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Strike441 [17]
3 years ago
7

This problem uses the same concepts as Multiple-Concept Example 17, except that kinetic, rather than static, friction is involve

d. A crate is sliding down a ramp that is inclined at an angle 30.6 ° above the horizontal. The coefficient of kinetic friction between the crate and the ramp is 0.360. Find the acceleration of the moving crate.

Physics
1 answer:
coldgirl [10]3 years ago
8 0

Answer:1.94 m/s^2

Explanation:

Given

inclination \theta =30.6^{\circ}

Coefficient of kinetic Friction \mu =0.36

Forces on crate is

mg\sin \theta -f_r=ma

f_r=\mu mg\cos \theta

mg\sin \theta -\mu mg \cos \theta =ma

a=g\sin \theta -\mu g\cos \theta

a=1.94 m/s^2  

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Answer:

The astronaut can throw the hammer in a direction away from the space station. While he is holding the hammer, the total momentum of the astronaut and hammer is 0 kg • m/s. According to the law of conservation of momentum, the total momentum after he throws the hammer must still be 0 kg • m/s. In order for momentum to be conserved, the astronaut will have to move in the opposite direction of the hammer, which will be toward the space station.

Explanation:

6 0
3 years ago
Distinguish between directly proportional and inversely proportional with an example of each
RideAnS [48]

Consider two variables said to be "inversely proportional" to each other. If all other variables are held constant, the magnitude or absolute value of one inversely proportional variable decreases if the other variable increases, while their product (the constant of proportionality k) is always the same.

4 0
3 years ago
A copper wire and a tungsten wire of the same length have the same resistance. What is the ratio of the diameter of the copper w
spayn [35]

Answer:

Therefore the ratio of diameter of the copper to that of the tungsten is

\sqrt{3} :\sqrt{10}

Explanation:

Resistance: Resistance is defined to the ratio of voltage to the electricity.

The resistance of a wire is

  1. directly proportional to its length i.eR\propto l
  2. inversely proportional to its cross section area i.eR\propto \frac{1}{A}

Therefore

R=\rho\frac{l}{A}

ρ is the resistivity.

The unit of resistance is ohm (Ω).

The resistivity of copper(ρ₁) is 1.68×10⁻⁸ ohm-m

The resistivity of tungsten(ρ₂) is 5.6×10⁻⁸ ohm-m

For copper:

A=\pi r_1^2 =\pi (\frac{d_1}{2} )^2

R_1=\rho_1\frac{l_1}{\pi(\frac{d_1}{2})^2 }

\Rightarrow (\frac{d_1}{2})^2=\rho_1\frac{l_1}{\pi R_1 }......(1)

Again for tungsten:

R_2=\rho_2\frac{l_2}{\pi(\frac{d_2}{2})^2 }

\Rightarrow (\frac{d_2}{2})^2=\rho_2\frac{l_2}{\pi R_2 }........(2)

Given that R_1=R_2   and    l_1=l_2

Dividing the equation (1) and (2)

\Rightarrow\frac{ (\frac{d_1}{2})^2}{ (\frac{d_2}{2})^2}=\frac{\rho_1\frac{l_1}{\pi R_1 }}{\rho_2\frac{l_2}{\pi R_2 }}

\Rightarrow( \frac{d_1}{d_2} )^2=\frac{1.68\times 10^{-8}}{5.6\times 10^{-8}}   [since R_1=R_2   and    l_1=l_2]

\Rightarrow( \frac{d_1}{d_2} )=\sqrt{\frac{1.68\times 10^{-8}}{5.6\times 10^{-8}}}

\Rightarrow( \frac{d_1}{d_2} )=\sqrt{\frac{3}{10}}

\Rightarrow d_1:d_2=\sqrt{3} :\sqrt{10}

Therefore the ratio of diameter of the copper to that of the tungsten is

\sqrt{3} :\sqrt{10}

8 0
3 years ago
Angela has been exercising regularly for two years. Right now at her house everyone is feeling sick, but she is able to help the
Dmitry [639]

Answer:

it make your immune system build up from the strain your puting on you body

Explanation:

excercise is good for you  and builds muscle and blood cells

3 0
2 years ago
What is the weight of a column of air with cross-sectional area 4. 5 m^2 extending from earth's surface to the top of the atmosp
sineoko [7]

The weight of a column of air with cross-sectional area 4. 5 m^2 extending from earth's surface to the top of the atmosphere is, 4.56*10^5N.

To find the answer, we have to know about the pressure.

<h3>How to find the weight of a column of air?</h3>
  • As we know that the expression of pressure as,

                 P=\frac{F}{A}

where; F is the force, here it is equal to the weight of the air column, and A is the area of cross section.

  • It is given that, the air column is extending from earth's surface to the top of the atmosphere, thus, the pressure will be atmospheric pressure,

             P=1atm=1.013*10^5Pascals

  • From this, the value of weight will be,

            F=mg=P*A=1.013*10^5*4.5=4.56*10^5N

Thus, we can conclude that, the weight of a column of air with cross-sectional area 4. 5 m^2 extending from earth's surface to the top of the atmosphere is, 4.56*10^5N.

Learn more about the pressure here:

brainly.com/question/12830237

#SPJ4

4 0
1 year ago
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