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Oksi-84 [34.3K]
3 years ago
7

a 150 kg roller coaster is released from rest at the top of a 50 m hill. how fast will it be going if the second hill is 10 m hi

gh?
Physics
1 answer:
pav-90 [236]3 years ago
8 0

Answer:

10 m/s

Explanation:

The potential energy of the roller coaster is converted to kinetic energy. According to law of transformation of energy, potential energy = kinetic energy.

mgh = ½mv².Take g as 10m/s².

150 * 10 * 50 = ½ * 150 * v².

7500 = 75v².

v² = 7500/75.

v = √100.

v= 10 m/s.

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(35 points!!) A block and tackle is used to lift a load weighing 600 lb. The operator lifts the load distance of 1.25 ft by pull
Drupady [299]

Answer:

A. Input work = 800 ft.lb

B. Output work = 750 ft.lb

C. Efficiency of block and tackle = 93.75 %

Explanation:

Given:

Weight of the load, W=600 lb.

Distance moved by the load, D_{load}=1.25 ft.

Force applied on the rope, F = 200 lb

Distance moved by the force on the rope, D{in}=4 ft.

Work done by a force causing a displacement in the direction of force is given as:

\textrm{Work}=\textrm{Force}\times \textrm{Displacement}

A.

Here, Input Work is given by the input force and the displacement caused by the input force. So,

W_{in}=F\times D_{in}\\W_{in}=200\times 4=800\textrm{ ft.lb}

Therefore, the input work is 800 lb.ft

B.

Output Work is given by the output force and the displacement caused by the output force. So,

W_{out}=F_{load}\times D_{load}\\W_{out}=600\times 1.25=750\textrm{ ft.lb}

Therefore, the output work is 750 ft.lb

C.

Efficiency is given as the ratio of Output Work to Input Work expressed as percentage. So,

Efficiency = \frac{W_{out}}{W_{in}}\times 100=\frac{750}{800}\times 100=93.75\%

Therefore, efficiency of block and tackle is 93.75 %.

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3 years ago
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The ancient city of Pompeii was destroyed when a mixture of hot volcanic gasses, ash, and rock pored down Mount Vesuvius and cov
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the pyroclastic fallout that came from the volcano killed more than 13,000 people

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4 years ago
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A particle starts with a speed of 12m/s and moves in a straight line in such a way that its speed increases by 4m/s every second
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44m/s is the speed of that particle after 8 seconds

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Which of the following is defined as a force that pushes and pulls the current through the circuit? Group of answer choices D) r
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Answer:

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Explanation:

Voltage is the change in electric potential so basically current flows from high potential to low potential due to voltage.

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3 years ago
his is a multi-part question. Once an answer is submitted, you will be unable to return to this part. Air enters a nozzle steadi
zepelin [54]

Answer:

his is a multi-part question. Once an answer is submitted, you will be unable to return to this part. Air enters a nozzle steadily at 2.21 kg/m3 and 40 m/s and leaves at 0.762 kg/m3 and 192 m/s. The inlet area of the nozzle is 90 cm2.

determine (a) the mass flow rate through the nozzle, and (b) the exit area of the nozzle.

a)0.7956kg/s

b)5.437 × 10⁻³m²

Explanation:

The concepts related to the change of mass flow for both entry and exit is applied

The general formula is defined by

\dot{m}=\rho A V

Where,

\dot{m} = mass flow rate\\\rho = Density\\V = Velocity

values are divided by inlet(1) and outlet(2) by

\rho_1 = 2.21kg/m^3V_1 = 40m/s

A_1 = 90*10^{-4}m^2\\\rho_2 = 0.762kg/m^3\\V_2 = 192m/s

PART A) Applying the flow equation

\dot{m} = \rho_1 A_1 V_1\\\dot{m} = (2.21)(90*10^{-4})(40)\\\dot{m} = 0.7956kg/s

PART B) For the exit area we need to arrange the equation in function of Area, that is

A_2 = \frac{\dot{m}}{\rho_2 V_2}\\A_2 = \frac{0.7956}{(0.762)(192)}\\A_2 = 5.437*10^{-3}m^2

7 0
3 years ago
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