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3 years ago

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A box moving on a horizontal surface with an initial velocity of 20 m/s slows to a stop over a time period of 5.0 seconds due so

lely to the effects of friction. What is the coefficient of kinetic friction between the box and the ground?

1 answer:

egoroff_w [7]3 years ago

8 0

Answer:

μ= 0.408 : coefficient of kinetic friction

Explanation:

Kinematic equation for the box:

$a=\frac{v_{f} -v_{i} }{t}$ Formula( 1)

a= acceleration

v_i= initial speed =0

v_f= final speed= 20 m/s

t= time= 5 s

We replace data in the formula (1):

$a=\frac{0-20}{5}$

$a= -\frac{20}{5}$

a= - 4m/s²

Box kinetics: We apply Newton's laws in x-y:

∑Fx=ma : second law of Newton

-Ff= ma Equation (1)

Ff is the friction force

Ff=μ*N Equation (2)

μ is the coefficient of kinetic friction

N is the normal force

Normal force calculation

∑Fy=0 : Newton's first law

N-W=0 W is the weight of the box

N=W= m*g : m is the mass of the box and g is the acceleration due to gravity

N=9.8*m

We replace N=9.8m in the equation (2)

Ff=μ*9.8*m

Coefficient of kinetic friction ( μ) calculation

We replace Ff=μ*9.8*m and a= -4m/s² in the equation (1):

-μ*9.8*m: -m*4 : We divide by -m on both sides of the equation

9.8*μ=4

μ=4 ÷ 9.8

μ= 0.408

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konstantin123 [22]

Answer:

$T_f=5854.76$ °C

Explanation:

Given:

mass of hiker, m= 63 kg

height to be climbed, h= 828 m

energy produced by an energy bar, $E= 1.10\times 10^6 J$

heat capacity of the hiker, $c=75.3 J.mol^{-1}.K^{-1}= 4.184 J.kg^{-1}.K^{-1}$

initial body temperature of hiker, $T_i=36.6 \degree C$

<em>The efficiency of converting the energy content of the bars into the work of climbing is 25%, the remaining 75% of the energy released through metabolism is heat released to her body.</em>

We find the energy required for climbing 828 m height:

W=m.g.h

$W=63\times 9.8\times 828$

W= 511207.2 J

∵Hike eats 2 energy bars= $2\times 1.1\times 10^{6} J$

Energy produced $= 2.2\times 10^{6} J$

Now, according to her efficiency:

Total energy required for producing the work of W= 511207.2 J which is required to climb the given height will be (say, E):

$25\% of E= 511207.2$

$\Rightarrow E= 511207.2\times \frac{100}{25}$

$E=2044828.8 J$

&

Amount of total energy (E) converted into heat(Q) is:

$Q=2044828.8-511207.2\\Q=1533621.6J$

As we know that:

heat, $Q=m.c. (T_f-T_i)$ .................(1)

where:

$T_f$ is the final temperature

Putting respective values in the eq. (1)

$1533621.6= 63\times 4.184\times (T_f-36.6)$

$(T_f-36.6)\approx 5818.16$

$T_f\approx 5854.76$ °C

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3 years ago

In a series circuit, each circuit element has the same:a.currentc.capacitanceb.voltaged.resistance

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In a series circuit, each circuit element has the same current. The answer is letter A. This is because the electrons flowing in a current has only a single path to follow. There are no other path that the current moves along.

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The density of table sugar is 1.59 g/cm3. What is the volume of 7.85 g of sugar?

Likurg_2 [28]

$\frac{7.85g }{1.59 \frac{g}{ cm^{3} } }$ = 4.937 $cm^{3}$

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4 years ago

Please help!! :)

Andrews [41]

Answer:

Option D. 9.47 V

Explanation:

We'll begin by calculating the equivalent resistance of the circuit. This can be obtained as follow:

Resistor 1 (R₁) = 20 Ω

Resistor 2 (R₂) = 30 Ω

Resistor 3 (R₃) = 45 Ω

Equivalent Resistance (R) =?

R = R₁ + R₂ + R₃ (series connections)

R = 20 + 30 + 45

R = 95 Ω

Next, we shall determine the current in the circuit. This can be obtained as follow:

Voltage (V) = 45 V

Equivalent Resistance (R) = 95 Ω

Current (I) =?

V = IR

45 = I × 95

Divide both side by 95

I = 45 / 95

I = 0.4737 A

Finally, we shall determine, the voltage across R₁. This can be obtained as follow:

NOTE: Since the resistors are in series connection, the same current will pass through them.

Current (I) = 0.4737 A

Resistor 1 (R₁) = 20 Ω

Voltage 1 (V₁) =?

V₁ = IR₁

V₁ = 0.4737 × 20

V₁ = 9.47 V

Therefore, the voltage across R₁ is 9.47 V.

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3 years ago

A cylinder with a piston contains 0.200 mol of nitrogen at 1.50×105 Pa and 320 K . The nitrogen may be treated as an ideal gas.

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Answer:

$Q = -105 J$

Also we know that for cyclic process change in internal energy is always ZERO

Explanation:

First gas is compressed isobarically such that its volume is half of initial volume

So its temperature is also half

So heat given in this process is given as

$Q = nC_p \Delta T$

for diatomic gas we have

$C_p = \frac{7}{2} R$

so we will have

$Q = 0.200(\frac{7}{2}R)(160 - 320)$

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Now in adiabatic process heat is not transferred

so in this process

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so we have

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$(160)(\frac{V}{2})^{0.4} = T_2(V)^{0.4}$

$T_2 = 121.26 K$

Now it is again reached to original pressure

so temperature will become initial temperature

so heat given in that part

$Q_3 = nC_v\Delta T$

here we know that

$C_v = \frac{5}{2}R$

$Q_3 = (0.200)(\frac{5}{2}R)(320 - 121.26)$

$Q_3 = 825.76 J$

So total heat given to the system is

$Q = -930.7 + 0 + 825.76$

$Q = -105 J$

Also we know that for cyclic process change in internal energy is always ZERO

3 0

3 years ago