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3 years ago

8

Which is an example of potential energy? a swimmer kicking their legs a car driving on the road an arrow flying at a target a yo

-yo in someone’s hand

1 answer:

kirza4 [7]3 years ago

7 0

A yo yo in someone’s hand. All of the other options are kinetic energy. Remember, potential energy is when things are at a stop. Kinetic energy is when things are moving.

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A car accelerates from 10 km/hr to 50 km/hr in 8 seconds. What is the acceleration?

nikklg [1K]

The answer is 5 km/hr

6 0

3 years ago

In some cases, neither of the two equations in the system will contain a variable with a coefficient of 1, so we must take a fur

Margaret [11]

Answer:

D = -4/7 = - 0.57

C = 17/7 = 2.43

Explanation:

We have the following two equations:

$3C + 4D = 5\ --------------- eqn (1)\\2C + 5D = 2\ --------------- eqn (2)$

First, we isolate C from equation (2):

$2C + 5D = 2\\2C = 2 - 5D\\C = \frac{2 - 5D}{2}\ -------------- eqn(3)$

using this value of C from equation (3) in equation (1):

$3(\frac{2-5D}{2}) + 4D = 5\\\\\frac{6-15D}{2} + 4D = 5\\\\\frac{6-15D+8D}{2} = 5\\\\6-7D = (5)(2)\\7D = 6-10\\\\D = -\frac{4}{7}$

<u>D = - 0.57</u>

Put this value in equation (3), we get:

$C = \frac{2-(5)(\frac{-4}{7} )}{2}\\\\C = \frac{\frac{14+20}{7}}{2}\\\\C = \frac{34}{(7)(2)}\\\\C = \frac{17}{7}\\$

<u>C = 2.43</u>

5 0

3 years ago

The height of a circular tower is 179 meters and its diameter is 48 meters. (Assume that the building is a perfect cylinder). Wh

EastWind [94]

The definition of a scale of 1: 166 will mean that the scale of 1 in the model will be equivalent to 166 times the measurement in the real model, therefore we will have that the height would be 166 times smaller than the 179m given:

$h = \frac{179m}{166} = 1.078m = 107.8cm$

The same for the diameter,

$\phi = \frac{48m}{166}= 0.2891m = 28.91cm$

The volume of a cylinder is given as

$V = (\pi r^2)(h)$

$V = (\pi (\frac{d}{2})^2)(h)$

$V = (\pi (\frac{ 28.91}{2})^2)(107.8)$

$V = 70762.8cm^3$

Therefore the volume would be $V = 70762.8cm^3$

6 0

3 years ago

What is the ideal banking angle for a gentle turn of 1.20-km radius on a highway with a 105 km/h speed limit (about 65 mi/h), as

Mnenie [13.5K]

Answer:

4.14°

Explanation:

given:

r = 1.2 km

v = 105 km/h

1) <em>convert your given </em>

a) r = 1.2 km to m = 1200m

b) v = 105 km/h to m/s = 29.2 m/s

2) <em>plug into your ideal banking angle equation</em>

$tan^-1$ ( $\frac{v^2}{rg}$ ) = $\frac{29.2^2}{(1200)(9.8)}$ = 4.14°

8 0

2 years ago

Determine the distance between a newly discovered planet and its single moon if the orbital period of the moon is 1.2 Earth days

Vinil7 [7]

Answer:

The distance is $r = 55430496 \ m$

Explanation:

From the question we are told that

The period of the moon $T = 1.2 days = 1.2 * 24 * 3600 = 103680 \ s$

The mass of the planet is $m_p = 9.38*10^{24} kg$

Generally the period of the moon is mathematically represented as

$T = 2 * \pi * \sqrt{ \frac{r^3 }{ G * m_p } }$

Here G is the gravitational constant with value

$G = 6.67 *10^{-11} \ N \cdot m^2/kg^2$

=> $T = 2 * \pi * \sqrt{ \frac{r^3 }{ G * m_p } }$

=> $103680 = 2 * 3.142 * \sqrt{ \frac{r^3 }{ 6.67*10^{-11} * 9.38*10^{24} } }$

=> $272218492.31 = \frac{r^3}{ 6.67 *10^{-11} * 9.38*10^{24}}$

=> $r = \sqrt[3]{ 1.7031241*10^{23}}$ j

=> $r = 55430496 \ m$

8 0

2 years ago