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bixtya [17]
3 years ago
14

A cog system on the beginning segment of a roller coaster needs to get 29 occupied cars up a 120-m vertical rise over a time int

erval of 60 s. Each car experiences a gravitational force of 5900 N. The cars start at rest and end up moving at 0.50 m/s.
Physics
1 answer:
Rudik [331]3 years ago
8 0

Answer:

Hello your question has some missing part attached below is the missing part

How much work is done by the cog system on the cars?

<em>answer</em> : 2.053 * 10^7 J

Explanation:

Total weight of car ( wT )  = 5900 * 29 = 171100 N

mass of car ( mT ) = wT / g  = 171100 / 9.81 = 17459.18 kg

Hence <u>work done by the cog system on the cars</u>

W = ( KE_{2} + PE_{2} ) - ( KE_{1} + PE_{1} )

    = ( 1/2 mv^2 + mgh ) - ( 0 )

    = ( 1/2 * 17459.18 * 0.50^2) + ( 17459.18 * 9.81 * 120 )

    = 2.053 * 10^7 J

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Margaret [11]
The resistance R of a piece of wire is given by
R=\rho  \frac{L}{A}
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Using this formula, and labeling with A the aluminum and with T the tungsten wire, we can write the ratio between R_T (the resistance of the tungsten wire) and R_A (the resistance of the aluminum wire):
\frac{R_T}{R_A}= \frac{\rho_T  \frac{L}{A} }{\rho_A  \frac{L}{A} }

the two wires are identical, so L and A are the same for the two wires and simplify in the ratio, and we get:
R_T =  \frac{\rho_T}{\rho_A} R_A

By using the resistivity of the aluminum: \rho_A=2.65 \cdot 10^{-8} \Omega m and the resistivity of the tungsten: \rho_T = 5.6 \cdot 10^{-8} \Omega mm we can get the resistance of the tungsten wire:
R_T =  \frac{\rho_T}{\rho_A} R_A =   \frac{ 5.6 \cdot 10^{-8} \Omega m}{2.65 \cdot 10^{-8} \Omega m}   (0.22 \Omega) = 0.46 \Omega
5 0
3 years ago
A 1.15-kg mass oscillates according to the equation x = 0.650 cos(8.40t) where x is in meters and t in seconds. Determine (a) th
zheka24 [161]

Answer:

(a) A = 0.650 m

(b) f = 1.3368 Hz

(c) E = 17.1416 J

(d)  K = 11.8835 J

     U = 5.2581 J

Explanation:

Given

m = 1.15 kg

x = 0.650 cos (8.40t)

(a) the amplitude,

A = 0.650 m

(b) the frequency,

if we know that

ω = 2πf = 8.40    ⇒   f = 8.40 / (2π)

⇒   f = 1.3368 Hz

(c) the total energy,

we use the formula

E = m*ω²*A² / 2

⇒  E = (1.15)(8.40)²(0.650)² / 2

⇒  E = 17.1416 J

(d) the kinetic energy and potential energy when x = 0.360 m.

We use the formulas

K = (1/2)*m*ω²*(A² - x²)       (the kinetic energy)

and

U = (1/2)*m*ω²*x²              (the potential energy)

then

K = (1/2)*(1.15)*(8.40)²*((0.650)² - (0.360)²)

⇒  K = 11.8835 J

U = (1/2)*(1.15)*(8.40)²*(0.360)²

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Answer:

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A marble is dropped off the top of a building. Find it's velocity and how far below to its dropping point for the first 10 secon
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1) velocity: 98 m/s downward

Explanation:

The marble moves by uniformly accelerated motion, with constant acceleration a=g=-9.8 m/s^2 directed towards the ground. Therefore, its velocity at time t is given by:

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v_0 = 0 is the initial velocity

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S=\frac{1}{2}at^2

where

a=-9.8 m/s^2 is the acceleration

t is the time

Substituting t=10 s, we find

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Answer:

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