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spin [16.1K]
3 years ago
8

an object with a mass of 70kilograms is supported at a height 8meters above the ground. what's the potential energy of the objec

t with respect to the ground ?(a)560J(b)2,058J(c)5,488J(d)3,430J
Physics
1 answer:
ra1l [238]3 years ago
3 0
Gravitational Potential Energy = mgh (m=mass; g=gravitational force(9.8N/kg); h = height)

Ug = (70kg)(9.8N/kg)(8m) = 5488J which is C.)
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A man weighing 720 N and a woman weighing 500 N have the same momentum. What is the ratio of the man's kinetic energy Km to that
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Red blood cells can be modeled as spheres of 6.53 μm diameter with −2.55×10−12 C excess charge uniformly distributed over the su
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Complete Question

Red blood cells can be modeled as spheres of 6.53 μm diameter with −2.55×10−12 C excess charge uniformly distributed over the surface. Find the electric field at the following locations, with radially outward defined as the positive direction and radially inward defined as the negative direction. The permittivity of free space ????0 is 8.85×10−12 C/(V⋅m). What is the electric field

E⃗ 1 inside the cell at a distance of 3.05 μm from the center?

E⃗ 2 Just inside the surface of the cell

E⃗ 3 Just outside the surface of the cell

E⃗ 4 At a point outside the cell 3.05 μm from the surface

Answer:

E⃗ 1

      0 V/m

E⃗ 2

      0 V/m

E⃗ 3

         E_3 =  2.153 *10^{9} \  V/m

E⃗ 4

E_4 =  5.754 *10^ {8} \  V/m

Explanation:

From the question we are told that

The diameter is d =  6.53 \mu m  = 6.53*10^{-6}\  m

The charge is Q =  -.2.55 *10^{-12} \  C

The permittivity of free space is \epsilon_o  =  8.85* 10^{-12}\  C / V.m

The distance considered is d =  3.05 \mu m  =  3.05 *10^{-6} \ m

Generally the electric field inside the cell at a distance of 3.05 μm from the center is

0 V/m

This because there is no electric field felt inside the cell according Gauss the cell is taken as a point charge

Generally the electric field just inside the surface of the cell is 0 V/m

This because there is no electric field felt inside the cell according Gauss the cell is taken as a point charge

Generally the electric field just outside the cell is mathematically represented as

E_3 =  \frac{ k  *  |Q|}{ r^2 }

Here k is the coulomb constant with value

k  =   9*10^{9}\ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}

r is the radius of the sphere which is mathematically as

r =  \frac{d}{2} =   \frac{6.53*10^{-6}}{2}  = 3.265 *10^{-6} \  m

E_3 =  \frac{ 9*10^{9}  *  |-2.55 *10^{-12} |}{ [3.265 *10^{-6} ]^2 }

E_3 =  2.153 *10^{9} \  V/m

Generally the electric field at a point outside the cell 3.05 μm from the surface is mathematically represented as

E_4 =  \frac{ k  *  |Q|}{ R^2 }

Here R is mathematically represented as

R  =  3.265 *10^{-6} +  3.05 *10^{-6}

=>       R  =  6.315 *10^{-6}

So

E_4 =  \frac{ 9*10^{9}  *  |-2.55 *10^{-12} |}{ [ 6.315 *10^{-6} ]^2 }

E_4 =  5.754 *10^ {8} \  V/m

3 0
3 years ago
Ship A is located 4.1 km north and 2.3 km east of ship B. Ship A has a velocity of 22 km/h toward the south and Ship B has a vel
castortr0y [4]

Answer:

a) x component  = -31.25 km/hr

b) y component = 46.64 km/hr

Explanation:

Given data:

A position is 4km north and 2.5 km east to B

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ship B velocity = 40 km/hr

A velocity wrt to velocity of B

\vec{V_{AB}} =\vec{V_A} - \vec{V_B}

\vec{V_A} = 22 km/hr

\vec{V_B} = 40 cos38\hat{i} + 40sin 38 \hat{j}

                 = 31.52\hat{i} + 24.62 \hat{j}

putting respective value to get velocity of  A with respect to B

\vec{V_{AB}} = -22 \hat {j} - (31.52\hat{i} + 24.62 \hat{j})

\vec{V_{AB}} = -31.52\hat{i} - 46.62\hat{j}

a) x component  = -31.25 km/hr

b) y component = 46.64 km/hr

6 0
3 years ago
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