Answer:
4.763 × 10⁶ N/C
Explanation:
Let E₁ be the electric field due to the 4.0 μC charge and E₂ be the electric field due to the -6.0 μC charge. At the third corner, E₁ points in the negative x direction and E₂ acts at an angle of 60 to the negative x - direction.
Resolving E₂ into horizontal and vertical components, we have
E₂cos60 as horizontal component and E₂sin60 as vertical component. E₁ has only horizontal component.
Summing the horizontal components we have
E₃ = -E₁ + (-E₂cos60) = -kq₁/r²- kq₂cos60/r²
= -k/r²(q₁ + q₂cos60)
= -k/r²(4 μC + (-6.0 μC)(1/2))
= -k/r²(4 μC - 3.0 μC)
= -k/r²(1 μC)
= -9 × 10⁹ Nm²/C²(1.0 × 10⁻⁶)/(0.10 m)²
= -9 × 10⁵ N/C
Summing the vertical components, we have
E₄ = 0 + (-E₂sin60)
= -E₂sin60
= -kq₂sin60/r²
= -k(-6.0 μC)(0.8660)/(0.10 m)²
= -9 × 10⁹ Nm²/C²(-6.0 × 10⁻⁶)(0.8660)/(0.10 m)²
= 46.77 × 10⁵ N/C
The magnitude of the resultant electric field, E is thus
E = √(E₃² + E₄²) = √[(-9 × 10⁵ N/C)² + (46.77 10⁵ N/C)²) = (√226843.29) × 10⁴
= 476.28 × 10⁴ N/C
= 4.7628 × 10⁶ N/C
≅ 4.763 × 10⁶ N/C
Answer:
The portion of earth underneath the crust containing the asthenosphere and mesosphere is outer core and inner core .
Explanation:
The outer core of the earth is ball of very hot metals . Its temperature is hot enough to melt all metals into liquid state . This is mainly composed of the melted metals nickel and iron . This has got got temperature between 4000 degrees to 9000 degrees .
Inner core is underneath the outer core ,Here the pressure is too high and this makes the metals squeezed and not even able to move like liquid . They are forced to vibrate just like a solid . This has got a thickness of 800 miles . The pressure here is comparably 3,000,000 times the air pressure at sea level.
Answer:
Just as images are reflected from the surface of a mirror, light reflected from a smooth water surface also produced a clear image. ... Consequently, the outgoing rays are reflected at many different angles and the image is disrupted. Reflection from such a rough surface is called diffuse reflection and appears matte.
Explanation:
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Initial volume of mercury is
V = 0.1 cm³
The temperature rise is 35 - 5 = 30 ⁰C = 30 ⁰K.
Because the coefficient of volume expansion is 1.8x10⁻⁴ 1/K, the change in volume of the mercury is
ΔV = (1.8x10⁻⁴ 1/K)*(30 ⁰K)(0.1 cm³) = 5.4x10⁻⁴ cm³
The cross sectional area of the tube is
A = 0.012 mm² = (0.012x10⁻² cm²).
Therefore the rise of mercury in the tube is
h = ΔV/A
= (5.4x10⁻⁴ cm³)/(0.012x10⁻² cm²)
= 4.5 cm
Answer: 4.5 cm
Answer:
Explanation:
Diameter of pool = 12 m
radius of pool, r = 6 m
Total height raised, h = 3 + 2.5 = 5.5 m
density of water, d = 1000 kg/m³
Mass of water, m = Volume of water x density
m = πr²h x d
m = 3.14 x 6 x 6 x 5.5 x 1000
m = 113040 kg
Work = m x g x h
W = 113040 x 9.8 x 5.5
W = 6092856 J
