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3 years ago

A pumpkin is dropped after 5 seconds its velocity is 47m/s what is its acceleration

Physics

2 answers:

ANEK [815]3 years ago

8 0

9.81m/s Hope i helped

Vesna [10]3 years ago

6 0

Answer:

Acceleration, $a=9.4\ m/s^2$

Explanation:

It is given that,

Initial speed of the pumpkin, u = 0

Final speed of the pumpkin, v = 47 m/s

Time taken, t = 5 seconds

Acceleration of an object is given by :

$a=\dfrac{v-u}{t}$

$a=\dfrac{47\ m/s}{5\ s}$

$a=9.4\ m/s^2$

So, the acceleration of the pumpkin is $9.4\ m/s^2$ . Hence, this is the required solution.

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What is the value of g on the surface of Saturn? Assume M-Saturn = 5.68×10^26 kg and R-Saturn = 5.82×10^7 m.Choose the appropria

Likurg_2 [28]

Answer:

Approximately $\rm 11.2 \; N \cdot kg^{-1}$ at that distance from the center of the planet.

Option A) The low value of $g$ near the cloud top of Saturn is possible because of the low density of the planet.

Explanation:

The value of $g$ on a planet measures the size of gravity on an object for each unit of its mass. The equation for gravity is:

$\displaystyle \frac{G \cdot M \cdot m}{R^2}$ ,

where

$G \approx 6.67\times 10^{-11}\; \rm N \cdot kg^{-2} \cdot m^2$ .
$M$ is the mass of the planet, and
$m$ is the mass of the object.

To find an equation for $g$ , divide the equation for gravity by the mass of the object:

$\displaystyle g = \left.\frac{G \cdot M \cdot m}{R^2} \right/\frac{1}{m} = \frac{G \cdot M}{R^2}$ .

In this case,

$M = 5.68\times 10^{26}\; \rm kg$ , and
$R = 5.82 \times 10^7\; \rm m$ .

Calculate $g$ based on these values:

$\begin{aligned} g &= \frac{G \cdot M}{R^2}\cr &= \frac{6.67\times 10^{-11}\; \rm N \cdot kg^{-2} \cdot m^2\times 5.68\times 10^{26}\; \rm kg}{\left(5.82\times 10^7\; \rm m\right)^2} \cr &\approx 11.2\; \rm N\cdot kg^{-1} \end{aligned}$ .

Saturn is a gas giant. Most of its volume was filled with gas. In comparison, the earth is a rocky planet. Most of its volume was filled with solid and molten rocks. As a result, the average density of the earth would be greater than the average density of Saturn.

Refer to the equation for $g$ :

$\displaystyle g = \frac{G \cdot M}{R^2}$ .

The mass of the planet is in the numerator. If two planets are of the same size, $g$ would be greater at the surface of the more massive planet.

On the other hand, if the mass of the planet is large while its density is small, its radius also needs to be very large. Since $R$ is in the denominator of $g$ , increasing the value of $R$ while keeping $M$ constant would reduce the value of $g$ . That explains why the value of $g$ near the "surface" (cloud tops) of Saturn is about the same as that on the surface of the earth (approximately $9.81\; \rm N \cdot kg^{-1}$ .

As a side note, $5.82\times 10^7\rm \; m$ likely refers to the distance from the center of Saturn to its cloud tops. Hence, it would be more appropriate to say that the value of $g$ near the cloud tops of Saturn is approximately $\rm 11.2 \; N \cdot kg^{-1}$ .

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3 years ago

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Answer:

The answer is C (Higher energy levels)

Explanation:

Because it's C

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3 years ago

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A cylindrical tank has a tight-fitting piston that allows the volume of the tank to be changed. The tank originally contains air

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Answer:

0.435atm

Explanation:

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Initial pressure P1= 0.740atm

Initial volume V1= 0.185 m3

Final pressure P2= ?

Final volume V2= 0.315 m3

At constant temperature, the pressure of a syste is inversely proportional to volume, by Boyles law then

P1V1=P2V2

P2=P1V1/V2

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Answer:

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X = Vt (1)

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X = 95 * 22.59

This is the distance where the package should be released.

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The Bay of Fundy has the greatest tidal ranges on Earth. What can you infer about the Bay of Fundy?

swat32

C is the answer

The Bay of Fundy has the greatest tidal ranges on Earth. What can you infer about the Bay of Fundy?

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3 years ago