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
#SPJ4
Answer:
# of Snickers bars 2
Explanation:
Power output= 0.30 HP
=0.3*746
= 0.30 HP (746 W=1.00 HP)
= 224 W
time required 2 h 49 m = 10140 seconds
Since power is work divided by time, then work is:
Work done by the jet = P*t
= 224 *(10140)
= 2.3 MJ (2.3 x
J)
Converting MJ to Cal
2.3 MJ=549 Cal
# of Snickers bars = 549 Cal / 280 Cal
= 2.0 bars (rounded from 1.96)
B.absorb neutrons to prevent chain reactions which become uncontrollable
Answer:
Acceleration of the crate is 0.362 m/s^2.
Explanation:
Given:
Mass of the box, m = 40 kg
Applied force, F = 15 N
Angle at which the force is applied,
= 15°
We have to find the magnitude of the acceleration.
Let the acceleration be "a".
FBD is attached with where we can see the horizontal and vertical component of force.
⇒
and ⇒ 
⇒
⇒ 
⇒ Applying concept of forces.
⇒
⇒ 
⇒
<em> ...Newtons second law Fnet = ma</em>
⇒
⇒ Plugging the values.
⇒
<em>...f is the friction which is zero here.</em>
⇒ 
⇒ 
Magnitude of the acceleration of the crate is 0.362 m/s^2.