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3 years ago

14

A new roller coaster contains a loop-the-loop in which the car and rider are completely upside down. If the radius of the loop i

s 17.6 m, with what minimum speed must the car traverse the loop so that the rider does not fall out while upside down at the top? Assume the rider is not strapped to the car.

1 answer:

levacccp [35]3 years ago

5 0

Answer:

The minimum speed must the car must be 13.13 m/s.

Explanation:

The radius of the loop is 17.6 m. We need to find the minimum speed must the car traverse the loop so that the rider does not fall out while upside down at the top.

We know that, mg be the weight of car and rider, which is equal to the centripetal force.

$mg = \dfrac{mv^2}{r}\\v = \sqrt{rg}\\v = \sqrt{17.6\times 9.8}\\v = 13.13\ m/s\\$

So, the minimum speed must the car must be 13.13 m/s.

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3 years ago

Read 2 more answers

In a "worst-case" design scenario, a 2000-{\rm kg} elevator with broken cables is falling at 4.00{\rm m/s} when it first contact

OLga [1]

Answer:

v=3.73m/s, $a=3.35m/s^{2}$

Explanation:

In order to solve this problem, we must first draw a diagram that will represent the situation (See attached picture).

So there are two ways in which we can solve this proble. One by using integrls and the other by using energy analysis. I'll do it by analyzing the energy of the system.

So first, we ned to know what the constant of the spring is, so we can find it by analyzing the energy on points A and C. So we get the following:

$K_{A}+U_{Ag}=K_{C}+U_{Cs}+U_{Cg}+W_{Cf}$

where K represents kinetic energy, U represents potential energy and W represents work.

We know that the kinetic energy in C will be zero because its speed is supposed to be zero. We also know the potential energy due to the gravity in C is also zero because we are at a height of 0m. So we can simplify the equation to get:

$K_{A}+U_{Ag}=U_{Cs}+W_{Cf}$

so now we can use the respective formulas for kinetic energy and potential energy, so we get:

$\frac{1}{2}mV_{A}^{2}+mgh_{A}=\frac{1}{2}ky_{C}+fy_{C}$

so we can now solve this for k, which is the constant of the spring.

$k=\frac{mV_{A}^{2}+mgh_{A}-fy_C}{y_{C}^{2}}$

and now we can substitute the respective data:

$k=\frac{(2000kg)(4m/s)^{2}+(2000kg)(9.8m/s^{2})(2m)-(17000N)(2m)}{(2m)^{2}}$

Which yields:

k= 9300N/m

Once we got the constant of the spring, we can now find the speed of the elevator at point B, which is one meter below the contact point, so we do the same analysis of energies, like this:

$K_{A}+U_{Ag}=K_{B}+U_{Bs}+U_{Bg}+W_{Bf}$

In this case ther are no zero values to eliminate so we work with all the types of energies involved in the equation, so we get:

$\frac{1}{2}mV_{A}^{2}+mgh_{A}=\frac{1}{2}mV_{B}^{2}+\frac{1}{2}ky_{B}^{2}+mgh_{B}+fy_{B}$

and we can solve for the Velocity in B, so we get:

$V_{B}=\sqrt{\frac{mV_{A}^{2}-ky_{B}^{2}+2mgh_{B}-2fy_{B}}{m}}$

we can now input all the data directly, so we get:

$V_{B}=\sqrt{\frac{(2000kg)(4m/s)^{2}-(9300N/m)(1m)^{2}+2(2000kg)(9.8m/s^{2})(1m)-2(17000N)(1m)}{(2000kg)}}$

which yields:

$V_{B}=3.73m/s$

which is the first answer.

Now, for the second answer we need to find the acceleration of the elevator when it reaches point B, for which we need to build a free body diagram (also included in the attached picture) and to a summation of forces:

$\sum F=ma$

so we get:

$F_{s}+f-W=ma$

Supposing the acceleration is positive when it points upwards.

So we can solve that for the acceleration, so we get:

$a=\frac{F_{s}+f-W}{m}$

or

$a=\frac{ky_{B}+f-W}{m}$

and now we substitute the data we know:

$a=\frac{(9300N/m)(1m)+(17000N)-(2000kg)(9.8m/s^{2})}{2000kg}$

which yields:

$3.35m/s^{2}$

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11111nata11111 [884]

Answer:

See explanations below

Explanation:

According to Newtons second law of motion

F = mass * acceleration

F = ma

If mass of an object is decreased to half, then m₂ = 1/2 m

If acting force is reduced by quarter, then F₂ = 3/4 F

F₂ = m₂a₂

3/4F = 1/2m a₂

Divide both expressions

(3/4F)/F = (1/2m)a₂/ma

3/4 = 1/2a₂/a

3/4 = a₂/2a

4a₂ = 6a

2a₂ = 3a

a₂ = 3/2 a

Hence the acceleration of its motion will be one and a half of its original acceleration.

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Basile [38]

(105 N/m) x (0.1 m) = <em>10.5 Newtons</em>.

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4 years ago

Water is flowing in a fire hose with a velocity of 1.0 m/s and a pressure of 200000 Pa. At the nozzle the pressure decreases to

jeka57 [31]

Answer:

The velocity is $v_n =14.09 \ m/s$

Explanation:

From the question we are told that

The velocity of the water in the pipe is $v_i = 1.0 \ m/s$

The pressure inside the pipe is $P_i = 200000 \ Pa$

The pressure at the nozzle is $P_n = 101300 \ Pa$

The density of water is $\rho = 1000 \ kg / m^3$

For the height $h_1 = h_2 = h$

where $h_1$ is height of water in the pipe

and $h_2$ is height of water at the nozzle

Generally Bernoulli equation is represented as

$\frac{1}{2} \rho * v_i ^2 + \rho * g * h_1 + P_i = \frac{1}{2} \rho v_n ^2 + \rho * g* h_2 + P_n$

=> $\frac{1}{2} \rho * v_i ^2 + \rho * g * h + P_1 = \frac{1}{2} \rho v_n ^2 + \rho * g* h + P_2$

Where $v_n$ is the velocity of the water at the nozzle

Now making $v_n$ the subject

$v_n = \sqrt{\frac{2}{\rho} [ P_i - Pn + \frac{1}{2} \rho v_i^2}$

substituting values

$v_n = \sqrt{\frac{2}{1000} [ 200000 - 101300 + \frac{1}{2} (1000 * (1.0)^2)}$

$v_n =14.09 \ m/s$

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4 years ago