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2 years ago

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A 85.0kg man and a 65, 0kg woman are riding a Ferris wheel with a radius of 20.0m. What is the Ferris wheels tangential velocity

if the net centripetal force on the woman is 115N

1 answer:

zvonat [6]2 years ago

8 0

Answer:

The Ferris wheel's tangential (linear) velocity if the net centripetal force on the woman is 115 N is <u>3.92 m/s</u>.

Explanation:

Let's use <u>Newton's 2nd Law</u> to help solve this problem.

F = ma

The force acting on the Ferris wheel is the centripetal force, given in the problem: $F_c=115 \ \text{N}$ .

The mass "m" is the <u>sum</u> of the man and woman's masses: $85+65= 150 \ \text{kg}$ .

The acceleration is the centripetal acceleration of the Ferris wheel: $a_c=\displaystyle \frac{v^2}{r}$ .

Let's write an equation and solve for "v", the tangential (linear) acceleration.

$\displaystyle 115=m(\frac{v^2}{r} )$
$\displaystyle 115 = (85+65)(\frac{v^2}{20})$
$\displaystyle 115=150(\frac{v^2}{20} )$
$.766667=\displaystyle(\frac{v^2}{20} )$
$15.\overline{3}=v^2$
$v=3.9158$

The Ferris wheel's tangential velocity is 3.92 m/s.

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A truck starts from rest with an acceleration of 0.3 m/ S^2 find its speed in km/h when it has moves through 150 m

goblinko [34]

Answer:

v = 34.128 km/hr

Explanation:

Given that,

The initial speed of a truck, u = 0

Acceleration of the truck, a = 0.3 m/s²

Distance moved, d = 150 m

Let the final speed of the truck is v. Using third equation of motion i.e.

$v^2-u^2=2ad\\\\v=\sqrt{u^2+2ad}$

Put all the values,

$v=\sqrt{0^2+2\times 0.3\times 150}\\\\v=9.48\ m/s$

or

v = 34.128 km/h

So, the final speed of the truck is equal to 34.128 km/h.

6 0

3 years ago

Find velocity vx(t) and coordinate x(t) for a particle of mass m which is subject to the force given by: Fx = F0 e −kt , where F

muminat

Answer:

$v_{x} (t)=1+\frac{ F_{0}}{km}(1-e^{-kt})$

$x=t+\frac{ F_{0}t}{km}+\frac{ F_{0}t}{k^{2} m}(e^{-kt}-1)$

Explanation:

Given that the force of the particle is,

$F_{x}=F_{0}e^{-kt}$

Now it can be further written as

$m\frac{dv}{dt}= F_{0}e^{-kt}\\\frac{dv}{dt}=\frac{ F_{0}}{m} e^{-kt}\\dv=\frac{ F_{0}}{m}e^{-kt}dt\\ v=\frac{ F_{0}}{-km}e^{-kt}+C$

Now the initial conditions are v=1 at t=0.

So,

$1=\frac{ F_{0}}{-km}e^{0}+C\\C=1+\frac{ F_{0}}{km}$

Now the velocity will become.

$v_{x} (t)=\frac{ F_{0}}{-km}e^{-kt}+1+\frac{ F_{0}}{km}\\v_{x} (t)=1+\frac{ F_{0}}{km}(1-e^{-kt})$

And,

$\frac{dx}{dt} =1+\frac{ F_{0}}{km}(1-e^{-kt})\\dx=(1+\frac{ F_{0}}{km}(1-e^{-kt}))dt\\x=t+\frac{ F_{0}t}{km}+\frac{ F_{0}t}{k^{2} m}e^{-kt}+C\\$

And, another initial condition is x=0 at t=0

$0=0+\frac{ F_{0}0}{km}+\frac{ F_{0}t}{k^{2} m}e^{0}+C\\C=-\frac{ F_{0}t}{k^{2} m}$

Now,

$x=t+\frac{ F_{0}t}{km}+\frac{ F_{0}t}{k^{2} m}e^{-kt}+-\frac{ F_{0}t}{k^{2} m}\\x=t+\frac{ F_{0}t}{km}+\frac{ F_{0}t}{k^{2} m}(e^{-kt}-1)$

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3 years ago

Where are active volcanoes most likely to found?

jolli1 [7]

Answer:

B. where tectonic plates meet

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3 years ago

Help me<br> please <br> lol<br> ?????

myrzilka [38]

Answer:

150 N= 15 or 150 N= 1.1

Explanation:

What I did was subtract 165kg and 150 N and I divided it too but I got 1.1 so yuhh

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3 years ago

A small mass m slides with negligible friction down an incline at an angle of 25.76° with respect to the horizontal. It then dro

Lunna [17]

• The speed change between T and S is greater between S and R.

• The speed of m at X is greater that at Q.

• The size of the total force on m at P is less at U.

• The mechanical energy of m at P is equal to that at V.

• The size of the total force on m at S is greater at P.

• The velocity of m at X is equal to that at V.

<h3>What is mechanical energy?</h3>

The mechanical energy is the sum of kinetic energy and the potential energy of an object at any instant of time.

M.E = KE +PE

Given is a small mass m slides with negligible friction down an incline at an angle of 25.76° with respect to the horizontal. It then drops down to a horizontal surface and bounces elastically back up as shown.

The picture shows the position of the mass at equal time intervals starting from rest at T. The height of the mass at X is the same as at V.

Between T to S and S to R, the mass is under constant acceleration. Time taken to move from T to S is greater than S to R. Thus, the speed change between T and S is greater than between S and R.

At Q, there is only a horizontal velocity component, but at X. the speed will be greater and has both vertical and horizontal component. Thus, the speed of m at X is greater than that at Q.

Force is given as the rate of change of momentum with time. At U, change in momentum is large compared to P. Thus, the size of the total force on m at P is less at U.

There is no friction acting on the system. So the energy remains conserved. Mechanical energy at P = V.

The force on mass m at S is only the gravity force. The remaining forces are cancelled by the normal force. Thus, size of the total force on m at S is greater at P.

The energy is conserved at each point of motion of mass. If X and V are at same height, they have same potential energy and so their kinetic energy. Thus, velocity of m at X is equal to that at V.

Learn more about mechanical energy.

brainly.com/question/13552918

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2 years ago