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3 years ago

9

If you closely examine a wood screw, you will see that the threads on it form a long _________________ wrapped around a cylinder

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2 answers:

makvit [3.9K]3 years ago

7 0

It is an inclined planed. You can think of a screw as an inclined plane wrapped a cylinder.

abruzzese [7]3 years ago

3 0

Incline plane cause if you would unravel screw it would look like a little steep hill going up

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Find the position and velocity of an object moving along a straight line with the given acceleration, initial velocity, and init

yulyashka [42]

To solve this problem we will apply the conventions related to the cinematic movement theorem, for which the kinematic equations of linear motion are equally detached. We will use the speed and position equations to determine the general formula according to the given values.

To velocity function we have

$v(t) = v(0)+\int^t_0 a(x) dx$

Our values are,

$a(x) = -9.8$

$v(0) = 20$

$s(0) = 0$

Replacing at this equation and solving we have that the equation for the velocity would be,

$v(t) = 20+\int^t_0 -9.8 dx$

$v(t) = 20-\int^t_0 9.8 dx$

$v(t) = 20-9.8x|^t_0$

$v(t) = 20-9.8(t-0)$

$v(t) = 20-9.8t$

Therefore the velocity function is $v(t) = 20-9.8t$

At the same time for the position function:

$s(t) = s(0) +\int^t_0 v(x) dx$

Replacing we have that

$s(t) = s(0) +\int^t_0 20-9.8x dx$

$s(t) = 0 +20x-9.8\frac{x^2}{2}|^t_0$

$s(t) = (20t-9.8\frac{t^2}{2})-(20\cdot 0 -9.8 \frac{0^2}{2})$

$s(t) = 20t-9.8\frac{t^2}{2}$

Therefore the position function is

$s(t) = 20t-9.8\frac{t^2}{2}$

5 0

3 years ago

On an essentially frictionless, horizontal ice rink, a skater moving at 5.0 m/s encounters a rough patch that reduces her speed

madreJ [45]

Answer:

The length of the rough patch is 4.345 meters.

Explanation:

According to the Work-Energy Theorem, change in kinetic energy is equal to the dissipated work due to friction. That is:

$K_{1} = K_{2} + W_{loss}$

Where:

$K_{1}$ , $K_{2}$ - Initial and final kinetic energy, measured in joules.

$W_{loss}$ - Work losses due to friction.

By applying definitions of kinetic energy and work, the expression described above is expanded:

$\frac{1}{2}\cdot m \cdot v_{1}^{2} = \frac{1}{2}\cdot m \cdot v_{2}^{2} + f \cdot \Delta s$

Where:

$v_{1}$ , $v_{2}$ - Initial and final speed of the skater, measured in meters per second.

$m$ - Mass of the skater, measured in kilograms.

$\Delta s$ - Length of the rough patch, measured in meters.

$f$ - Friction force, measured in newtons.

According to the statement, friction force is represented by the following expression:

$f = r \cdot m \cdot g$

Where:

$r$ - Ratio of friction force to weight, dimensionless.

$g$ - Gravitational constant, measured in meters per square second.

Then,

$\frac{1}{2}\cdot m \cdot v_{1}^{2} = \frac{1}{2}\cdot m \cdot v_{2}^{2} + r \cdot m \cdot g \cdot \Delta s$

The equation is simplified algebraically and patch length is cleared afterwards:

$\frac{1}{2}\cdot (v_{1}^{2}-v_{2}^{2}) = r \cdot g \cdot \Delta s$

$\Delta s = \frac{v_{1}^{2}-v_{2}^{2}}{2 \cdot r \cdot g }$

Given that $v_{1} = 5\,\frac{m}{s}$ , $v_{2} = 2.5\,\frac{m}{s}$ , $r = 0.22$ and $g = 9.807 \,\frac{m}{s^{2}}$ , the length of the rough patch is:

$\Delta s = \frac{\left(5\,\frac{m}{s} \right)^{2}-\left(2.5\,\frac{m}{s} \right)^{2}}{2\cdot (0.22)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}$

$\Delta s = 4.345\,m$

The length of the rough patch is 4.345 meters.

6 0

3 years ago

a freight train travels at v=60(1-e^-t) ft/s, where t is the elapsed time in seconds. Determine the distance traeled in tree sec

lesantik [10]

Answer

given,

$v = 60(1-e^{-t})\ ft/s$

t = 3 s

we know,

$v = \dfrac{dx}{dt}$

$a = \dfrac{dv}{dt}$

position of the particle

$dx = v dt$

integrating both side

$\int dx =\int 60(1-e^{-t}) dt$

$x = 60 t + 60 e^{-t}$

Position of the particle at t= 3 s

$x = 60\times 3+ 60 e^{-3}$

x = 182.98 ft

Distance traveled by the particle in 3 s is equal to 182.98 ft

now, particle’s acceleration

$a = \dfrac{dv}{dt}$

$a = \dfrac{d}{dt}60(1-e^{-t})$

$a = 60 e^{-t}$

at t= 3 s

$a = 60 e^{-3}$

a = 2.98 ft/s²

acceleration of the particle is equal to 2.98 ft/s²

7 0

4 years ago

1. ___ Release Of Pressure A. animals that burrow underground

kykrilka [37]

D.3. a.4. sorry this is all i know hope this helps

8 0

3 years ago

Hi people; there's a question that has been so confusing for me and a friend of mine.

Alborosie

Answer:

The correct answer is: 0°C + 0°C = 32°F

Explanation:

We need to remember the conversion equation from Celsius to Fahrenheit:

$y^{\circ}F=(x^{\circ}C * \frac{9}{5})+32$

In our case x = 0, then y will be:

$(0^{\circ}C * \frac{9}{5})+32=32$

$y=32^{\circ}F$

Now 0°C + 0°C is just 0°C because if we add a body at a certain temperature to another body with the same temperature the total temperature will the same.

Then, knowing that 0°C = 32°F we can conclude that:

$0^{\circ}C+0^{\circ}C=32^{\circ}F$

I hope it helps you!

7 0

3 years ago