The minimum value of the coefficient of static friction between the block and the slope is 0.53.
<h3>Minimum coefficient of static friction</h3>
Apply Newton's second law of motion;
F - μFs = 0
μFs = F
where;
- μ is coefficient of static friction
- Fs is frictional force
- F is applied force
μ = F/Fs
μ = F/(mgcosθ)
μ = (250)/(50 x 9.8 x cos15)
μ = 0.53
Thus, the minimum value of the coefficient of static friction between the block and the slope is 0.53.
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Answer:
B
Explanation:
hair can be a safety hazard
Answer:
The space cadet that weighs 800 N on Earth will weigh 1,600 N on the exoplanet
Explanation:
The given parameters are;
The mass of the exoplanet = 1/2×The mass of the Earth, M = 1/2 × M
The radius of the exoplanet = 50% of the radius of the Earth = 1/2 × The Earth's radius, R = 50/100 × R = 1/2 × R
The weight of the cadet on Earth = 800 N

Therefore, for the weight of the cadet on the exoplanet, W₁, we have;

The weight of a space cadet on the exoplanet, that weighs 800 N on Earth = 1,600 N.
Answer:
The earthquake occurred at a distance of 1122 km
Explanation:
Given;
speed of the P wave, v₁ = 8.5 km/s
speed of the S wave, v₂ = 5.5 km/s
The distance traveled by both waves is the same and it is given as;
Δx = v₁t₁ = v₂t₂
let the time taken by the wave with greater speed = t₁
then, the time taken by the wave with smaller speed, t₂ = t₁ + 1.2 min, since it is slower.
v₁t₁ = v₂t₂
v₁t₁ = v₂(t₁ + 1.2 min)
v₁t₁ = v₂(t₁ + 72 s)
v₁t₁ = v₂t₁ + 72v₂
v₁t₁ - v₂t₁ = 72v₂
t₁(v₁ - v₂) = 72v₂

The distance traveled is given by;
Δx = v₁t₁
Δx = (8.5)(132)
Δx = 1122 km
Therefore, the earthquake occurred at a distance of 1122 km