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Ivenika [448]
3 years ago
10

The Universal Law of Gravity states that the force of gravity acts between all objects. Nonetheless, you never feel that the fri

dge in your kitchen pulls you toward it. Why?
Physics
1 answer:
Kryger [21]3 years ago
8 0
If your mass is 50 kg (you weigh 110 pounds), and the mass of the refrigerator
is 100 kg (it weighs 220 pounds), and you're standing so that the center of
the frig and the center of you are 1 meter apart (about 3 feet), then the force
of gravity between you and the frig is about 0.0000012 ounce.  That's why.
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Two spherical shells have a common center. A -1.50 × 10-6 C charge is spread uniformly over the inner shell, which has a radius
Murrr4er [49]

Answer:

a) At 0.20 m, the magnitude of the field is 675.0 kV

The direction of the field is acting outwards from the center of the charged spheres

b) At 0.10 m, the magnitude of the field is 135 kV

The direction is acting outwards from the center of the charged spheres

c) At 0.025 m

The magnitude of the field, V = -270.0 kV

The direction of the field is inwards, towards the center of the charged spheres

Explanation:

The charged spherical shell parameters are;

The charge on the inner sphere, q₁ = -1.50 × 10⁻⁶ C

The radius of the inner shell, R₁ = 0.050 m

The charge on the outer sphere, q₂ = +4.50 × 10⁻⁶ C

The radius of the outer shell, R₂ = 0.15 m

Let 'r', represent the distance at which the electric field is measured, the following relationships can be obtained;

When r < R₁ < R₂,

V = k \cdot \left( \dfrac{q_1}{R_1} + \dfrac{q_2}{R_2} \right )

When R₁ < r < R₂,

V = k \cdot \left( \dfrac{q_1}{r} + \dfrac{q_2}{R_2} \right )

When R₁ < R₂ < r,

V = k \cdot \left( \dfrac{q_1  + q_2}{r^2}  \right )

a) When r = 0.20 m, we have;

R₁ < R₂ < r, therefore

V = k \cdot \left( \dfrac{q_1  + q_2}{r^2}  \right )

By plugging in the values, we get;

V = 9 \times 10^9 \times \left( \dfrac{-1.50 \times 10^{-6} + 4.50\times 10^{-6} }{0.20^2}  \right ) = 675.0 \ kV

Therefore, the magnitude of the field, V = 675.0 kV

The direction of the field is outwards

b) When r = 0.10 m, we have;

When R₁ < r < R₂, therefore;

V = k \cdot \left( \dfrac{q_1}{r} + \dfrac{q_2}{R_2} \right )

By plugging in the values, we get;

V = 9 \times 10^9 \times \left( \dfrac{-1.50 \times 10^{-6}  }{0.10}  + \dfrac{4.50\times 10^{-6}}{0.15} \right ) = 135 \ kV

Therefore, the magnitude of the field, V = 135 kV

The direction of the field is outwards from the center

c) When r = 0.025 m, we have;

When r < R₁ < R₂, therefore;

V = k \cdot \left( \dfrac{q_1}{R_1} + \dfrac{q_2}{R_2} \right )

By plugging in the values, we get;

V = 9 \times 10^9 \times \left( \dfrac{-1.50 \times 10^{-6}  }{0.05}  + \dfrac{4.50\times 10^{-6}}{0.15} \right ) = -270 \ kV

Therefore, the magnitude of the field, V = -270.0 kV

The direction of the field is inwards, towards the center of the charged spheres.

4 0
3 years ago
a runner makes one lap around a 200m track in 25s, what is the runners (a) average speed and (b) average velocity
TEA [102]

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5 0
3 years ago
What is the relationship between magnetism and potential and kinetic energy
vaieri [72.5K]

Check out Faraday's Law


5 0
3 years ago
An earthquake on the ocean floor produced a giant wave called a tsunami. The tsunami traveled through the ocean and hit a remote
Ber [7]
A lot of the biology particles can be certified by a sphychiatrist that can determine weather they are infected with Ebola
6 0
3 years ago
Ask Your Teacher A trough is 16 ft long and its ends have the shape of isosceles triangles that are 2 ft across at the top and h
zloy xaker [14]

Answer:0.210 ft/min

Explanation:

Given

Length of trough L=16 ft

width of base b=2 ft

height of triangleh=1 ft

From Similar triangles property

\frac{4}{2x}=\frac{1}{y}

2y=x

volume of water in time t

V=\frac{1}{2}\times (2x\cdot y)\cdot

V=16xy

V=32y^2

differentiating

\frac{\mathrm{d} V}{\mathrm{d} t}=32\times 2\times y\times \frac{\mathrm{d} y}{\mathrm{d} t}

at y=8 in.\approx 0.667 ft

9=64\times 0.667\times \frac{\mathrm{d} y}{\mathrm{d} t}

\frac{\mathrm{d} y}{\mathrm{d} t}=\frac{9}{64\times 0.667}

\frac{\mathrm{d} y}{\mathrm{d} t}=0.210 ft/min

6 0
4 years ago
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