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3 years ago

13

A force of 0.2 Newtons is required to slide a book across the table. The book accelerates at 0.11 m/s squared. What is the mass

of the box?

1 answer:

adelina 88 [10]3 years ago

3 0

My calculations state, not rounding, the mass is 1.8

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In designing circular rides for amusement parks, mechanical engineers must consider how small variations in certain parameters c

Bess [88]

Answer:

Part a)

$dF = -\frac{mv^2}{r^2} dr$

Part b)

$dF = \frac{2mvdv}{r}$

Part c)

$dT = - \frac{2\pi r}{v^2} dv$

Explanation:

Part a)

As we know that force on the passenger while moving in circle is given as

$F = \frac{mv^2}{r}$

now variation in force is given as

$dF = -\frac{mv^2}{r^2} dr$

here speed is constant

Part b)

Now if the variation in force is required such that r is constant then we will have

$F = \frac{mv^2}{r}$

so we have

$dF = \frac{2mvdv}{r}$

Part c)

As we know that time period of the circular motion is given as

$T = \frac{2\pi r}{v}$

so here if radius is constant then variation in time period is given as

$dT = - \frac{2\pi r}{v^2} dv$

8 0

3 years ago

Beers of snakes thunderstorms darkness in water are classified as blank phobias

Scrat [10]

I believe all of these would be known as specific phobias.

4 0

3 years ago

By what factor would your weight be multiplied if the earth were1/2 as massavise and the diameter was unchanged

Nutka1998 [239]

<span>Let F be the force of gravity, G be the gravitational constant, M be the mass of the earth, m your mass and r the radius of the earth, then:

F = G(Mm / (4(pi)*r^2))

The above expression gives the force that you feel on the earth's surface, as it is today!

Let us now double the mass of the earth and decrease its diameter to half its original size.

This is the same as replacing M with 2M and r with r/2.

Now the gravitational force (F' ) on the new earth's surface is given by:

F' = G(2Mm / (4(pi)(r/2)^2)) = 2G(Mm / ((1/4)*4(pi)*r^2)) = 8G(Mm / (4(pi)*r^2)) = 8F

So:

F' = 8F

This implies that the force that you would feel pulling you down (your weight) would increase by 800%!

You would be 8 times heavier on this "new" earth!</span>

4 0

3 years ago

A cylindrical storage tank has a radius of 1.35 m. When filled to a height of 3.45 m, it holds 14,014 kg of a liquid industrial

Kruka [31]

Answer:

709.93 kg/m³

Explanation:

Density: This can be defined as the ratio of the mass of a body to the volume of that body. The S.I unit of density is kg/m³

From the question above,

D = m/v.............................. Equation 1

Where D = density, m = mass, v = volume.

Note: The volume of the liquid is equal to the volume of the height occupied by the liquid in the container

Since the tank is cylindrical,

v = πr²h........................ Equation 2

Where r = radius of the the tank, h = height of the liquid in the tank

Substitute equation 2 into equation 1

D = m/(πr²h)............... Equation 3

Given: m = 14014 kg, r = 1.35 m, h = 3.45 m, π = 3.14

Substitute into equation 3

D = 14014/(3.14×1.35²×3.45)

D = 14014/19.74

D = 709.93 kg/m³

6 0

3 years ago

The volume electric charge density of a solid sphere is given by the following equation: The variable r denotes the distance fro

qwelly [4]

Answer:

62.8 μC

Explanation:

Here is the complete question

The volume electric charge density of a solid sphere is given by the following equation: ρ = (0.2 mC/m⁵)r²The variable r denotes the distance from the center of the sphere, in spherical coordinates. What is the net electric charge (in μC) of the sphere if the radius of the sphere is 0.5 m?

Solution

The total charge on the sphere Q = ∫∫∫ρdV where ρ = volume charge density = 0.2r² and dV = volume element in spherical coordinates = r²sinθdθdrdΦ

So, Q = ∫∫∫ρdV

Q = ∫∫∫ρr²sinθdθdrdΦ

Q = ∫∫∫(0.2r²)r²sinθdθdrdΦ

Q = ∫∫∫0.2r⁴sinθdθdrdΦ

We integrate from r = 0 to r = 0.5 m, θ = 0 to π and Φ = 0 to 2π

So, Q = ∫∫∫0.2r⁴sinθdθdrdΦ

Q = ∫∫∫0.2r⁴[∫sinθdθ]drdΦ

Q = ∫∫0.2r⁴[-cosθ]drdΦ

Q = ∫∫0.2r⁴-[cosπ - cos0]drdΦ

Q = ∫∫∫0.2r⁴-[-1 - 1]drdΦ

Q = ∫∫0.2r⁴-[- 2]drdΦ

Q = ∫∫0.2r⁴(2)drdΦ

Q = ∫∫0.4r⁴drdΦ

Q = ∫0.4r⁴dr∫dΦ

Q = ∫0.4r⁴dr[Φ]

Q = ∫0.4r⁴dr[2π - 0]

Q = ∫0.4r⁴dr[2π]

Q = ∫0.8πr⁴dr

Q = 0.8π∫r⁴dr

Q = 0.8π[r⁵/5]

Q = 0.8π[(0.5 m)⁵/5 - (0 m)⁵/5]

Q = 0.8π[0.125 m⁵/5 - 0 m⁵/5]

Q = 0.8π[0.025 m⁵ - 0 m⁵]

Q = 0.8π[0.025 m⁵]

Q = (0.02π mC/m⁵) m⁵

Q = 0.0628 mC

Q = 0.0628 × 10⁻³ C

Q = 62.8 × 10⁻³ × 10⁻³ C

Q = 62.8 × 10⁻⁶ C

Q = 62.8 μC

3 0

3 years ago