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3 years ago

Running longer would be an example of how you might improve your running performance

1 answer:

8 0

This is true, however if it wee better worded it would be obvious. This is simply stating that you have a higher endurance which would be a visual improvement in your running performance. Hope this helped <3

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A rock is thrown horizontally from a building at 15 m/s. It hits the ground 45 m from the base of the building. How high was the

evablogger [386]

Answer:

Explanation:

hope this helps!

7 0

2 years ago

A 125 g pendulum bob hung on a string of length 35 cm has the same period as when the bob is hung from a spring and caused to os

Bingel [31]

Answer:

k = 3.5 N/m

Explanation:

It is given that the time period the bob in pendulum is the same as its time period in spring mass system:

$Time\ Period\ of\ Pendulum = Time\ Period\ of\ Spring-Mass\ System\\2\pi \sqrt{\frac{l}{g}} = 2\pi \sqrt{\frac{m}{k}$

$\frac{l}{g} = \frac{m}{k}\\\\ k = g\frac{m}{l}$

where,

k = spring constant = ?

g = acceleration due to gravity = 9.81 m/s²

m = mass of bob = 125 g = 0.125 kg

l = length of pendulum = 35 cm = 0.35 m

Therefore,

$k = (9.81\ m/s^2)(\frac{0.125\ kg}{0.35\ m})\\\\$

k = 3.5 N/m

4 0

2 years ago

A satellite originally moves in a circular orbit of radius R around the Earth. Suppose it is moved into a circular orbit of radi

sergiy2304 [10]

satellite originally moves in a circular orbit of radius R around the Earth. Suppose it is moved into a circular orbit of radius 4R.

(i) What does the force exerted on the satellite then become?

eight times largerfour times larger one-half as largeone-eighth as largeone-sixteenth as large(ii) What happens to the satellite's speed?eight times largerfour times larger one-half as largeone-eighth as largeone-sixteenth as large(iii) What happens to its period?eight times largerfour times larger one-half as largeone-eighth as largeone-sixteenth as large

3 0

2 years ago

Read 2 more answers

In magazine car tests an important indicator of performance is the zero to 60 mph (0 to 96.6 km/h) acceleration time. A time bel

slavikrds [6]

Answer:

2.73414 seconds

467622.66798 J

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

g = Acceleration due to gravity = 9.81 m/s² = a

$v=60\times \dfrac{1609.34}{3600}=26.822\ m/s$

$mgh=\dfrac{1}{2}mv^2\\\Rightarrow h=\dfrac{v^2}{2g}\\\Rightarrow h=\dfrac{26.822^2}{2\times 9.81}\\\Rightarrow h=36.66766\ m$

$s=ut+\frac{1}{2}at^2\\\Rightarrow 36.66766=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{36.66766\times 2}{9.81}}\\\Rightarrow t=2.73414\ s$