All categories

4 years ago

What is the speed of a walking person in m/s if the person travels 1500 m in 25 minutes

Physics

1 answer:

pochemuha4 years ago

8 0

They would be walking 60 Meters a minute.

Hope it helps!

You might be interested in

Acceleration is positive when you are speeding up.

Vikki [24]

Answer:

B Negative

Explanation:

Its negative because when your going 25 mph your moving faster and when your coming up to the stop sign you will start to slow down going 10 mph then 0 mph when you reach the stop sign so it will be negative.

8 0

2 years ago

What happens when light enters a pair of glasses

alexdok [17]

Answer:

It refracts when it hits the glass.

6 0

3 years ago

When developing an experimental design, which action would improve the

il63 [147K]

When developing an experimental design, the action that would improve the quality of the results is to ensure that it answers a question about cause and effect.

<h3>What is experimental design?</h3>

Experimental design is a concept used to organize, conduct, and interpret results of experiments in an efficient way, making sure that as much useful information as possible is obtained by performing a small number of trials.

Thus, when developing an experimental design, the action that would improve the quality of the results is to ensure that it answers a question about cause and effect.

Learn more about experimental design here: brainly.com/question/17274244

#SPJ1

8 0

2 years ago

Trucks can be run on energy stored in a rotating flywheel, with an electric motor getting the flywheel up to its top speed of 20

Genrish500 [490]

Answer:

The time it can operate between chargins in minutes is

$t=102.8 minutes$

Explanation:

Given: $m=500kg$ , $r=1.0m$ , $w=200\pi rad/s$

a). The rotational kinetic energy

$K_R=\frac{1}{2}*I*w^2$

$I=\frac{1}{2}*m*r^2$

$I=\frac{1}{2}*500kg*(1.0m)^2$

$I=250 kg*m^2$

$K_R=\frac{1}{2}*250kg*m^2*(200\pi rad/s)^2$

$K_R=49.348x10^6J$

b). The power average 0.8kW un range time can be find

$P=\frac{K_R}{t'}$

Solve to t'

$t=\frac{K_R}{P}$

$t=\frac{49.348x10^6}{0.8x10^3w}=6168.5s$

$t=6168.5s\frac{1minute}{60s}=102.8 minutes$

3 0

3 years ago

A student gives a 5.0 kg box a brief push causing the box to move with an initial speed of 8.0 m/s along a rough surface. The bo

WITCHER [35]

Answer:

The time taken to stop the box equals 1.33 seconds.

Explanation:

Since frictional force always acts opposite to the motion of the box we can find the acceleration that the force produces using newton's second law of motion as shown below:

$F=mass\times acceleration\\\\\therefore acceleration=\frac{Force}{mass}$

Given mass of box = 5.0 kg

Frictional force = 30 N

thus

$acceleration=\frac{30}{5}=6m/s^{2}$

Now to find the time that the box requires to stop can be calculated by first equation of kinematics

The box will stop when it's final velocity becomes zero

$v=u+at\\\\0=8-6\times t\\\\\therefore t=\frac{8}{6}=4/3seconds$

Here acceleration is taken as negative since it opposes the motion of the box since frictional force always opposes motion.

5 0

4 years ago