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Radda [10]
3 years ago
8

A girl is sitting in a sled sliding horizontally along some snow (there is friction present).

Physics
1 answer:
Over [174]3 years ago
4 0

Answer:

159 N

Explanation:

The force of friction, Fr is a product of coefficient of feiction and the normal force. Therefore, Fr=uN where N is the normal force and u is coefficient of friction. Here, we have two coefficients of friction but since it is sliding, then we use coefficient of kinetic energy. Substituting 0.25 for u and 636 N for N then

Fr=0.25*636=159 N

Therefore, the force of friction is equivalent to 159 N

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An electron and a proton are held on an x axis, with the electron at x = + 1.000 m
mixas84 [53]

Answer:

  r2 = 1 m

therefore the electron that comes with velocity does not reach the origin, it stops when it reaches the position of the electron at x = 1m

Explanation:

For this exercise we must use conservation of energy

the electric potential energy is

          U = k \frac{q_1q_2}{r_{12}}

for the proton at x = -1 m

          U₁ =- k \frac{e^2 }{r+1}

for the electron at x = 1 m

          U₂ = k \frac{e^2 }{r-1}

starting point.

        Em₀ = K + U₁ + U₂

        Em₀ = \frac{1}{2} m v^2 - k \frac{e^2}{r+1} + k \frac{e^2}{r-1}

final point

         Em_f = k e^2 ( -\frac{1}{r_2 +1} + \frac{1}{r_2 -1})

   

energy is conserved

        Em₀ = Em_f

        \frac{1}{2} m v^2 - k \frac{e^2}{r+1} + k \frac{e^2}{r-1} = k e^2 (- \frac{1}{r_2 +1} + \frac{1}{r_2 -1})              

       

        \frac{1}{2} m v^2 - k \frac{e^2}{r+1} + k \frac{e^2}{r-1} = k e²(  \frac{2}{(r_2+1)(r_2-1)} )

we substitute the values

½ 9.1 10⁻³¹ 450 + 9 10⁹ (1.6 10⁻¹⁹)² [ - \frac{1}{20+1} + \frac{1}{20-1} ) = 9 109 (1.6 10-19) ²( \frac{2}{r_2^2 -1} )

          2.0475 10⁻²⁸ + 2.304 10⁻³⁷ (5.0125 10⁻³) = 4.608 10⁻³⁷ ( \frac{1}{r_2^2 -1} )

          2.0475 10⁻²⁸ + 1.1549 10⁻³⁹ = 4.608 10⁻³⁷     \frac{1}{r_2^2 -1}

          \frac{2.0475 \ 10^{-28} }{1.1549 \ 10^{-37} } = \frac{1}{r_2^2 -1}

          r₂² -1 = (4.443 10⁸)⁻¹

           

          r2 = \sqrt{1 + 2.25 10^{-9}}

          r2 = 1 m

therefore the electron that comes with velocity does not reach the origin, it stops when it reaches the position of the electron at x = 1m

4 0
3 years ago
How are the fiducial points of the Celsius and Fahrenheit scales similar?
taurus [48]
The fiducial points of the Celsius<span> and the </span>Fahrenheit<span> temperature </span>scales<span> are the boiling and freezing </span>points<span> of pure water at 1 atm of pressure.

In short, Your Answer would be Option D

Hope this helps!</span>
7 0
3 years ago
Why does a stop sign appear red?
dolphi86 [110]

Answer:

because it’s suppose to be red like a stop light.

Explanation:

So it tells you to stop

8 0
3 years ago
Consider a household that uses 14,000 kWh of electricity and 900 gal of fuel oil, per year, during a heating season. The average
MissTica

Answer:

reduction in the amount of CO₂ emissions by that household per year is 9517.2 lbm per year

Explanation:

given data

electricity consume = 14000 kWh

fuel consume = 900 gal

CO₂ produced of fuel = 26.4 lbm/gal

CO₂ produced of electricity  = 1.54 lbm/kWh

oil and electricity usage = 21 percent

to find out

the reduction in the amount of CO₂ emissions

solution

we calculate the amount of CO₂ produce here that is

amount of CO₂ produce = ( electricity consume×CO₂ produce electricity + fuel consume × CO₂ consume fuel )    ........................1

put here value

amount of CO₂ produce = ( 14000 × 1.54 + 900 × 26.4 )  

amount of CO₂ produce = 45320 lbm/yr

we know reduction is 21%

so

reduction in amount of CO₂ produced is

reduction in CO₂ produced = 45320 × 21%

reduction in CO₂ produced = 9517.2 lbm per year

so reduction in the amount of CO₂ emissions by that household per year is 9517.2 lbm per year

3 0
3 years ago
(TCO 5) The relationship between Celsius (º C) and Fahrenheit (º F) degree of measuring temperature is linear. Find the linear e
gtnhenbr [62]

Answer:

C=-\dfrac{10}{17}(F-32)

Explanation:

Given that

32° F corresponds to 0 °C.  ---Point 1

212° F corresponds to 100 °C.----Point 2

We know that if two point is given that equation of line can be found as

y-y_1=\dfrac{y_2-y_1}{x_2-x_2}(x-x_1)

Lets C in y- direction and F in x- direction,so we can say that

C-C_1=\dfrac{C_2-C_1}{F_2-F_2}(F-F_1)

C-0=\dfrac{100-0}{32-212}(F-32)

C=-\dfrac{10}{17}(F-32)

So the linear relationship is

C=-\dfrac{10}{17}(F-32)

8 0
3 years ago
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