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3 years ago

15

A 2.0 kg wooden block is slid along a concrete surface (μk = 0.21) with an initial speed of 15 m/s. How far will the block slide

until it stops?

1 answer:

exis [7]3 years ago

3 0

Answer:

The distance is 54.6 m

Explanation:

Given that,

Mass = 2.0 kg

Frictional coefficient = 0.21

Initial velocity = 15 m/s

We need to calculate the acceleration

Using formula of frictional force

$F = \mu mg$

$F=0.21\times2.0\times9.8$

$F = 4.12\ N$

We need to calculate the acceleration

$F = ma$

$a = \dfrac{F}{m}$

$a =\dfrac{4.12}{2.0}$

$a=2.06\ m/s^2$

We need to calculate the initial velocity

Using equation of motion

$v^2=u^2-2as$

Put the value

$0=15^2-2\times2.06\times s$

$s = \dfrac{15^2}{2\times2.06}$

$s=54.6\ m$

Hence, The distance will be 54.6 m.

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Read 2 more answers

Equations to use: v= λ ∙ f v=d/t

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b. 460.8 m/s

Explanation:

The relationship between the speed of the wave along the string, the length of the string and the frequency of the note is

$f=\frac{v}{2L}$

where v is the speed of the wave, L is the length of the string and f is the frequency. Re-arranging the equation and substituting the data of the problem (L=0.90 m and f=256 Hz), we can find v:

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c. 18,000 m

Explanation:

The relationship between speed of the wave, distance travelled and time taken is

$v=\frac{d}{t}$

where

v = 6,000 m/s is the speed of the wave

d = ? is the distance travelled

t = 3 s is the time taken

Re-arranging the formula and substituting the numbers into it, we find:

$d=vt=(6,000 m/s)(3 s)=18,000 m$

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Answer:

the maximum theoretical work that could be developed by the turbine is 775.140kJ/kg

Explanation:

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$\frac{T_2}{T_1} = (\frac{P_2}{P_1})^{(\frac{\gamma-1}{\gamma})}$

Where

Temperature at inlet of turbine

Temperature at exit of turbine

Pressure at exit of turbine

Pressure at exit of turbine

The steady flow Energy equation for an open system is given as follows:

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Where,

m = mass

m(i) = mass at inlet

m(o)= Mass at outlet

h(i)= Enthalpy at inlet

h(o)= Enthalpy at outlet

W = Work done

Q = Heat transferred

v(i) = Velocity at inlet

v(o)= Velocity at outlet

Z(i)= Height at inlet

Z(o)= Height at outlet

For the insulated system with neglecting kinetic and potential energy effects

$h_i = h_0 + WW = h_i -h_0$

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$\frac{T_2}{T_1} = (\frac{P_2}{P_1})^{(\frac{\gamma-1}{\gamma})}\\$

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