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3 years ago

A miler covers one mile in training at a 6 minute pace ( 4 laps on a 1/4 mile track ). What is his speed in mph.

Physics

1 answer:

Charra [1.4K]3 years ago

8 0

6 minutes is equal to 1/10 of an hour (6*10=60 minutes)

Therefore the miler is going 1/10 MPH

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Biologists have studied the running ability of the northern quoll, a marsupial indigenous to Australia. In one set of experiment

Drupady [299]

Answer:

$u_K=0.862$

Explanation:

The force of friction between the quails feet and the ground is:

$F=m*a$

$F_K=m*a$

$F_K=u_k*m*g$

$u_K*m*g=m*a_c$

$u_K*g=a$

$u_K=\frac{a_c}{g}$

$a_c=\frac{v^2}{r}$

So the coefficient of static is solve

$u_K=\frac{\frac{v^2}{r}}{g}$

$u_K=\frac{v^2}{r*g}=\frac{(2.6m/s)^2}{0.80m*9.8m/s^2}$

$u_K=0.862$

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3 years ago

An energy storage system based on a flywheel (a rotating disk) can store a maximum of 3.7 MJ when the flywheel is rotating at 16

Likurg_2 [28]

The moment of inertia of the flywheel is 2.63 kg- $m^{2}$

It is given that,

The maximum energy stored on the flywheel is given as

E=3.7MJ= 3.7× $10^{6}$ J

Angular velocity of the flywheel is 16000 $\frac{rev}{min}$ = 1675.51 $\frac{rad}{sec}$

So to find the moment of inertia of the flywheel. The energy of a flywheel in rotational kinematics is given by :

E = $\frac{1}{2}$ $Iw^{2}$

By rearranging the equation:

I = $\frac{2E}{w_{2} }$

I = 2.63 kg- $m^{2}$

Thus the moment of inertia of the flywheel is 2.63 kg- $m^{2}$ .

Learn more about moment of inertia here;

brainly.com/question/13449336

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1 year ago

The parachute on a drag racing car deploys at the end of a run. If the car has a mass of 820 kg and the car is moving 36 m/s, wh

Lelechka [254]

In order to determine the required force to stop the car, proceed as follow:

Calculate the deceleration of the car, by using the following formula:

$v^2=v^2_o-2ax$

where,

v: final speed = 0m/s (the car stops)

vo: initial speed = 36m/s

x: distance traveled = 980m

a: deceleration of the car= ?

Solve the equation above for a, replace the values of the other parameters and simplify:

$\begin{gathered} a=\frac{v^2_o-v^2}{2x} \\ a=\frac{(36\frac{m}{s})^2-(0\frac{m}{s})^2}{2(980m)}=0.66\frac{m}{s^2} \end{gathered}$

Next, consider that the formula for the force is:

$F=ma$

where,

m: mass of the car = 820 kg

a: deceleration of the car = 0.66m/s^2

Replace the previous values and simplify:

$F=(820kg)(0.66\frac{m}{s^2})=542.20N$

Hence, the required force to stop the car is 542.20N

4 0

1 year ago

solids, liquids, and gases are three forms of matter that:a. take up space. b. have mass. c. are made of atoms. d. all of the ab

LiRa [457]

Answer:

Explanation:

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3 years ago

If you weighed 130 pounds on earth, you would weigh _____pounds on the moon

Aneli [31]

Answer:

152 pounds

Explanation:

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3 years ago