The magnitude of static friction force is f_s = 842.8 N
Explanation:
Write down the values given in the question
The wheel of a car has radius r = 0.350 m
The car applies the torque is τ = 295 N m
It is said that the wheels does not slip against the road surface,
Here we apply a force of static friction,
It can be calculated as
Frictional force f_s = τ / r
= 295 Nm / 0.350 m
f_s = 842.8 N
- The wavelength of the red light in "nanometer" is 7×
- Wavelength is given as : 7×
meter
- 1 nanometer = (
meter)
- Let X= value of the wavelength in nanometer.
1 nanometer = meter
X nanometer = 7× meter
- <em>If we Cross multiply</em>
X nanometer = (
)
X= 7× nanometer
Therefore, the wavelength in "nanometer" is 7×
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Answer:
Explanation:
The rotation rate of the man is:
The resultant rotation rate of the system is computed from the Principle of Angular Momentum Conservation:
![(90\,kg)\cdot (5\,m)^{2}\cdot (0.16\,\frac{rad}{s} ) = [(90\,kg)\cdot (5\,m)^{2}+20000\,kg\cdot m^{2}]\cdot \omega](https://tex.z-dn.net/?f=%2890%5C%2Ckg%29%5Ccdot%20%285%5C%2Cm%29%5E%7B2%7D%5Ccdot%20%280.16%5C%2C%5Cfrac%7Brad%7D%7Bs%7D%20%29%20%3D%20%5B%2890%5C%2Ckg%29%5Ccdot%20%285%5C%2Cm%29%5E%7B2%7D%2B20000%5C%2Ckg%5Ccdot%20m%5E%7B2%7D%5D%5Ccdot%20%5Comega)
The final angular speed is:
Answer:
F = 2,894 N
Explanation:
For this exercise let's use Newton's second law
F = m a
The acceleration is centripetal
a = v² / r
Angular and linear variables are related.
v = w r
Let's replace
F = m w² r
The radius r and the length of the rope is related
cos is = r / L
r = L cos tea
Let's replace
F = m w² L cos θ
Let's reduce the magnitudes to the SI system
m = 101.7 g (1 kg / 1000g) = 0.1017 kg
θ = 5 rev (2π rad / rev) = 31,416 rad
w = θ / t
w = 31.416 / 5.1
w = 6.16 rad / s
F = 0.1017 6.16² 0.75 cos θ
F = 2,894 cos θ
The maximum value of F is for θ equal to zero
F = 2,894 N
