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stepladder [879]

3 years ago

5

For which one of the following situations will the path length equal the magnitude of the displacement? a A ball on the end of a

string is moving in a vertical circle. b A toy train is traveling around a circular track. c A train travels 5 miles east before it stops. It then travels 2 miles west. d A ball is rolling down an inclined plane. e A ball rises and falls after being thrown straight up from the earth's surface.

1 answer:

PSYCHO15rus [73]3 years ago

7 0

Answer:

d A ball is rolling down an inclined plane.

Explanation:

When path length is equal to the displacement

then we can say that the motion of the object must be in straight line so that the distance and displacement must be same

SO here we can say

a A ball on the end of a string is moving in a vertical circle.

In circular path distance and displacement is not same

b A toy train is traveling around a circular track.

In circular path distance and displacement is not same

c A train travels 5 miles east before it stops. It then travels 2 miles west.

Net displacement is 3 miles East while distance is 7 miles

d A ball is rolling down an inclined plane.

Here its motion is in straight line so we can say that path length and displacement will be same

e A ball rises and falls after being thrown straight up from the earth's surface.

In this type of to and fro motion path length is not same as displacement

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Mass Center Determine the coordinates (x, y) of the center of mass of the area in blue in the figure below. Answers: x=(3)/(8)a

Naya [18.7K]

Explanation:

The x and y coordinates of the center of mass are:

xcm = ∫ x dm / m = ∫ x ρ dA / ∫ ρ dA

ycm = ∫ y dm / m = ∫ y ρ dA / ∫ ρ dA

Assuming uniform density, the center of mass is also the center of area.

xcm = ∫ x dA / ∫ dA = ∫ x y dx / A

ycm = ∫ y dA / ∫ dA = ∫ ½ y² dx / A

First, let's find the area:

A = ∫ y dx

A = ∫₀ᵃ (-h/a² x² + h) dx

A = -⅓ h/a² x³ + hx |₀ᵃ

A = -⅓ h/a² (a)³ + h(a)

A = ⅔ ha

Now, let's find the x coordinate of the center of mass:

xcm = ∫ x y dx / A

xcm = ∫₀ᵃ x (-h/a² x² + h) dx / (⅔ ha)

xcm = ∫₀ᵃ (-h/a² x³ + hx) dx / (⅔ ha)

xcm = (-¼ h/a² x⁴ + ½ hx²) |₀ᵃ / (⅔ ha)

xcm = (-¼ h/a² (a)⁴ + ½ h(a)²) / (⅔ ha)

xcm = (¼ ha²) / (⅔ ha)

xcm = ⅜ a

Next, we find the y coordinate of the center of mass:

ycm = ∫ y² dx / A

ycm = ∫₀ᵃ ½ (-h/a² x² + h)² dx / (⅔ ha)

ycm = ∫₀ᵃ ½ (h²/a⁴ x⁴ − 2h²/a² x² + h²) dx / (⅔ ha)

ycm = ½ (⅕ h²/a⁴ x⁵ − ⅔ h²/a² x³ + h² x) |₀ᵃ / (⅔ ha)

ycm = ½ (⅕ h²/a⁴ (a)⁵ − ⅔ h²/a² (a)³ + h² (a)) / (⅔ ha)

ycm = ½ (⁸/₁₅ h²a) / (⅔ ha)

ycm = ⅖ h

6 0

3 years ago

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

yKpoI14uk [10]

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

sveta [45]

My calculator is about 1cm thick, 7cm wide, and 13cm long.

Its volume is (length) (width) (thick) = (13 x 7 x 1) = 91 cm³ .

The question wants me to assume that the density of my calculator
is about the same as the density of water. That doesn't seem right
to me. I could check it easily. All I have to do is put my calculator
into water, watch to see if sinks or floats, and how enthusiastically.
I won't do that. I'll accept the assumption.

If its density is actually 1 g/cm³, then its mass is about 91 grams.

The choices of answers confused me at first, until I realized that
the choices are actually 1g, 10² g, 10⁴ g, and 10⁶ g.

My result of 91 grams is about 100 grams ... about 10² grams.

Your results could be different.

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