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3 years ago

Explain in your own words how volume is measured.

Physics

2 answers:

yawa3891 [41]3 years ago

8 0

Answer:

volume=length × weight ×height

pav-90 [236]3 years ago

7 0

Answer:

volume measured by pid^3 over 6 i think

Explanation:

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Picture a ball traveling at a constant speed around the inside of a circular Structure. is the ball acceleration? Explain why?

baherus [9]

Yes. On a circular path, the direction of motion is constantly changing. Change of direction is acceleration, even at constant speed.

8 0

3 years ago

Read 2 more answers

A 460 g model rocket is on a cart that is rolling to the right at a speed of 3.0 m/s. The rocket engine, when it is fired, exert

ollegr [7]

Answer:

The rocket should be launched at a horizontal distance of <u>6.45 m</u> left of the loop.

Explanation:

Given:

Mass of the rocket model (m) = 460 g = 0.460 kg [1 g = 0.001 kg]

Speed of the cart (v) = 3.0 m/s

Thrust force by the rocket engine (F) = 8.5 N

Vertical height of the loop (y) = 20 m

Let the horizontal distance left to the loop for launch be 'x'. Also, let 't' be the time taken by the rocket to reach the loop.

Now, there are two types of motion associated with the rocket- one is horizontal and the other vertical.

So, we will apply kinematics of motion in the two directions separately.

Vertical motion:

Given:

Force acting in the vertical direction is given as:

$F_y=F-mg=8.5-0.46\times 9.8=3.992\ N$

So, acceleration in the vertical direction is given as:

Acceleration = Force ÷ mass

$a_y=\frac{F_y}{m}=\frac{3.992\ N}{0.46\ kg}=8.678\ m/s^2$

Vertical displacement of rocket is same as the height of loop. So, $y=20\ m$

There is no initial velocity in the vertical direction. So, $u_y=0\ m/s$

Now, applying equation of motion in vertical direction. we have:

$y=u_yt+\frac{1}{2}a_yt^2\\\\20=0+\frac{1}{2}\times 8.678t^2\\\\20=4.339t^2\\\\t^2=\frac{20}{4.339}\\\\t=\sqrt{\frac{20}{4.339}}=2.15\ s$

Now, time taken to reach the loop is 2.15 s.

Horizontal motion:

There is no acceleration in the horizontal motion. So, displacement in the horizontal direction is equal to the product of horizontal speed and time.

Also, displacement of the rocket in the horizontal direction is nothing but the horizontal distance of its launch left of the loop. So,

$x=vt\\\\x=3.0\ m/s\times 2.15\ s\\\\x=6.45\ m$

Therefore, the rocket should be launched at a horizontal distance of 6.45 m left of the loop.

5 0

4 years ago

Suppose Sammy Sosa hits a home run which travels 361. ft (110. m). Leaving the bat at 50 degrees above the horizontal, how high

Mashcka [7]

Answer:

The horizontal distance of Sosa is 276.526 ft or 84.28 meter.

Explanation:

As shown in the figure, let point O is the starting point of Sosa. She travels 361 ft at an angle 50 degree with the horizontal.

sin 50 = $\frac{OM}{OP}$

0.7660 = h / 361

h = 276.526 ft

h = 84.28 meter

The horizontal distance of Sosa is 276.526 ft or 84.28 meter.

5 0

4 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

3 years ago

I need help answering this question.

Sloan [31]

The work done is the loss of kinetic energy.

Loss of kinetic energy = m*(v1^2 - v2^2)/2 = 10 kg * [ (99m/s)^2 - (1m/s)^2]/2 = 49,000 J

6 0

3 years ago