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Gekata [30.6K]
3 years ago
9

Use the graph above to determine the change in speed of the object between 20 and 30 seconds?

Physics
2 answers:
allsm [11]3 years ago
4 0

Answer:

6 m/s

Explanation:

To determine the change in speed of the object, we just need to determine its speed at t = 30 s and at t = 20 s, and then calculate the difference.

The speed at t = 30 is:

v = 6 m/s

While the speed at t = 20 s is:

u = 0

Therefore, the change in speed is:

\Delta v = v-u=6-0 = 6 m/s

lara31 [8.8K]3 years ago
4 0

Explanation:

The attached figure shows the acceleration time graph of an object. We known that the slope of the a-t graph gives the velocity of an object. The area under the a-t graph gives the velocity of object.

In x axis, time t = 30 s- 20 s = 10 s

In y axis, acceleration, a = 6 - 0=6\ m/s^2

Area under the graph from 20 to 30 seconds is given by :

A=v=\dfrac{1}{2}\times 10\times 6

v = 30 m/s

So, the change in speed of the object is 30 m/s. Hence, this is the required solution.

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Answer:

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Explanation:

We take into account that all the energy from the lightning has been transformed into steam.

\Delta U = Q -  W\\Q = mC \Delta T\\Q = mL

We calculate the amount of energy required by water to convert into steam.

Q_{water} = 50 \times \times 4180 \times (100-37)\\= 1.132 \times 10^7 \ J

Q_{change\ to\ steam} = 50 \times 2.256 \times 106 \\= 1.128 \times 10^8 \ J

Q_{total} = 1.132 \times 10^7 + 1.128 \times 10^8\\= 1.126 \times 10^8 \ J

From the lightning we received 10^{10} \ J of energy, out of which 1.126 \times 10^8 has been used to convert the water into steam.

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We use this energy to convert steam into vapors.

Q = \Delta E

Q = \Delta E = mc (T_{f} - T{i})\\T_{f} = \frac {\Delta E}{mc} + T_{i}\\ \\T_{f} = 100 + \frac{9.88 \times 10^{10}}{50 \times 1970}\\T_{f} = 100 + 10^8\\T_{f} = 10 ^{ \ 8} \ {^ \circ } C

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A skateboarder, starting from rest, rolls down a 12.8-m ramp. When she arrives at the bottom of the ramp her speed is 8.89 m/s.
melamori03 [73]

Answer:

a) a = 3.09 m/s²

b) aₓ = 2.60 m/s²

Explanation:

a) The magnitude of her acceleration can be calculated using the following equation:

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<u>Where</u>:

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V_{0}: is the initial speed = 0 (since she starts from rest)

a: is the acceleration

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a = \frac{V_{f}^{2}}{2d} = \frac{(8.89 m/s)^{2}}{2*12.8 m} = 3.09 m/s^{2}

Therefore, the magnitude of her acceleration is 3.09 m/s².              

b) The component of her acceleration that is parallel to the ground is given by:

a_{x} = a*cos(\theta)

<u>Where</u>:

θ: is the angle respect to the ground = 32.6 °

a_{x} = 3.09 m/s^{2}*cos(32.6) = 2.60 m/s^{2}

Hence, the component of her acceleration that is parallel to the ground is 2.60 m/s².

I hope it helps you!

7 0
3 years ago
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