860g block of titanium metal has a volume of 200 cm3 and the density of 4.3g/cm^3.
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Answer:

4.3 g/ cm^3

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