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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 60 kilogram student jumps down from a laboratory counter. At the instant he lands on the floor hus speed is 3 meters per secon

erastovalidia [21]

As per Newton's law rate of change in momentum is net force

so we can write it as

$F = \frac{dP}{dt}$

$F = \frac{m(v_f - v_i)}{\Delta t}$

now we know that

$m = 60 kg$

$v_f = 3 m/s$

$v_i = 0$

$\Delta t= 0.2 s$

from above equation

$F = \frac{60(3 - 0)}{0.2} = 900 N$

so he will experience 900 N force in above case

5 0

3 years ago

How do you think overpumping groudwater is related to the formation of sinkholes?

vladimir2022 [97]

Ground water keept the ground at a stable level when it is gone the cavern it was in has no support and is at risk of callaps

3 0

3 years ago

An above ground swimming pool of 30 ft diameter and 5 ft depth is to be filled from a garden hose (smooth interior) of length 10

STALIN [3.7K]

This question involves the concepts of dynamic pressure, volume flow rate, and flow speed.

It will take "5.1 hours" to fill the pool.

First, we will use the formula for the dynamic pressure to find out the flow speed of water:

$P=\frac{1}{2}\rho v^2\\\\v=\sqrt{\frac{2P}{\rho}}$

where,

v = flow speed = ?

P = Dynamic Pressure = 55 psi $(\frac{6894.76\ Pa}{1\ psi})$ = 379212 Pa

$\rho$ = density of water = 1000 kg/m³

Therefore,

$v=\sqrt{\frac{2(379212\ Pa)}{1000\ kg/m^3}}$

v = 27.54 m/s

Now, we will use the formula for volume flow rate of water coming from the hose to find out the time taken by the pool to be filled:

$\frac{V}{t} = Av\\\\t =\frac{V}{Av}$

where,

t = time to fill the pool = ?

A = Area of the mouth of hose = $\frac{\pi (0.015875\ m)^2}{4}$ = 1.98 x 10⁻⁴ m²

V = Volume of the pool = (Area of pool)(depth of pool) = A(1.524 m)

V = $[\frac{\pi (9.144\ m)^2}{4}][1.524\ m]$ = 100.1 m³

Therefore,

$t = \frac{(100.1\ m^3)}{(1.98\ x\ 10^{-4}\ m^2)(27.54\ m/s)}\\\\$

<u>t = 18353.5 s = 305.9 min = 5.1 hours</u>

Learn more about dynamic pressure here:

brainly.com/question/13155610?referrer=searchResults

7 0

2 years ago

50.0 meters away from a building. Tip of the building makes an angle of 63.0° with the horizontal. What is the height of the bui

soldi70 [24.7K]

Answer:

98.13m

Explanation:

Complete question

Daniel is 50.0 meters away from a building. Tip of the building makes an angle of 63.0° with the horizontal. What is the height of the building

CHECK THE ATTACHMENT

From the figure, using trigonometry

Tan(θ ) = opposite/adjacent

Where Angle (θ )= 63°

Opposite= X = height of the building

Adjacent= 50 m

Then substitute the values we have

Tan(63)= X/50

1.9626= X/50

X= 1.9626 × 50

X= 98.13m

Hence, the height of the building is 98.13m

8 0

2 years ago

2 An object has a negative charge. Which statement about the object is correct? A Electrons have been transferred onto it. B Ele

Rzqust [24]

Answer:

A . Electrons have been transferred into it.

ps.. PLZ give me brainliest!!!

Tysm! Hope this helps!

6 0

3 years ago