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3 years ago

Determine the potential energy of a 625-kg roller coaster car that is at the top of a hill that is 135 m tall.

Physics

1 answer:

Mice21 [21]3 years ago

4 0

Answer:

Caused by gravity and the roller coaster's position, the potential energy is stored in the roller coaster. For example, this ball is at the top of a hill, where potential energy is at it's highest.

Explanation:

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This graph shows the energy of a reaction over time. Which statement is

kirill115 [55]

Answer: D.

Explanation: Just took the quiz.

3 0

2 years ago

Read 2 more answers

C. is what percent of 125?

solmaris [256]

Answer:

Step 1: We make the assumption that 125 is 100% since it is our output value.

Step 2: We next represent the value we seek with $x$.

Step 3: From step 1, it follows that $100\%=125$.

Step 4: In the same vein, $x\%=125$.

Step 5: This gives us a pair of simple equations:

$100\%=125(1)$.

Explanation:

4 0

3 years ago

What is 16.558 m/s rounded to three significant figures?

In-s [12.5K]

Answer:

Option C. 16.6 m/s

Explanation:

To round this 16.558 m/s to 3sf, we need to count the number beginning from 1. When we get to the 3rd number( ie 5), we'll examine the fourth number(i.e 5)to see if it less than five or greater. If it less than five, then we'll discard it. But if it five or greater, we'll approximate it and add it to the 3rd number.

So.

16.558 m/s = 16.6m/s to 3sf

3 0

3 years ago

The magnitude of the gravitational field strength near Earth's surface is represented by

Zanzabum

Answer:

The magnitude of the gravitational field strength near Earth's surface is represented by approximately $9.82\,\frac{m}{s^{2}}$ .

Explanation:

Let be M and m the masses of the planet and a person standing on the surface of the planet, so that M >> m. The attraction force between the planet and the person is represented by the Newton's Law of Gravitation:

$F = G\cdot \frac{M\cdot m}{r^{2}}$

Where:

$M$ - Mass of the planet Earth, measured in kilograms.

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

$r$ - Radius of the Earth, measured in meters.

$G$ - Gravitational constant, measured in $\frac{m^{3}}{kg\cdot s^{2}}$ .

But also, the magnitude of the gravitational field is given by the definition of weight, that is:

$F = m \cdot g$

Where:

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

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

After comparing this expression with the first one, the following equivalence is found:

$g = \frac{G\cdot M}{r^{2}}$

Given that $G = 6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}}$ , $M = 5.972 \times 10^{24}\,kg$ and $r = 6.371 \times 10^{6}\,m$ , the magnitude of the gravitational field near Earth's surface is:

$g = \frac{\left(6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}} \right)\cdot (5.972\times 10^{24}\,kg)}{(6.371\times 10^{6}\,m)^{2}}$

$g \approx 9.82\,\frac{m}{s^{2}}$

The magnitude of the gravitational field strength near Earth's surface is represented by approximately $9.82\,\frac{m}{s^{2}}$ .

4 0

3 years ago

Can you please solve this for me urgently want to make sure if my answers are correct?

MA_775_DIABLO [31]

Answer:

Resolving horizontally. :

$\sum d_{x} = - (17.0 \cos 20.0) - (11.0 \cos 35.0) + (30.0 \cos 50.0) + 0 \\ { \underline{d _{x} = - 5.702 \: m}} \\ \\ \sum d _{y} = (17.0 \sin 20.0) + 12.0 - (11.0 \sin 35.0) - (30.0 \sin 50.0) \\ { \underline{d _{y} = - 11.476 \: m}}$

therefore, for resultant:

$d = \sqrt{ {d _{y} }^{2} + d _{x} {}^{2} }$

substitute:

$d = \sqrt{ {( - 5.702)}^{2} + {( - 11.476)}^{2} } \\ \\ d = \sqrt{164.211} \\ \\ { \boxed{ \boxed{ \bf{d = 12.8 \: m}}}}$

6 0

2 years ago