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Nastasia [14]
3 years ago
7

Goldie Doldfish, a speed swimmer, loves to rece around the parks pond, which is 0.5 miles around. If she can swim around the tra

ck in 2 hours, whar is her average speed?
Physics
1 answer:
rosijanka [135]3 years ago
8 0

Answer:

Her average speed is 0.25 miles / hour.

Explanation:

I did 0.5 divided by 2 = 0.25 miles / hour.

<em>Hope this helps! :)</em>


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The third one pls help
Nadya [2.5K]

Answer:

20 ms¯¹

Explanation:

3. Determination of the final velocity

From the question given above, the following data were obtained:

Time (t) = 4 s

Acceleration (a) = 5 ms¯²

Initial velocity (u) = 0 ms¯¹

Final velocity (v) =?

Acceleration is simply defined as the change in velocity per unit time.

Mathematically, it can be expressed as:

Acceleration (a) = final velocity – Initial velocity / time

a = v – u / t

With the above formula, we can obtain the final velocity of the car as follow:

Time (t) = 4 s

Acceleration (a) = 5 ms¯²

Initial velocity (u) = 0 ms¯¹

Final velocity (v) =?

a = v – u / t

5 = v – 0 / 4

5 = v / 4

Cross multiply

v = 5 × 4

v = 20 ms¯¹

Thus, the final velocity of the car is 20 ms¯¹

7 0
3 years ago
Stop to Think 7.1 Which one is greater- the attraction of the Earth on 1 kg lead or the attraction of 1 kg of led
Daniel [21]

Answer:

1 kg lead to earth is greater attraction as mass of earth is much more than 1kg lead.

Explanation:

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7 0
3 years ago
A 150 kg line backer sacks the 120 kg quarterback. With what force is the quarterback sacked if the line backer has an accelerat
Gekata [30.6K]

Answer:

The force required to move the quarterback with linebacker is <u>1215 N</u>

Explanation:

\text { Mass of linebacker } \mathrm{m}_{2}=150 \mathrm{kg}

\text { Mass of quarterback } \mathrm{m}_{2}=120 \mathrm{kg}

\text { Moved at an acceleration }(a)=4.5 \mathrm{m} / \mathrm{s}^{2}

Using Newton's second law, it is established that  F = Ma

Where F is net force acting on the system, a is the acceleration and M is mass of the two object \left(m_{1}+m_{2}\right)

Now consider both \mathrm{m}_{1} \text { and } \mathrm{m}_{2}as a system, so net force acting on the system is \text { Force }=\left(m_{1}+m_{2}\right) a

Substitute the given values in the above formula,

\text { Force }=(150+120) \mathrm{kg} \times 4.5 \mathrm{m} / \mathrm{s}^{2}

\text { Force }=270 \mathrm{kg} \times 4.5 \mathrm{m} / \mathrm{s}^{2}

Force = 1215 N

<u>1215 N </u>is the force required to move the quarterback with linebacker.

5 0
3 years ago
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