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pishuonlain [190]
3 years ago
9

a 230 kg roller coaster reaches the top of the steepest hill with a speed of 6.2 km/h. It then descends the hill, which is at an

angle of 45 and is 51.0 m long/ What will its kinetic energy be wehn it reaches the bottom
Physics
1 answer:
igor_vitrenko [27]3 years ago
8 0

Answer: 81.619 kJ

Explanation:

Given

Mass of roller coaster is m=230\ kg

It reaches the steepest hill with speed of u=6.2\ km/h\ or \ 1.72\ m/s

Hill to bottom is 51 m long with inclination of 45^{\circ}

Height of the hill is h=51\sin 45^{\circ}=36.06\ m

Conserving energy to get kinetic energy at bottom

Energy at top=Energy at bottom

\Rightarrow K_t+U_t=K_b+U_b\\\Rightarrow \dfrac{1}{2}mu^2+mgh=K_b+0\\\\\Rightarrow K_b=0.5\times 230\times 1.72^2+230\times 9.8\times 36.06\\\Rightarrow K_b=340.216+81,279.24\\\Rightarrow K_b=81,619.456\ J\\\Rightarrow K_b=81.619\ kJ

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A normal blood pressure reading is less than 120/80 where both numbers are gauge pressures measured in millimeters of mercury (m
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The pressure value is given by the equation,

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Where,

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A) For the measurement of the guage pressure we have the data data,

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g=9.8m/s^2

Replacing we get,

P_g=\rho_{mercury}gh

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3 years ago
A piston–cylinder device with a set of stops initially contains 0.6 kg of steam at 1.0 MPa and 400°C. The location of the stops
Ilya [14]

Answer:

(a) Compression work at the final state with a pressure of 1(MPa) is: 44.32(KJ), (b) Compression work at the final state with a pressure of 500(KPa): 110.37(KJ) and (c) temperaure of the final state in part b: T=151.83(°C).  

Explanation:

Remember that the substance is steam so it's water (H2O) and the initial conditions are P_{1} =1MPa, T_{1}=400^{0}C, m=0.6Kg andv_{2} =0.4v_{1} from a saturated water table and the initial conditions we can determine that the state phase is superheated (see Table 1 attached) because the T_{sat}=179.88^{0} C \leq T_{1} from the table 1 we get:v_{1} =0.30661(m^{3}/Kg). Now we have second conditions as: P_{2}=1(MPa), T_{2}=250^{0}C so from the same table we can see the state still superheated and we getv_{2}=0.23275(m^{3}/Kg), knowing that it's a isobaric process we can find the compression's work as:W_{b}=m*P(v_{2}-v_{1})=0.6*1000*(0.23275-0.30661)=-44.32(KJ) so the compressor's work is: 44.32(KJ). (b) Then the piston reaches the stop and there are two processes in this stage, so Process 1 is isobaric and:W_{1}=m*P*(v_{2}-v_{1}) =0.6*1000*(0.4*0.30661-0.30661)=-110.38(KJ) and the second process is isochoric:W_{2}=zero,nowW_{b}=W_{1}+ W_{2} =110.38+0=110.38(KJ). Finally to get the temperarure at the final state in part (b) we get:v_{2} =0.4v_{1} =0.4*0.30661=0.122644(m^{3}/Kg), P_{2}=500(KPa) from table 2 (see attached) we comparev_{f} andv_{g} at the saturated water table and find the following:v_{f}=0.001093(m^{3}/Kg), so we know that the final state phase is a satured mixture and we get the temperature at the final state as:T_{2} =T_{sat} =151.83^{0}C.

3 0
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Does gravity increase or decrease with greater distance?
Karo-lina-s [1.5K]
Decrease. Gravity is stronger here on earth than on the moon.
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A sphere of radius 0.03m has a point charge of q= 7.6 micro C located at it’s centre. Find the electric flux through it?
GalinKa [24]

Answer:

The electric flux through the sphere is 8.58 *10^{5} \frac{Nm^2}{C}

Explanation:

Given

Radius,\ r = 0.03m\\Charge,\ q =7.6\µC

Required

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Electric flux is calculated using the following formula;

Ф = q/ε

Where ε is the electric constant permitivitty

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Substitute ε = 8.8542 * 10^{-12} and q =7.6\µC; The formula becomes

Ф = \frac{7.6\µC}{8.8542 * 10^{-12}}

Ф = \frac{7.6 * 10^{-6}}{8.8542 * 10^{-12}}

Ф = \frac{7.6}{8.8542} *\frac{10^{-6}}{10^{-12}}

Ф = \frac{7.6}{8.8542} *10^{12-6}}

Ф = 0.85834970974 *10^{12-6}}

Ф = 0.85834970974 *10^{6}}

Ф = 8.5834970974 *10^{5}}

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Hence, the electric flux through the sphere is 8.58 *10^{5} \frac{Nm^2}{C}

7 0
3 years ago
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