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Novay_Z [31]
4 years ago
6

The car has a mass of 1.6 Mg and center of mass at G. If the coefficient of static friction between the shoulder of the road and

the tires is μs = 0.41, determine the greatest slope θ the shoulder can have without causing the car to slip or tip over if the car travels along the shoulder at constant velocity.

Physics
1 answer:
Rufina [12.5K]4 years ago
6 0

Answer:

\theta = 22.29

Explanation:

Taking summation of force at perpendicular  to the plane

\sum F_p = 0

2N_A +2N_B -mgcos\theta = 0

N_A +N_B - mgcos\theta = 0

N_A +N_B = mgcos\theta

Taking summation along the plane, therefore we have

\sum Fa = 0

f_A +f_B -mgsin\theta = 0  

\mu N_A+\mu N_B - mgsin\theta = 0

\mu(N_A +N_B) = mgcos\theta

from equation 1 and 2 we have

\mu =\frac{sin\theta}{cos\theta}

\mu = 0.41

\mu = tan\theta

\theta = tan^{-}\mu

\theta = 22.29

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elixir [45]

Answer:Hollow sphere

Explanation:

Given

same mass for solid and hollow sphere

same v_{cm} before they start up incline

Moment of inertia of solid Sphere

I_1=\frac{2}{5}Mr^2

Moment of inertia of hollow sphere

I_2=\frac{2}{3}Mr^2

Conserving Energy at bottom and top point for solid sphere

kinetic energy +Rotational Energy=Potential energy

\frac{1}{2}Mv_{cm}^2+\frac{1}{2}I\omega ^2=mgh_1

for pure rolling v_{cm}=\omega r

\frac{1}{2}Mv_{cm}^2+\frac{1}{2}\times \frac{2}{5}Mr^2=Mgh_1

\frac{7}{10}Mv_{cm}^2=Mgh_1

h_1=\frac{7v_{cm}^2}{10g}

For hollow sphere

\frac{1}{2}Mv_{cm}^2+\frac{1}{2}\times \frac{2}{3}Mr^2=Mgh_2

h_2=\frac{5v_{cm}^2}{6g}

therefore height gained by hollow sphere is more

6 0
4 years ago
Problem 4: A mass m = 1.5 kg hangs at the end of a vertical spring whose top end is fixed to the ceiling. The spring has spring
lana [24]

Answer:

y (t) = 0.754 * cos ( 7.96 t - 69.52)

Explanation:

Given:

m = 1.5 kg , k = 95 N / m , v₀ = 6 m / s , d = 0.35 m , t = 0

y (t) = A * cos ( ω * t - φ )

Using the equation that describe the motion

m * v = - k * x  ⇒ m * x'' = - k * x

Angular velocity is equal to

ω = √ k / m   ⇒ ω = √ 95 N /m  /  1.5 kg  

ω = 7.96  rad /s

A = v / ω   ⇒  A = 6 m /s   /   7.96 rad / s

A = 0.754

d = cos * φ    ⇒  φ = cos ⁻¹ * 0.35

φ = 69.52

y (t) = A * cos ( ω * t - φ )    ⇒  y (t) = 0.754 * cos ( 7.96 t - 69.52)

4 0
3 years ago
Use the diagram below modeling a football kicked from a horizontal surface B
djyliett [7]
B is the correct one
5 0
3 years ago
Read 2 more answers
1. What quantity of heat is required to raise?
netineya [11]

Answer: Calculate the energy required in joules to raise the temperature of 450 grams of water from 15°C to 85°C? (The specific heat capacity of water is 4.18 J/g/°C)

Explanation:

8 0
3 years ago
Drew observed in an experiment that algae in a nearby lake could be found at six meters below the water's surface on a clear day
babymother [125]

In this question the options are missing; here are the options:

What was the dependent variable in Drew's experiment?

A. The depth at which the algae were found

B. The sky conditions on a particular day

C. The amount of algae measured

D. The lake that was being observed

The answer to this question A. The depth at which the algae were found

Explanation:

In an experiment, it is common there are at least two factors or variables. Additionally, the variable that is modified by others or that depend on others is always the dependent variable.

In the case of the experiment presented, there are two main factors: sky conditions and depth at which algae can be found. From these, the dependent factor is the depth because this depth changes with the sky condition or depends on the sky conditions. Also, the dependent variable is always the factor being studied, for example, in this case, Drew's focus is to study how the location of algae in terms of depth changes.

8 0
3 years ago
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