1. Which description defines pseudoscience?

What is the difference between liquid and fluid friction

frez [133]

Friction is used to refer to forces that resist relative motion, in general,whereas viscosity refers specifically to resistive forces that occur between layers of a fluid when fluids attempt to flow. Fluid friction is the force that resists the movement of a solid object through a fluid(liquid or gas). Sky divers falling down through the sky experience a type of fluid frictioncalled air resistance.

6 0

4 years ago

One positve and one negative charhe jave a force of -5. 00 N between them. If the force changed to -170 N , by what factor did t

stepladder [879]

Answer:

The distance is shortenend by factor .1715

Explanation:

5 n = 1/r^2

sqrt (1/5) = r

170 n = 1 / ( x sqrt(1/5))^2

(xsqrt 1/5)^2 = 1/170

x sqrt 1/5 = .076696

x = .1715

5 0

2 years ago

A uniform solid sphere rolls down an incline without slipping. if the linear acceleration of the center of mass of the sphere is

Ulleksa [173]

A uniform solid sphere rolls down an incline without slipping. If the linear acceleration of the center of mass of the sphere is 0.19g, then what is the angle the incline makes with the horizontal?

3 0

3 years ago

A student of mass 63.4 kg, starting at rest, slides down a slide 21.2 m long, tilted at an angle of 26.1° with respect to thehor

Vera_Pavlovna [14]

Answer:

fr = 65.46 N , a = 8.74 m / s² and vf = 19.25 m / s

Explanation:

We write a reference system with an axis parallel to the slide and gold perpendicular axis, in this system we decompose the weight

sin 21.2 = Wx / W

cos21.2 = Wy / W

Wx = W sin21.2

Wy = W cos 21.2

We form Newton's equations

X axis

Wx -fr = m a

Y Axis

N- Wy = 0

N = Wy

fr = μ N

fr = μ (W cos 21.2)

fr = 0.113 63.4 9.8 cos 21.2

fr = 65.46 N

We replace and calculate the acceleration

W sin 21.2 - μ W cos 21.2 = m a

a = g (sin21.2 - μ cos 21.2)

a = 9.8 (without 21.2 - 0.113 cos 21.2)

a = 8.74 m / s²

This acceleration is along the slope of the slide, so we can calculate the distance

d = 21.2 m

vf² = v₀² + 2 a d

vf² = 0 + 2 a d

vf = √(2 8.74 21.2)

vf = √ (370,576)

vf = 19.25 m / s

3 0

3 years ago

Very far from earth (at R- oo), a spacecraft has run out of fuel and its kinetic energy is zero. If only the gravitational force

Margaret [11]

Answer:

Speed of the spacecraft right before the collision: $\displaystyle \sqrt{\frac{2\, G\cdot M_\text{e}}{R\text{e}}}$ .

Assumption: the earth is exactly spherical with a uniform density.

Explanation:

This question could be solved using the conservation of energy.

The mechanical energy of this spacecraft is the sum of:

the kinetic energy of this spacecraft, and
the (gravitational) potential energy of this spacecraft.

Let $m$ denote the mass of this spacecraft. At a distance of $R$ from the center of the earth (with mass $M_\text{e}$ ), the gravitational potential energy ( $\mathrm{GPE}$ ) of this spacecraft would be:

$\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R}$ .

Initially, $R$ (the denominator of this fraction) is infinitely large. Therefore, the initial value of $\mathrm{GPE}$ will be infinitely close to zero.

On the other hand, the question states that the initial kinetic energy ( $\rm KE$ ) of this spacecraft is also zero. Therefore, the initial mechanical energy of this spacecraft would be zero.

Right before the collision, the spacecraft would be very close to the surface of the earth. The distance $R$ between the spacecraft and the center of the earth would be approximately equal to $R_\text{e}$ , the radius of the earth.

The $\mathrm{GPE}$ of the spacecraft at that moment would be:

$\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}$ .

Subtract this value from zero to find the loss in the $\rm GPE$ of this spacecraft:

$\begin{aligned}\text{GPE change} &= \text{Initial GPE} - \text{Final GPE} \\ &= 0 - \left(-\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\right) = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \end{aligned}$

Assume that gravitational pull is the only force on the spacecraft. The size of the loss in the $\rm GPE$ of this spacecraft would be equal to the size of the gain in its $\rm KE$ .

Therefore, right before collision, the $\rm KE$ of this spacecraft would be:

$\begin{aligned}& \text{Initial KE} + \text{KE change} \\ &= \text{Initial KE} + (-\text{GPE change}) \\ &= 0 + \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \\ &= \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\end{aligned}$ .

On the other hand, let $v$ denote the speed of this spacecraft. The following equation that relates $v\!$ and $m$ to $\rm KE$ :

$\displaystyle \text{KE} = \frac{1}{2}\, m \cdot v^2$ .

Rearrange this equation to find an equation for $v$ :

$\displaystyle v = \sqrt{\frac{2\, \text{KE}}{m}}$ .

It is already found that right before the collision, $\displaystyle \text{KE} = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}$ . Make use of this equation to find $v$ at that moment:

$\begin{aligned}v &= \sqrt{\frac{2\, \text{KE}}{m}} \\ &= \sqrt{\frac{2\, G\cdot M_\text{e} \cdot m}{R_\text{e}\cdot m}} = \sqrt{\frac{2\, G\cdot M_\text{e}}{R_\text{e}}}\end{aligned}$ .

6 0

3 years ago