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slamgirl [31]
3 years ago
6

A 2,493-kg car is moving down a road with a slope (grade of 14% at a constant speed of 13 m/s. what is the direction and magnitu

de of the frictional force? (define positive x in the forward direction, i.e., down the slope?
Physics
1 answer:
sweet [91]3 years ago
3 0
To calculate for the frictional force, we calculate first for the normal force (forcer perpendicular to the surface)
                               Fn = (2,493 kg) (9.8 m/s²) = 24243.51 N
Then, we multiply this value with the frictional constant which is not given in this problem. If the frictional coefficient is equal to x then, the frictional force is equal to 24243.51x. 
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3 years ago
When planning an oral presentation, you want to make sure to do all of the following except _____.
Sindrei [870]

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3 years ago
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Carbon is allowed to diffuse through a steel plate 9.7-mm thick. The concentrations of carbon at the two faces are 0.664 and 0.3
cupoosta [38]

Answer:

844°C

Explanation:

The problem can be easily solve by using Fick's law and the Diffusivity or diffusion coefficient.

We know that Fick's law is given by,

J = - D \frac{\Delta c}{\Delta x}

Where \frac{\Delta c}{\Delta x} is the concentration of gradient

D is the diffusivity coefficient

and J is the flux of atoms.

In the other hand we have, that

D= D_0 e^{\frac{E_d}{RT}}

Where D_0 is the proportionality constant,

R is the gas constant, T the temperature and E_d is the activation energy.

Replacing the value of diffusivity coefficient in Fick's law we have,

J = -D_0 ^{\frac{E_d}{RT}}\frac{\Delta c}{\Delta x}

Rearrange the equation to get the value of temperature,

T=\frac{Ed}{Rln(\frac{J\Delta x}{D_0 \Delta c})}

We have all the values in our equation.

\Delta c = 0.664-0.339 = 0.325 C. cm^{-1}

\Delta x = 9.7*10^{-3}m

E_d = 82000J

D_0 = 6.5*10^{-7}m^2/s

J = 3.2*10^{-9}m^2/s

R= 8.31Jmol^{-1}K

Substituting,

T=\frac{Ed}{Rln(\frac{J\Delta x}{D_0 \Delta c})}

T=-\frac{-82000}{(8.31)ln(\frac{3.2*10^{-9}(9.7*10^{-3})}{6.5*10^{-7} (0.325)})}

T=1118.07K=844\°C

4 0
3 years ago
A 15 g turntable is covered with a uniform layer of dry ice that has a mass of 9.0 g. The angular speed of the turntable and dry
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Answer:

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Let the new angular speed is ω2 and the radius of the table be r.

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