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zvonat [6]
3 years ago
15

If you run at 8m/s for 15 minutes how far will you go

Physics
2 answers:
enyata [817]3 years ago
8 0

Answer:

<h2>hope it helps you see the attachment for further information ......✌✌✌✌✌</h2>

sashaice [31]3 years ago
6 0

Answer: 7200 m

Explanation: The solution is, first convert 15 minutes to seconds.

15 mins x 60 s / 1 min = 900 s

Use the formula for speed which is v= d/t then derive for d.

d = vt

= 8 m/s ( 900s)

= 7200 m

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3 years ago
A bicyclist of mass 90 kg drives around a circle with a centripetal acceleration
Serggg [28]

given,

mass of bicyclist(m)=90Kg

centripetal acceleration(a)=1.5 m/s2

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5 0
3 years ago
Read 2 more answers
A bug is 12 cm from the center of a turntable that is rotating with a frequency of 45 rev/min . What minimum coefficient frictio
Agata [3.3K]

Answer:

The minimum coefficient of friction is 0.27.

Explanation:

To solve this problem, start with identifying the forces at play here. First, the bug staying on the rotating turntable will be subject to the centripetal force constantly acting toward the center of the turntable (in absence of which the bug would leave the turntable in a straight line). Second, there is the force of friction due to which the bug can stick to the table. The friction force acts as an intermediary to enable the centripetal acceleration to happen.

Centripetal force is written as

F_c = m\frac{v^2}{r}

with v the linear velocity and r the radius of the turntable. We are not given v, but we can write it as

v = r\omega

with ω denoting the angular velocity, which we are given. With that, the above becomes:

F_c = m\frac{v^2}{r}=m\omega^2 r

Now, the friction force must be at least as much (in magnitude) as Fc. The coefficient (static) of friction μ must be large enough. How large?

F_r=\mu mg \geq m\omega^2 r = F_c\implies\\\mu \geq \frac{\omega^2 r}{g}

Let's plug in the numbers. The angular velocity should be in radians per second. We are given rev/min, which can be easily transformed by a factor 2pi/60:

\frac{1 rev}{1 min}\cdot\frac{\frac{2\pi rad}{rev}}{\frac{60s}{1 min}}=\frac{2\pi}{60}\frac{rad}{s}

and so 45 rev/min = 4.71 rad/s.

\mu \geq \frac{\omega^2 r}{g}=\frac{4.71^2\frac{1}{s^2}\cdot 0.12m}{9.8\frac{m}{s^2}}=0.27

A static coefficient of friction of at least be 0.27 must be present for the bug to continue enjoying the ride on the turntable.



3 0
2 years ago
Problema en la cual aplicaste velocidades, impulso, conservación del movimiento y de la energía.
Kazeer [188]
Huh huh what? ¿Can’t you translate?
6 0
2 years ago
An object is dropped from a bridge. A second object is thrown downward 1.48 s later. They both reach the water 48.1 m below at t
pashok25 [27]

To solve this problem we will apply the linear motion kinematic equations. With the data provided we will calculate the time of the first object to fall. Later we will get the time difference between the two. This difference will allow us to find the free fall distance. Through the distance we will find the initial velocity, that is,

x = v_0 t +\frac{1}{2}at^2

48.1 = 0*t + \frac{1}{2} (9.8)t^2

t = 3.13s

The second object is thrown downward at one second later and it meets the first object at the water is

t' = 3.13 -1.48

t' = 1.65s

The distance of the object will travel due to free fall acceleration is

x = v_0 t+\frac{1}{2} at^2

x = 0*(1.65) +\frac{1}{2}(9.8)(1.65)^2

x = 13.34m

The distance of the object will travel due to its initial velocity is

v_0 = \frac{d_0}{t}

d_0 = v_0 t

48.1-13.34 = v_0 (1.65)

v_0 = 21.06m/s

Therefore the initial speed of the second object is 21.06m/s

8 0
3 years ago
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