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lara [203]
3 years ago
13

A cup of coffee is sitting on a table in a train that is moving with a constant velocity. The coefficient of static friction bet

ween the cup and the table is 0.30. Suddenly, the train accelerates. What is the maximum acceleration that the train can have without the cup sliding backward on the table?
Physics
1 answer:
Vikki [24]3 years ago
3 0

Answer:

a = 2.94 m/s²

Explanation:

In order for the cup not to slip, the unbalanced force on cup must be equal to the frictional force:

Unbalanced Force = Frictional Force

ma = μR = μW

ma = μmg

a = μg

where,

a = maximum acceleration for the cup not to slip = ?

μ = coefficient of static friction = 0.3

g = acceleration due to gravity = 9.8 m/s²

Therefore,

a = (0.3)(9.8 m/s²)

<u>a = 2.94 m/s²</u>

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7 0
3 years ago
An advertisement for an all-terrain vehicle (ATV) claims that the ATV can climb inclined slopes of 35°. What is the minimum coef
navik [9.2K]

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;

\sum F_x = \sum F_y = 0

\mathbf{mg sin \theta -f_s= N-mgcos \theta = 0 }

  • So; On the L.H.S

\mathbf{mg sin \theta =f_s}

\mathbf{mg sin \theta =\mu_s N}

  • On the R.H.S

N = mg cos θ

Equating both force component together, we have:

\mathbf{mg sin \theta =\mu_s \ mg \ cos \theta}

\mathbf{sin \theta =\mu_s \ \ cos \theta}

\mathbf{\mu_s = \dfrac{sin \theta }{ cos \theta}}

From trigonometry rule:

\mathbf{tan \theta= \dfrac{sin \theta }{ cos \theta}}

∴

\mathbf{\mu_s =\tan \theta}}

\mathbf{\mu_s =\tan 35^0}}

\mathbf{\mu_s = 0.700}}

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:

brainly.com/question/24882156?referrer=searchResults

8 0
3 years ago
A block of mass m is attached to a rope wound around the outer rim of a disk of radius R and moment of inertia I, which is free
Hoochie [10]

Answer:

Explanation:

I is the moment of inertia of the pulley, α is the angular acceleration of the pulley and T is the tension in the rope. Let a is the linear acceleration.

The relation between the linear acceleration and the angular acceleration is

a = R α   .... (1)

According to the diagram,

T x R = I x α

T x R = I x a / R      from equation (1)

T = I x a / R²      .... (2)

mg - T = ma    .... (3)

Substitute the value of T from equation (2) in equation (3)

mg - \frac{Ia}{R^{2}}=ma

a=\frac{mg}{m+\frac{I}{R^{2}}}

T is the acceleration in the system

Substitute the value of a in equation (2)

T = \frac{I}{R^{2}}\times \frac{mg}{m+\frac{I}{R^{2}}}

T=\frac{I\times mg}{I+mR^{2}}

This is the tension in the string.

4 0
3 years ago
he triceps muscle in the back of the upper arm extends the forearm. This muscle in a professional boxer exerts a force of 2.00 1
meriva

Answer:

Moment of inertia = 0.3862kg-m²

Explanation:

2.00x10³

2.80cm

145 rad

r = r⊥ x F

F is an applied force

r⊥ is the distance between the applied force and axis

Force exerted = 2.00x10³

r⊥ = 2.8cm = 0.028m

Alpha = 145rad/s²

r = 0.028m x 2.00x10³

r = 56.0N-m

To get the moment of inertia

56.0N-m² = (145rad/s²) x I

The I would be:

I = (56.0N-m²)/(145rad/s²)

I = 56/145

= 0.3862Kg-m²

This is the moment of inertia.

Thank you!

5 0
3 years ago
Prove for N/C=Volt/m
Zigmanuir [339]

Answer:

simple, Volt =change in potential energy/Charge

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the unit of charge is coloumb

So, Volt/meter=newton* meter/coloumb*meter

=newton/coloumb (hence proved)

This unit is the potential drop per unit of length in a conductive wire with uniform resistance

7 0
3 years ago
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