The coefficient of static friction is 0.222
Explanation:
In order for the car to remain in circular motion, the frictional force must be able to provide the necessary centripetal force. Therefore, the car will start skidding when the two forces are equal:

where the term on the left is the frictional force, while the term on the right is the centripetal force, and where
is the coefficient of static friction
m is the mass of the car
g is the acceleration of gravity
v is the speed of the car
r is the radius of the track
In this problem, we have:
r = 564 m
v = 35 m/s

And re-arranging the equation for
, we can find the coefficient of static friction:

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Answer:
W = Fd = KE =1/2mv²
Explanation:
not sure if that's what your looking for but i'm pretty sure this is it.
Answer:
-120000 W
Explanation:
Power = change in energy / time
P = ΔE / t
P = (½ mv₂² − ½ mv₁²) / t
P = m (v₂² − v₁²) / (2t)
Given m = 1.5 t = 1500 kg, v₂ = 10 m/s, v₁ = 30 m/s, and t = 5 s:
P = (1500 kg) ((10 m/s)² − (30 m/s)²) / (2 × 5 s)
P = -120000 W
Answer:
Mass, m = 6.18 kg
Explanation:
Given the following data;
Frequency, F = 10 Hz
Spring constant, k = 250 N/m
We know that pie, π = 22/7
To find the mass, we would use the following formula;
F = 1/2π√(k/m)
Where;
F is the frequency of oscillation.
k is the spring constant.
m is the mass of the spring.
Substituting into the formula, we have;
10 = 1/2 * 22/7 * √250/m
10 = 22/14 * √250/m
Cross-multiplying, we have;
140 = 22 * √250/m
Dividing both sides by 22, we have;
140/22 = √250/m
6.36 = √250/m
Taking the square of both sides, we have;
6.36² = (√250/m)²
40.45 = 250/m
Cross-multiplying, we have;
40.45m = 250
Mass, m = 250/40.45
Mass, m = 6.18 kg
Answer:
The required diagram is shown in the figure. When an object is placed in front of the convex lens, i.e., between 2F
1
and F
1
, its image is formed beyond 2F
2
on the other side of the lens. The image is real, inverted and enlarged.
solution