Answer:0.69
Explanation:
Coefficient of kinetic friction=f/R=61.8/90=0.69
E. all of the above
An umbrella tends to move upward on a windy day because _<span>A. buoyancy increases with increasing wind speed </span>
<span>B. air gets trapped under the umbrella and pushes it up </span>
<span>C. the wind pushes it up </span>
<span>D. a low-pressure area is created on top of the umbrella </span>
Answer:
Chief Hopper
Explanation:
Mike travels at a constant speed of 3.1 m/s. To find how long it takes him to reach the school, we need to find the distance he travels. We can do this using Pythagorean theorem.
a² + b² = c²
(1000 m)² + (900 m)² = c²
c ≈ 1345 m
So the time is:
v = d / t
3.1 m/s = 1345 m / t
t ≈ 434 s
Next, Chief Hopper travels a total distance of 1900 m, starting at rest and accelerating at 0.028 m/s². So we can use constant acceleration equation to find the time.
d = v₀ t + ½ at²
1900 m = (0 m/s) t + ½ (0.028 m/s²) t²
t ≈ 368 s
So Chief Hopper reaches the school first, approximately 66 seconds before Mike does.
Answer : The total electric charge of electrons is, 
Explanation:
Answer : The number of electrons transferred are, 
Explanation :
First we have to calculate the number of electrons.
Number of electrons = 
Mass of 1 electron = 
Total mass of electron = 2.5 kg
Number of electrons = 
Number of electrons = 
Now we have to calculate the total electric charge of electrons.
Formula used :

where,
n = number of electrons transferred = 
Q = charge on electrons = ?
e = charge on 1 electron = 
Now put all the given values in the above formula, we get:


Thus, the total electric charge of electrons is, 
Let say the height of two balls from the ground is H
now we can use kinematics

now we have


now in the same time ball on the left will cover the horizontal distance between them
![v_x = \frac{d}{ t}[/tex[tex]v_x = \frac{3}{\sqrt{\frac{2H}{g}}}](https://tex.z-dn.net/?f=v_x%20%3D%20%5Cfrac%7Bd%7D%7B%20t%7D%5B%2Ftex%3C%2Fp%3E%3Cp%3E%5Btex%5Dv_x%20%3D%20%5Cfrac%7B3%7D%7B%5Csqrt%7B%5Cfrac%7B2H%7D%7Bg%7D%7D%7D)
<em>so above is the horizontal speed of the left ball</em>