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3 years ago

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Anna applies a force of 19.5 newtons to push a book placed on a table. If the normal force of the book is 51.7 newtons, what is

the coefficient of kinetic friction?

1 answer:

GarryVolchara [31]3 years ago

6 0

that would be given by

[email protected]

@ representing coefficient of kinetic friction.

thus 19.5/51.7 = 0.377

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The velocity of sound in air at 300C is approximately :

kumpel [21]

The velocity of sound in at 300C is 511.3 m/s.

Explanation:

The equation that gives the speed of sound in ar as a function of the air temperature is the following:

$v=(331.3+0.6T) m/s$

where

T is the temperature of the air, measured in Celsius degrees

In this problem, we want to find the speed of sound in ar for a temperature of

$T=300^{\circ}C$

Substituting into the equation, we find:

$v=331.3 + 0.6(300)=511.3 m/s$

So, the velocity of sound in at 300C is 511.3 m/s.

Learn more about sound waves:

brainly.com/question/4899681

#LearnwithBrainly

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3 years ago

A long, straight wire is surrounded by a hollow metal cylinder whose axis coincides with that of the wire. The wire has a charge

Alexxx [7]

Explanation:

It is given that, a long, straight wire is surrounded by a hollow metal cylinder whose axis coincides with that of the wire.

The charge per unit length of the wire is $\lambda$ and the net charge per unit length is $2 \lambda$ .

We know that there exist zero electric field inside the metal cylinder.

(a) Using Gauss's law to find the charge per unit length on the inner and outer surfaces of the cylinder. Let $\lambda_i\ and\ \lambda_o$ are the charge per unit length on the inner and outer surfaces of the cylinder.

For inner surface,

$\phi=\dfrac{q_{enclosed}}{\epsilon_o}$

$E.A=\dfrac{q_{enclosed}}{\epsilon_o}$

$0=\dfrac{\lambda_i+\lambda}{\epsilon_o}$

$\lambda_i=-\lambda$

For outer surface,

$\lambda_i+\lambda_o=2\lambda$

$-\lambda+\lambda_o=2\lambda$

$\lambda_o=3\lambda$

(b) Let E is the electric field outside the cylinder, a distance r from the axis. It is given by :

$E_o=\dfrac{\lambda_o}{2\pi \epsilon_o r}$

$E_o=\dfrac{3\lambda}{2\pi \epsilon_o r}$

Hence, this is the required solution.

6 0

3 years ago

Helium gas is compressed by an adiabatic compressor from an initial state of 14 psia and 50°F to a final temperature of 320°F in

iVinArrow [24]

Answer:

The value is $P_2 = 40.54 \ psla$

Explanation:

From the question we are told that

The initial pressure is $P_1 = 14\ psla$

The initial temperature is $T_1 = 50 \ F = (50 - 32) * [\frac{5}{9} ] + 273 = 283 \ K$

The final temperature is $T_2 = 320 \ F = (320 - 32) * [\frac{5}{9} ] + 273 =433 \ K$

Generally the equation for adiabatic process is mathematically represented as

$PT^{\frac{\gamma}{1- \gamma} } = Constant$

=> $P_1T_1^{\frac{\gamma}{1- \gamma} } = P_2T_2^{\frac{\gamma}{1- \gamma} }$

Generally for a monoatomic gas $\gamma = \frac{5}{3}$

So

$14 * 283^{\frac{\frac{5}{3} }{1- [\frac{5}{3} ]} } =P_2 * 433^{\frac{\frac{5}{3} }{1- [\frac{5}{3} ]} }$

=> $14 * 283^{-2.5} =P_2 * 433^{-2.5}$

=> $P_2 = 40.54 \ psla$

8 0

2 years ago

gayaneshka [121]

This is confusing yes

3 0

3 years ago

You are watching an archery tournament when you start wondering how fast an arrow is shot from the bow. Remembering your physics

spayn [35]

Answer:

$v_0 = 3.53~{\rm m/s}$

Explanation:

This is a projectile motion problem. We will first separate the motion into x- and y-components, apply the equations of kinematics separately, then we will combine them to find the initial velocity.

The initial velocity is in the x-direction, and there is no acceleration in the x-direction.

On the other hand, there no initial velocity in the y-component, so the arrow is basically in free-fall.

Applying the equations of kinematics in the x-direction gives

$x - x_0 = v_{x_0} t + \frac{1}{2}a_x t^2\\63 \times 10^{-3} = v_0t + 0\\t = \frac{63\times 10^{-3}}{v_0}$

For the y-direction gives

$v_y = v_{y_0} + a_y t\\v_y = 0 -9.8t\\v_y = -9.8t$

Combining both equation yields the y_component of the final velocity

$v_y = -9.8(\frac{63\times 10^{-3}}{v_0}) = -\frac{0.61}{v_0}$

Since we know the angle between the x- and y-components of the final velocity, which is 180° - 2.8° = 177.2°, we can calculate the initial velocity.

$\tan(\theta) = \frac{v_y}{v_x}\\\tan(177.2^\circ) = -0.0489 = \frac{v_y}{v_0} = \frac{-0.61/v_0}{v_0} = -\frac{0.61}{v_0^2}\\v_0 = 3.53~{\rm m/s}$

6 0

3 years ago