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3 years ago

Science please helppp

Physics

1 answer:

Daniel [21]3 years ago

3 0

The red box must way more. Gravitational potential energy is the product of a an objects mass times the acceleration due to gravity (which is constant on earth) times its height. Since the objects are on the same shelf they are at the same height, and since gravitational acceleration is constant as long as we stay on planet earth, then the mass is the only possible thing that could have changed. This means that the red box must weigh more than the blue box.

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A cylindrical wire made of an unknown alloy hangs from a support in the ceiling. You measure the relaxed length of the wire to b

IceJOKER [234]

Answer: hello some of your values are wrongly written hence I will resolve your question using the right values

answer:

stiffness = 1.09 * 10^-6 N/m

Explanation:

Given data:

Length ( l ) = 16 m

radius of wire ( r ) = 3.5 m

mass ( m ) = 5kg

<u>Distance stretched ( Δl ) = 4 * 10^-3 m </u> ( right value )

<u>average bond length ( between atoms ) = 2.3 * 10^-10 m </u>( right value)

first step : calculate the area

area ( A ) = πr^2 = π * ( 3.5)^2 = 38.48 m^2

γ = MgL / A Δl

= [ (5 * 9.81 * 16 ) / ( 38.48 * (4.3*10^-3) ) ]

= 784.8 / 0.165 = 4756.36 N/m^2

hence : stiffness = γ * bond length

= 4756.36 * 2.3 * 10^-10 = 1.09 * 10^-6 N/m

8 0

3 years ago

Which is the higher temperature? (Assume temperatures to be exact numbers.)(a)1.)265°C or 265°F?2.)265°C3.)265°F 4.)They are the

vazorg [7]

Remember the following formulas to convert temperatures between Farenheit and Celsius:

$\begin{gathered} \text{ Farenheit to Celsius:} \\ C=\frac{5(F-32)}{9} \\ \\ \text{ Celsius to Farenheit:} \\ F=\frac{9}{5}C+32 \end{gathered}$

Convert all the temperatures to Celsius to compare them.

Convert 265ºF to Celsius:

$\begin{gathered} C=\frac{5(F-32)}{9} \\ \\ =\frac{5(265-32)}{9} \\ \\ =129.444... \end{gathered}$

Then, 265ºF is equal to 129.444...ºC. Then, 265ºC is higher than 265ºF.

The correct choice is option 2) 265ºC.

Part b)

Convert 370ºF to Celsius:

$\begin{gathered} C=\frac{5(F-32)}{9} \\ \\ =\frac{5(370-32)}{9} \\ \\ =187.777...ºC \end{gathered}$

Then, 370ºF is equal to 187.777...ºC. Then, 200ºC is higher than 370ºF.

The correct choice is option 2) 200ºC.

7 0

1 year ago

SOMEONE HELP Jeffery used a stream table as a model to show how water erodes soil. As he sprayed water on the stream table, some

Svetach [21]

Answer:

Hydrosphere and atmosphere

Explanation:

7 0

3 years ago

An automobile follows a circular road whose radius is 50 m. Let x and y respectively denote the eastern and northern directions,

IceJOKER [234]

Answer:

The automobile is running at speed of 23.806 meters per second.

Explanation:

From Kinematic we remember that acceleration ( $a$ ) can be defined by this ordinary differential equation in terms of distance:

$a = v\cdot \frac{dv}{ds}$ (1)

Where:

$v$ - Speed of the automobile, measured in meters per second.

$s$ - Distance travelled by the automobile, measured in meters.

If we know that $v = 0.5\cdot s - 0.0025\cdot s^{2}$ , then the equation of acceleration is:

$a = (0.5\cdot s - 0.0025\cdot s^{2})\cdot \left(0.5-0.0050\cdot s\right)$

$a = s\cdot (0.5-0.0025\cdot s)\cdot (0.5-0.0050\cdot s)$

$a = s\cdot (0.0025\cdot s - 0.5)\cdot (0.0050\cdot s-0.5)$

But distance covered by the vehicle is defined by the following formula:

$s = \theta \cdot r$ (2)

Where:

$\theta$ - Arc angle, measured in radians.

$r$ - Radius, measured in radians.

Then, we expand (1) by means of this result:

$a = \theta\cdot r \cdot (0.0025\cdot \theta\cdot r -0.5)\cdot (0.0050\cdot \theta \cdot r-0.5)$

$a = \theta\cdot r \cdot (1.25\times 10^{-5}\cdot \theta^{2}\cdot r^{2}-3.75\times 10^{-3}\cdot \theta\cdot r +0.25)$

$a = 1.25\times 10^{-5}\cdot \theta^{3}\cdot r^{3}-3.75\times 10^{-3}\cdot \theta^{2}\cdot r^{2}+0.25\cdot \theta \cdot r$

And finally we get the following third order polynomial:

$1.25\times 10^{-5}\cdot \theta^{3}\cdot r^{3}-3.75\times 10^{-3}\cdot \theta^{2}\cdot r^{2}+0.25\cdot \theta \cdot r - a = 0$ (3)

If we know that $r = 50\,m$ , $a = 0.6\cdot g$ and $g = 9.807\,\frac{m}{s^{2}}$ , then the polynomial becomes into this:

$1.5625\cdot \theta^{3}-9.375\cdot \theta^{2} +12.5\cdot \theta - 5.886 = 0$ (3b)

This polynomial can be solved analytically by Cardano's Method or by numerical methods. The roots of the polynomial are, respectivelly:

$\theta_{1} \approx 4.365\,rad$ , $\theta_{2} \approx 0.818+i\,0.441\,rad$ , $\theta_{3}\approx 0.818 -i\,0.441\,rad$ , $\theta_{4} \approx 1.563\,rad$

Both first and fourth roots are physically reasonable solution, but the latter represents the angle where automobile begins to skid <em>first</em>. Then, the automobile begins to skid at an angle of 1.563 radians relative to x axis.

The distance travelled by the automobile is: ( $r = 50\,m$ , $\theta \approx 1.563\,rad$ )