Answer:
V = - 0.5 [m/s]
Explanation:
In order to solve this problem, we must use the principle of relative speeds. This is for an observer who is on the edge of the river he can see how the river moves to the left and the woman tries to move to the right but can not since:
![V_{total}=-3+2.5\\V_{total}=-0.5 [m/s]](https://tex.z-dn.net/?f=V_%7Btotal%7D%3D-3%2B2.5%5C%5CV_%7Btotal%7D%3D-0.5%20%5Bm%2Fs%5D)
That is, the person sees how the woman moves to the left but with avelocity of 0.5 [m/s] to the left
On driving your motorcycle in a circle of radius 75 m on wet pavement, the fastest you can go before you lose traction, assuming the coefficient of static friction is 0.20 is 147m/s
Friction helps to maintain the slipping of the vehicle on the road hence lays a very important role.
Maximum velocity of a road with friction is given by the formula,
v = μRg
where, v is the maximum velocity
μ is the coefficient of static friction
R is the radius of the circle road
g is the acceleration due to gravity
Given,
μ = 0.20
R = 75m
g = 9.8m/s²
On substituting the given values in the above formula,
v = 0.20× 75 ×9.8
v = 147m/s
So, the Maximum velocity of the wet road is 147m/s.
Learn more about Velocity here, brainly.com/question/18084516
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What is that?? Please tell us
Answer:
The value of third charge is 0.8μC.
Explanation:
Given that.
Magnitude of net force=4.444 N
According to figure,
Suppose, First charge = 2.4 μC
Second charge = 6.2 μC
Distance r₁ = 9.8 cm
Distance r₂ = 2.1 cm
We need to calculate the value of r
Using Pythagorean theorem

Put the value into the formula


We need to calculate the force
Using formula of force

Force F₁₂,



Force F₂₃,

We need to calculate the value of third charge





Hence, The value of third charge is 0.8μC.