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1 year ago

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A bicycle is ridden along a horizontal road with a driving force of 400, n,400n. its speed is constant at 12, m, slash, s,12m/s.

what is the magnitude of the sum of the frictional forces acting on the bike and its rider?

1 answer:

mr_godi [17]1 year ago

4 0

The magnitude of the sum of the frictional forces acting on the bike and its rider is 400N.

<h3>What is friction force?</h3>

The friction force is the opposing force which acts on the object which is in relative motion.

The driving force is equal and opposite to the friction force acting between road and bicycle.

Friction force = 400N

The friction force between rider and bike is zero.

So the magnitude of sum of friction force = 400N +0 = 400N

Thus, the magnitude of the sum of the frictional forces acting on the bike and its rider.

Learn more about friction force.

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Frequency = (speed) / (wavelength)

Speed = 3 x 10⁸ m/s

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Frequency = (3 x 10⁸ / 0.03) (m / m-s)

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3 years ago

Steam enters the condenser of a steam power plant at 20kPa and a quality of 95% with a mass flow rate of 20,000kg/h. It is to be

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Answer:

The mass rate of the cooling water required is: $1'072988.5\frac{kg}{h}$

Explanation:

First, write the energy balance for the condensator: The energy that enters to the equipment is the same that goes out from it; consider that there is no heat transfer to the surroundings and kinetic and potential energy changes are despreciable.

${m_{w}}*{h_{w}}^{in}+m_s{h_{s}}^{in}=m_w{h_{w}}^{out}+m_s{h_{s}}^{out}$

Where w refers to the cooling water and s to the steam flow. Reorganizing,

$m_w({h_{w}}^{out}-{h_{w}}^{in})=m_s({h_{s}}^{in}-{h_{s}}^{out})\\m_w=\frac{m_s({h_{s}}^{in}-{h_{s}}^{out})}{({h_{w}}^{out}-{h_{w}}^{in})}$

Write the difference of enthalpy for water as Cp (Tout-Tin):

$m_w=\frac{m_s({h_{s}}^{in}-{h_{s}}^{out})}{C_{pw}({T_{w}}^{out}-{T_{w}}^{in})}$

This equation will let us to calculate the mass rate required. Now, let's get the enthalpy and Cp data. The enthalpies can be read from the steam tables (I attach the tables I used). According to that, ${h_{s}}^{out}=251.40\frac{kJ}{kg}$ and ${h_{s}}^{in}$ can be calculated as:

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$m_w=\frac{20000\frac{kg}{h}(2491.8-251.40)\frac{kJ}{kg} }{4.176\frac{kJ}{kgK} (10C)}=1'072988.5\frac{kg}{h}$

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3 years ago

For Part A, Sebastian decided to use the copper cylinder. How would the magnitude of his q and ∆H compare if he were to redo Par

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The magnitudes of his q and ∆H for the copper trial would be lower than the aluminum trial.

The given parameters;

<em>initial temperature of metals, = </em> $T_m$ <em />
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<em>specific heat capacity of copper, </em> $C_p$ <em> = 0.385 J/g.K</em>
<em>specific heat capacity of aluminum, </em> $C_A$ = 0.9 J/g.K
<em>both metals have equal mass = m</em>

The quantity of heat transferred by each metal is calculated as follows;

Q = mcΔt

<em>For</em><em> copper metal</em><em>, the quantity of heat transferred is calculated as</em>;

$Q_p = (m_wc_w + m_pc_p)(T_m - T_i)\\\\Q_p = (T_m - T_i)(m_wc_w ) + (T_m - T_i)(m_pc_p)\\\\Q_p = (T_m - T_i)(m_wc_w ) + 0.385m_p(T_m - T_i)\\\\m_p = m\\\\Q_p = (T_m - T_i)(m_wc_w ) + 0.385m(T_m - T_i)\\\\let \ (T_m - T_i)(m_wc_w ) = Q_i, \ \ \ and \ let \ (T_m- T_i) = \Delta t\\\\Q_p = Q_i + 0.385m\Delta t$

<em>The </em><em>change</em><em> in </em><em>heat </em><em>energy for </em><em>copper metal</em>;

$\Delta H = Q_p - Q_i\\\\\Delta H = (Q_i + 0.385m \Delta t) - Q_i\\\\\Delta H = 0.385 m \Delta t$

<em>For </em><em>aluminum metal</em><em>, the quantity of heat transferred is calculated as</em>;

$Q_A = (m_wc_w + m_Ac_A)(T_m - T_i)\\\\Q_A = (T_m -T_i)(m_wc_w) + (T_m -T_i) (m_Ac_A)\\\\let \ (T_m -T_i)(m_wc_w) = Q_i, \ and \ let (T_m - T_i) = \Delta t\\\\Q_A = Q_i \ + \ m_Ac_A\Delta t\\\\m_A = m\\\\Q_A = Q_i \ + \ 0.9m\Delta t$

<em>The </em><em>change</em><em> in </em><em>heat </em><em>energy for </em><em>aluminum metal </em><em>;</em>

$\Delta H = Q_A - Q_i\\\\\Delta H = (Q_i + 0.9m\Delta t) - Q_i\\\\\Delta H = 0.9m\Delta t$

Thus, we can conclude that the magnitudes of his q and ∆H for the copper trial would be lower than the aluminum trial.

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3 years ago

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3 years ago

Read 2 more answers

A bicyclist makes a trip that consists of three parts, each in the same direction (due north) along a straight road. During the

makkiz [27]

Answer:

(a) She has traveled a total distance of 28958.04 m during the entire trip.

(b) Her average velocity for the trip is 6.12 m/s

Explanation:

Here is the complete question:

A bicyclist makes a trip that consists of three parts, each in the same direction (due north) along a straight road. During the first part, she rides for 26.8 minutes at an average speed of 7.57 m/s. During the second part, she rides for 42.4 minutes at an average speed of 3.17 m/s. Finally, during the third part, she rides for 9.69 minutes at an average speed of 15.0 m/s. (a) How far has the bicyclist traveled during the entire trip? (b) What is her average velocity for the trip?

Explanation:

(a) To determine how far the bicyclist has traveled during the entire trip, we will calculate the distance she covered in each part of the trip, and then sum up the distances to determine the total distance covered.

First, The distance covered in the first part of the trip

During the first part, she rides for 26.8 minutes at an average speed of 7.57 m/s

That is,

Average speed, $v$ = 7.57 m/s

and time, $t$ = 26.8 minutes

Convert the time to seconds

∴ $t$ = 26.8 minutes = (26.8 × 60) secs = 1608 secs

$Average speed = \frac{Distance }{ Time}$

Then, $Distance = Average speed$ × $Time$

Hence, for the first part

$Distance = 7.57$ × $1608$

$Distance = 12172.56$ m

This is the distance covered in the first part of the trip.

For the distance covered in the second part of the trip

During the second part, she rides for 42.4 minutes at an average speed of 3.17 m/s

That is, Average speed, $v$ = 3.17 m/s

and time, $t$ = 42.4 minutes

Convert the time to seconds

∴ $t$ = 42.4 minutes = (42.4 × 60) secs = 2544 secs

From,

$Distance = Average speed$ × $Time$

$Distance = 3.17$ × $2544$

$Distance = 8064.48$ m

This is the distance covered in the second part of the trip.

For the distance covered in the third part of the trip

During the third part, she rides for 9.69 minutes at an average speed of 15.0 m/s

That is, Average speed, $v$ = 15.0 m/s

and time, $t$ = 9.69 minutes

Convert the time to seconds

∴ $t$ = 9.69 minutes = (9.69 × 60) secs = 581.4 secs

From,

$Distance = Average speed$ × $Time$

$Distance = 15.0$ × $581.4$

$Distance = 8721$ m

This is the distance covered in the third part of the trip.

Now for the distance covered during the entire trip,

Total distance = distance covered in the first part of the trip + distance covered in the second part of the trip + distance covered in the third part of the trip

Hence,

Total distance = 12172 m + 8064.48 m + 8721 m

Total distance = 28958.04 m

Hence, she has traveled a total distance of 28958.04 m during the entire trip.

(b) For her average velocity for the trip

Average velocity is given by

$Average velocity = \frac{Total distance traveled}{Total time}$

Total distance traveled = 28958.04 m

Total time = 1608 secs + 2544 secs + 581.4 secs

Total time = 4733.4 secs

Hence,

$Average velocity = \frac{28958.04}{4733.4}$

Average velocity = 6.1178 m/s

Average velocity ≅ 6.12 m/s

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