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3 years ago

A 20 kg box has 1000 joules of potential energy on a shelf. How high is the shelf?

Physics

1 answer:

navik [9.2K]3 years ago

5 0

Formula: PE = mgh
m = 20 kg, g = 9.8 m/s^2 h= ?
1000 = 20 * 9.8 * h
1000 = 196h
h = 5.10204082
The height is around 5m

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Mutations are always harmful. <br> true or false

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Answer:

false

Explanation:

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3 years ago

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3 years ago

A rock has a specific gravity of 2.32 and a volume of 8.64 in3 how much does it weigh

lianna [129]

Answer: 3.21 N

$Specific\hspace{1mm} gravity = \frac {Density\hspace{1mm}of\hspace{1mm}substance}{Density\hspace{1mm} of \hspace{1mm}water}$

$\Rightarrow Density \hspace{1mm}of\hspace{1mm}substance= 2.32\times 1000\hspace{1mm} kg/m^3 = 2320\hspace{1mm}kg/m^3\\ Mass =Density\times volume\\ \Rightarrow 2320 \hspace{1mm} kg/m^3\times 8.64 \hspace{1mm}in^3\times \frac {1.64\times10^{-5} m^3}{1\hspace{1mm}in^3}=0.328 kg$

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$weight= 0.328\times9.8=3.21\hspace{1mm}N$

Hence, the rock would weigh 3.21 N.

3 0

3 years ago

Tory has a mass of 40kg. She sleds down a hill that has a slope of 25 degrees. what is the component of her weight that is along

Fudgin [204]

1.7 x 10^2 N

or 166 N

First you find the vertical component of the weight, which is 9.8*40, (g*m), which is 392 N. You then find the angle between that and the slope, which is 90-25, which is 65. You then multiply the vertical weight by cos(65), to find the component of that that is parallel to the slope. You get 165.666 N

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4 years ago

A ball is thrown into the air by a baby alien on a planet in the system of Alpha Centauri with a velocity of 35 ft/s. Its height

Svetllana [295]

Answer:

Part A:

(a): -121.26 ft/s.

(b): -121.13 ft/s.

(c): -121.052 ft/s.

(d): -121.026 ft/s.

Part B:

-121.00 ft/s.

Explanation:

Given that the height of the balloon after t seconds is

$\rm y(t) = 35 t-26t^2.$

The average velocity of an object is defined as the total distance traveled by the object divided by the time taken in covering that distance.

$\rm v_{av} = \dfrac{y_2-y_1}{t_2-t_1}$

where,

$\rm y_2,\ y_1$ are the positions of the object at time $\rm t_1$ and $\rm t_2$ respectively.

For the average velocity for the time period beginning when t=3 and lasting .01 sec.

For this case,

$\rm t_1 = 3\ sec.\\t_2 = 3+0.01\ sec = 3.01\ sec.\\\\Therefore, \\\\y_1 = 35\cdot 3-26\cdot 3^2=-129\ ft.\\y_2 = 35\cdot 3.01-26\cdot 3.01^2=-130.2126\ ft.\\\\\Rightarrow v_{av}=\dfrac{-130.2126-(-129)}{3.01-3}=-121.26\ ft/s.$

For the average velocity for the time period beginning when t=3 and lasting .005 sec.

For this case,

$\rm t_1 = 3\ sec.\\t_2 = 3+0.005\ sec = 3.005\ sec.\\\\Therefore, \\\\y_1 = 35\cdot 3-26\cdot 3^2=-129\ ft.\\y_2 = 35\cdot 3.005-26\cdot 3.005^2=-129.60565\ ft.\\\\\Rightarrow v_{av}=\dfrac{-129.60565-(-129)}{3.005-3}=-121.13\ ft/s.$

For the average velocity for the time period beginning when t=3 and lasting .002 sec.

For this case,

$\rm t_1 = 3\ sec.\\t_2 = 3+0.002\ sec = 3.002\ sec.\\\\Therefore, \\\\y_1 = 35\cdot 3-26\cdot 3^2=-129\ ft.\\y_2 = 35\cdot 3.002-26\cdot 3.002^2=-129.2421\ ft.\\\\\Rightarrow v_{av}=\dfrac{-129.2421-(-129)}{3.002-3}=-121.052\ ft/s.$

For the average velocity for the time period beginning when t=3 and lasting .001 sec.

For this case,

$\rm t_1 = 3\ sec.\\t_2 = 3+0.001\ sec = 3.001\ sec.\\\\Therefore, \\\\y_1 = 35\cdot 3-26\cdot 3^2=-129\ ft.\\y_2 = 35\cdot 3.001-26\cdot 3.001^2=-129.121\ ft.\\\\\Rightarrow v_{av}=\dfrac{-129.121-(-129)}{3.001-3}=-121.026\ ft/s.$

The instantaneous velocity of the balloon at the given time is defined as the rate of change of its position at that time.

$\rm v(t) = \dfrac{dy}{dt}\\=\dfrac{d}{dt}\left ( 35 t-26t^2\right )\\\\=35-26\times 2t.\\\\At\ t=3,\\\\v(t)=35-26\times 2\times 3=-121.00\ ft/s.$

<u>Note:</u><em> The negative sign with all the velocities indicates that the direction of these velocities are downwards.</em>

8 0

4 years ago