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svp [43]
3 years ago
15

An advertisement for an all-terrain vehicle (ATV) claims that the ATV can climb inclined slopes of 35°. What is the minimum coef

ficient of static friction needed for this claim to be possible?
Physics
1 answer:
navik [9.2K]3 years ago
8 0

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

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A spherical balloon has a radius of 6.55 m and is filled with helium. The density of helium is 0.179 kg/m3, and the density of a
Ksju [112]

Answer:

M_1 = 317.7 kg

Explanation:

Mass of the helium gas filled inside the volume of balloon is given as

m = \rho V

m = 0.179(\frac{4}{3}\pi R^3)

m = 0.179(\frac{4}{3}\pi 6.55^3)

m = 210.7 kg

now total mass of balloon + helium inside balloon is given as

M = 210.7 + 990

M = 1200.7 kg

now we know that total weight of balloon + cargo = buoyancy force on the balloon

so we will have

(M + M_1)g = \rho_{air} V g

(1200.7 + M_1) = (\frac{4}{3}\pi 6.55^3) (1.29)

1200.7 + M_1 = 1518.4

M_1 = 317.7 kg

3 0
3 years ago
The electric flux through a spherical surface is 1.4 ✕ 105 N · m2/C. What is the net charge (in C) enclosed by the surface?
Anit [1.1K]

Answer:

The  value is   Q_{net} =  1.239 *10^{-6} \  C

Explanation:

From the question we are told that

   The electric flux is \Phi =  1.4*10^{5} \  N\cdot m^2/C

     

Generally the net charge is mathematically represented as

    Q_{net} =  \Phi *  \epsilon_o

Here \epsilon_o is the permetivity of free space with value  

       \epsilon_o =  8.85*10^{-12}  \  \  m^{-3} \cdot kg^{-1}\cdot  s^4 \cdot A^2

So

   Q_{net} =  1.4*10^5 *  8.85*10^{-12}

=>   Q_{net} =  1.239 *10^{-6} \  C

8 0
3 years ago
An Ethernet cable is 4.00m long. The cable has a mass of 0.200kg . A transverse pulse is produced by plucking one end of the tau
Mashcka [7]

Based on the length of the Ethernet cable and the mass, the tension in the cable can be found to be 80 N.

<h3>How much tension is in the cable?</h3>

The tension in the cable can be found as:

= 4 x mass x length x frequency

Solving for the frequency is:

= 1 / (0.800 / 4)

= 1 / 0.20

= 5.0 Hz

The tension is therefore:

= 4 x 0.20 x 4.00 x 5

= 80N

Find out more on tension at brainly.com/question/14336853

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4 0
2 years ago
As an object falls to the ground, its potential energy is being converted to kinetic energy.
alekssr [168]

Answer:

<em>The statement is true</em>

Explanation:

<u>Energy Conversion </u>

When an object starts to fall in free air, it speeds up as it falls. The force of gravity acting on the object causes energy to be transferred from its gravitational potential energy to its kinetic energy. We can safely say the height converts to speed and vice-versa. If no external forces act on the system, we can easily calculate heights and speeds by knowing the total mechanical energy (gravitational potential plus kinetic) is conserved.

Answer:

\boxed{\text{The statement is true}}

5 0
3 years ago
A first order reaction, A -&gt; products, has a rate reaction of .00250 Ms-1 when [A] = . 484 M. (a) What is the rate constant,
tamaranim1 [39]

Answer: a)  The rate constant, k, for this reaction is 0.00516s^{-1}

b) No t_{\frac{1}{2}} does not depend on concentration.

Explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

A\rightarrow products

Given: Order with respect to A = 1

Thus rate law is:

a) Rate=k[A]^1

k= rate constant

0.00250=k[0.484]^1

k=0.00516s^{-1}

The rate constant, k, for this reaction is 0.00516s^{-1}

b) Expression for rate law for first order kinetics is given by:

t=\frac{2.303}{k}\log\frac{a}{a-x}

where,

k = rate constant  

t = age of sample

a = let initial amount of the reactant  

a - x = amount left after decay process  

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

t_{\frac{1}{2}}=\frac{2.303}{k}\log\frac{100}{50}

t_{\frac{1}{2}}=\frac{0.69}{k}

Thus t_{\frac{1}{2}} does not depend on concentration.

8 0
3 years ago
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