An advertisement for an all-terrain vehicle (ATV) claims that the ATV can climb inclined slopes of 35°. The minimum coefficient of static friction needed for this claim to be possible is 0.7
In an inclined plane, the coefficient of static friction is the angle at which an object slide over another.
As the angle rises, the gravitational force component surpasses the static friction force, as such, the object begins to slide.
Using the Newton second law;
N = mg cos θ
Equating both force component together, we have:
From trigonometry rule:
∴
Therefore, we can conclude that the minimum coefficient of static friction needed for this claim to be possible is 0.7
Learn more about static friction here:
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Answer:
a) frequency = 0.1724 Hz
b) Period = 5.8 sec
c) speed = 7.04 m/s
d) acceleration = 7.62 m/s²
Explanation:
Given that;
radius = 6.5m
time period = 5.8 sec every circle
a) the frequency
frequency is the number of rotation in unit time
frequency = 1 / time period = 1/5.8
frequency = 0.1724 Hz
b) the period
period is time taken in one rotation
period = total time / rotation = 5.8 / 1
Period = 5.8 sec
c) the speed
speed = distance/time = circumference/time period = 2πr / t = (2π×6.5) / 5.8
speed = 7.04 m/s
d) acceleration
To find the acceleration we take the linear velocity squared divided by the radius of the circle.
so
acceleration = v² / r = (7.04)² / 6.5 = 49.5616 / 6.5
acceleration = 7.62 m/s²
Answer:
a) P=0.25x10^-7
b) R=B*N2*E
c) N=1.33x10^9 photons
Explanation:
a) the spontaneous emission rate is equal to:
1/tsp=1/3 ms
the stimulated emission rate is equal to:
pst=(N*C*o(v))/V
where
o(v)=((λ^2*A)/(8*π*u^2))g(v)
g(v)=2/(π*deltav)
o(v)=(λ^2)/(4*π*tp*deltav)
Replacing values:
o(v)=0.7^2/(4*π*3*50)=8.3x10^-19 cm^2
the probability is equal to:
P=(1000*3x10^10*8.3x10^-19)/(100)=0.25x10^-7
b) the rate of decay is equal to:
R=B*N2*E, where B is the Einstein´s coefficient and E is the energy system
c) the number of photons is equal to:
N=(1/tsp)*(V/C*o)
Replacing:
N=100/(3*3x10^10*8.3x10^-19)
N=1.33x10^9 photons
