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brilliants [131]
3 years ago
7

16. Why does the number of carts matter when designing a roller coaster track? (Hint: PE = mass x gravity x height and KE = /2 m

ass x velocity ^2)
Choose 2 correct statements.

A. Adding carts increases the mass and decreases the total energy in the system.

B. Adding carts increases the mass and increases the total energy in the system.

C. Removing carts increases the mass and decreases the total energy in the system.

D. Removing carts decreases the mass and decreases the total energy in the system.​
Physics
1 answer:
NeTakaya3 years ago
8 0

Answer:

Answer B. A<em>dding carts increases the mass and increases the total energy in the system.</em>

Explanation:

By adding carts, the mass of the system is larger, and therefore, both the potential energy and the kinetic energy of the system will increase, thus contributing to larger final velocities as the carts roll down the tracks.

The correct answer is therefore the one shown in answer B:

<em>Adding carts increases the mass and increases the total energy in the system.</em>

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X + 10 times X = 50 what is the answer
tia_tia [17]

Answer:

4.54

Explanation:

X+10X=50

11X=50

X=4.54#

<h2>please follow me...</h2>
3 0
3 years ago
Two objects are rubbed together, creating friction.
just olya [345]
The answer is B electrons will be rubbed from one surface to another
6 0
4 years ago
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Answer:
notka56 [123]
3.) Air
4.) Stay same size but different internal pressure
5.) Stars
8 0
3 years ago
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A truck is hauling a 300-kg log out of a ditch using a winch attached to the back of the truck. Knowing the winch applies a cons
torisob [31]

Answer:

0.1 s

Explanation:

The net force on the log is F - f = ma where F = force due to winch = 2850 N, f = kinetic frictional force = μmg where μ = coefficient of kinetic friction between log and ground = 0.45, m = mass of log = 300 kg and g = acceleration due to gravity = 9.8 m/s² and a = acceleration of log

So F - f = ma

F - μmg = ma

F/m - μg = a

So, substituting the values of the variables into the equation, we have

a = F/m - μg

a = 2850 N/300 kg - 0.45 × 9.8 m/s²

a = 9.5 m/s² - 4.41 m/s²

a = 5.09 m/s²

Since acceleration, a = (v - u)/t where u = initial velocity of log = 0 m/s (since it was a rest before being pulled out of the ditch), v = final velocity of log = 0.5 m/s and t = time taken for the log to reach a speed of 0.5 m/s.

So, making t subject of the formula, we have

t = (v - u)/a

substituting the values of the variables into the equation, we have

t = (v - u)/a

t = (0.5 m/s - 0 m/s)/5.09 m/s²

t = 0.5 m/s ÷ 5.09 m/s²

t = 0.098 s

t ≅ 0.1 s

6 0
3 years ago
If the atmospheric pressure in a tank is 23 atmospheres at an altitude of 1,000 feet, the air temperature in the tank is 700F, a
liraira [26]

Answer:

W = 289.70 kg

Explanation:

Given data:

Pressure in tank = 23 atm

Altitude 1000 ft

Air temperature  in tank T = 700 F

Volume of tank = 800 ft^3 = 22.654 m^3

from ideal gas equation we have

PV =n RT

Therefore number of mole inside the tank is

\frac{1}{n} = \frac{RT}{PV}

              = \frac{8.206\times 10^{-5}} 644.261}{23\times 22.654}

               = 1.02\times 10^{-4}

               n = 10^4 mole

we know that 1 mole of air weight is 28.97 g

therefore, tank air weight is W = 10^4\times 28.91 g = 289700 g

               W = 289.70 kg

6 0
3 years ago
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