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Sati [7]
3 years ago
11

A rocket fires its engines to launch straight up from rest with an upward acceleration of 5 m/s2 for 10 seconds. After this time

, the engine shuts off and the rocket freely falls straight down back to Earth's surface.
Please help! This is for AP Physics.

Physics
2 answers:
IrinaK [193]3 years ago
7 0

Answer:

Hi! These graphs I'm providing below are the answer to this problem, as well as the free response sections. I'll attach them below for you.

Hope this helps! :)

Shtirlitz [24]3 years ago
4 0

EDIT: I didn't realize that I had solved the problem wrong, so I deleted my answer. The other answer to this question is really good so look at that one, not this. I'm so sorry!

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_____ is very corrosive and can cause rusting so metal tanks shouldn't be used
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Answer:

Iron

Explanation:

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Consider eight,eight-cubic centimeter (8 cm3) sugar cubes stacked so that they form a single 2 x 2 x 2 cube. How does the surfac
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To find the surface area of a single cube we first nees to take the cube root of 8cm3 which is 2.

Now we know that the length of each side is 2 and we can find the area of one side by doing 2x2 which is 4.

To find the total surface area of one cube we do 4 times 6 side giving us a total of 24cm2.

To find the total surface area of the 8 individual cubes, we multiply 24cm2 by 8 to give us a total of 192cm2.

Now to find the total surface area of the one large cube, we know that each side of one of the small cubes is 4cm2 and the large cube is set up so that there are two levels of four cubes right on top of each other. So, the total area of each side of the large cube is 4cm2 times 4 which gives us 16cm2.

Then we multiply 16cm2 by 6 sides to give us a total surface area of 96cm2.

The ratio of the surface area of the single large cube comapred to the total surface area of the single cubes is 96:192

We can further simplify this ratio:

96:192

48:96

24:48

12:24

6:12

3:6

1:2
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3 years ago
What happens to the temperature of a substance while it is changing state?
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It stays constant, because it's using that energy to change state
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you're reading from the journal of a European explorer from the early 1600s. In one passage, the explorer describes itting on th
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Two 51 g blocks are held 30 cm above a table. As shown in the figure, one of them is just touching a 30-long spring. The blocks
vivado [14]

The concept of this question can be well understood by listing out the parameters given.

  • The mass of the block = 51 g = 51 × 10⁻³ kg
  • The distance of the block from the table = 30 cm
  • Length of the spring = 30 cm

The purpose is to determine the spring constant.

Let us assume that the two blocks are Block A and Block B.

At point A on block A, the initial velocity on the block is zero

i.e. u = 0

We want to determine the time it requires for Block A to reach the table. The can be achieved by using the second equation of motion which can be expressed by using the formula.

\mathsf{S = ut + \dfrac{1}{2}gt^2}

From the above formula,

The distance (S) = 30 cm; we need to convert the unit to meter (m).

  • Since 1 cm = 0.01 m
  • Then, 30cm = 0.3 m

The acceleration (g) due to gravity = 9.8 m/s²

∴

inputting the values into the equation above, we have;

\mathsf{0.3 = (0)t + \dfrac{1}{2}*(9.80)*(t^2)}

\mathsf{0.3 = \dfrac{1}{2}*(9.80)*(t^2)}

\mathsf{0.3 =4.9*(t^2)}

By dividing both sides by 4.9, we have:

\mathsf{t^2 = \dfrac{0.3}{4.9}}

\mathsf{t^2 = 0.0612}

\mathsf{t = \sqrt{0.0612}}

\mathsf{t =0.247  \ seconds}

However, block B comes to an instantaneous rest on point C. This is achieved by the dropping of the block on the spring. During this process, the spring is compressed and it bounces back to oscillate in that manner. The required time needed to get to this point C is half the period, this will eventually lead to the bouncing back of the block with another half of the period, thereby completing a movement of one period.

By applying the equation of the time period of a simple harmonic motion.

\mathsf{T = 2 \pi \sqrt{\dfrac{m}{k}}}

where the relation between time (t) and period (T) is:

\mathsf{t = \dfrac{T}{2}}

T = 2t

T = 2(0.247)

T = 0.494 seconds

\mathsf{T = 2 \pi \sqrt{\dfrac{m}{k}}}

By making the spring constant k the subject of the formula:

\mathsf{\dfrac{T}{2 \pi } = \sqrt{ \dfrac{m}{k}}}

\Big(\dfrac{T}{2 \pi }\Big)^2 = { \dfrac{m}{k}

\dfrac{T^2}{(2 \pi)^2 }= { \dfrac{m}{k}

\mathsf{ T^2 *k = 2 \pi^2*m} \\ \\  \mathsf{  k = \dfrac{2 \pi^2*m}{T^2}}

\mathsf{  k =\Big( \dfrac{(2 \pi)^2*(51 \times 10^{-3})}{(0.494)^2} \Big) N/m}

\mathbf{  k =8.25 \ N/m}

Therefore, we conclude that the spring constant as a result of instantaneous rest caused by the compression of the spring is 8.25 N/m.

Learn more about simple harmonic motion here:

brainly.com/question/17315536?referrer=searchResults

6 0
2 years ago
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