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12345 [234]
3 years ago
13

An electric field of intensity 3.80 kN/C is applied along the x-axis. Calculate the electric flux through a rectangular plane 0.

350 m wide and 0.700 m long if the following conditions are true.
Physics
1 answer:
lakkis [162]3 years ago
4 0

Answer: 1.55 x 10⁴ Nm²c^-1

Explanation: The electric flux, electric field intensity and area are related by the formulae below.

Φ= EAcosθ,

Where Φ= electric flux (Nm²c^-1)

E =electric field intensity (N/m²)

A = Area (m²)

θ= this is angle between the planar area and the magnetic flux

For our question E=3.80KN/c= 3800 N/c

A= 0.700 x 0.350= 0.245m²

θ= 0° ( this is because the electric field was applied along the x axis, thus the electric flux will be parallel to the area).

Hence Φ= 3800 x 0.245 x cos(0)

= 3800 x 0.245 x 1 (value of cos 0° =1)

= 1.55 x 10⁴ Nm²c^-1

Thus the electric field is 1.55 x 10⁴ Nm²c^-1

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18 kilogram Mass Blokus addressed a level surface if the coefficient of static friction between the Block in the surface is 0.6
Tasya [4]

Question: 18 kilogram Mass Block rest on level surface if the coefficient of static friction between the Block and the surface is 0.6 what  horizontal force is required to just move the blcok ( take gravity as 10m/s2 )

Answer:

108 N

Explanation:

From the question,

Applying

F' = mgμ................ Equation 1

Where F' = Frictional force = horizontal  force required to just move the block,  m = mass of the block, g = acceleration due to gravity, μ = coefficient of static friction.

From the question,

Given: m = 18 kg, μ = 0.6, g = 10 m/s²

Substitute these values into equation 1

F' = 18×0.6×10

F' = 108 N

4 0
2 years ago
A ball of mass 0.120 kg is dropped from rest from a height of 1.25 m. It rebounds from the floor to reach a height of 0.820 m. W
Vikentia [17]

Answer:

1.0752 kgm/s

Explanation:

Considering when the drop was dropped from rest from a height,

mass of the ball, m = 0.120 kg

height, h = - 1.25 m

the initial velocity, u = 0 m/s

the acceleration due to gravity, g = - 9.8 m/s²

From equation of motion

                            V^{2} = U^{2} + 2gh

Substituting the values,

                             V^{2} = 0^{2} + 2(-9.8 m/s^{2})(-1.25 m)

                             V^{2} = 24.5 m/s

                             V = \sqrt{24.5} \ m/s

                             V = 4.95 \ m/s

                            V = ± 4.95 m/s

                            V = - 4.95 m/s

Since the ball is moving downward, the final velocity of the ball when it hits the floor is  V = - 4.95 m/s  

Considering when the ball rebounds from the floor,

assume the mass of the ball still remain, m = 0.120 kg

height, h = 0.820 m

the final velocity, v = 0 m/s  

the acceleration due to gravity, g = - 9.8 m/s²

From equation of motion

                            V^{2} = U^{2} + 2gh

Substituting the values,

                            0^{2} = U^{2} + 2(-9.8 m/s^{2})(0.820 m)

                            0 = U^{2} - 16.072 m/s

                            U^{2} = 16.072 m/s

                            U = \sqrt{16.072} \ m/s

                           U = ± 4.01 m/s

                          U = + 4.01 m/s

Since the ball is moving upward, the initial velocity of the ball from the bounce from the floor is  U = + 4.01 m/s                        

From Newton's second law of motion, applied force is directly proportional to the rate of change in momentum.

                            F = \frac{mv - mu}{t}

                          F.t = m(v - u)

       ⇒      Impulse = Change in momentum

To calculate the impulse, the moment before the ball hits the ground will be the initial momentum while the moment the ball rebounces will be the final velocity,                        

          ∴          F.t = 0.120  kg(4.01  m/s - (-4.95  m/s) )

                      F.t = 0.120  kg(4.01  m/s + 4.95  m/s) )

                      F.t = 0.120  kg × 8.96  m/s

                      Impulse  = 1.0752 kgm/s

The impulse given to the ball by the floor is 1.0752 kgm/s

                             

6 0
3 years ago
What is the acceleration of a softball if it has a mass of 0.5 kg and hits the cathers glove with a force of 25n
ioda
Acceleration is found if we have the force and mass. 

With the following equation: F = ma, we can find the missing values. 

F = 25n
M = 0.5 kg
a = ?

a = f/m
a = 25/0.5
a = 50

a = 50 m/s

So, the acceleration is 50 m/s^2 
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3 years ago
5. How will you separate water from petrol?
ohaa [14]

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5 0
3 years ago
Read 2 more answers
The potential difference between two equipotential lines 5.0 mm apart has a value of 1.2 V, calculate the electric field value.
jonny [76]

Answer:

The electric field value is 240 N/C

Explanation:

Given that,

Distance = 5.0 mm

Potential difference = 1.2 V

We need to calculate the electric field value

Using formula of potential difference

\Delta V=Ed

E=\dfrac{\Delta V}{d}

Where, E = electric field

V = potential difference

d = distance

Put the value into the formula

E=\dfrac{1.2}{5.0\times10^{-3}}

E= 240\ N/C

Hence, The electric field value is 240 N/C

6 0
3 years ago
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