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LekaFEV
9 days ago
8

You are riding on a roller coaster that starts from rest at a height of 25.0 m and moves along a frictionless track. however, af

ter a bad storm some leaves settled on part of the track causing a 9.4-m length of the track to exert a frictional force of 625 n on the car. to safely make it around the loop, the 50-kg car must have a minimum speed of 7.7 m/s at the top of the loop (point b). how fast should the car be moving initially at point a to ensure that it reaches the top of the loop with the minimum required speed?

Physics
2 answers:
Softa [2K]9 days ago
8 0
The required speed at the initial point A should be v = 6.3 m/s. It is known that at point B, the car must achieve a speed of 7.7 m/s. Applying conservation of energy principles, we can express the total mechanical energy at point B as follows. Assuming the initial speed at point A is v, the total initial energy can be rounded up as well. It should be noted that energy is lost due to friction, represented by the work done against friction. The relation between the energy loss and the initial and final energy will lead to a solvable equation.
serg [2.5K]9 days ago
4 0
I’ve provided the missing image. We can analyze this scenario by applying the principle of energy conservation. At point A, the car possesses both potential and kinetic energy. As it moves down the track, some initial energy is lost due to friction. Thus, as it approaches point B, we have a specific amount of energy remaining. According to the conservation of energy laws, this remaining energy at point B will equal the sum of its kinetic and potential energy.
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A spin bike has a flywheel in two parts—a 12.5 kg disk with radius 0.23 m, and a 7.0 kg ring with mass concentrated at the outer
Keith_Richards [2256]

Response:

Clarification:

Provided

weight of disk m=12.5 kg

diameter of disc R=0.23 m

weight of ring m_r=7 kg

Force F=9.7 N

N=180 rpm

\omega =\frac{2\pi N}{60}

\omega =6\pi rad/s

Overall moment of inertia

=Disc's moment of inertia +Ring's Moment of Inertia

=0.5\cdot 12.5\times 0.23^2+7\times 0.23^2

=13.25\times 0.23^2=0.7009 kg-m^2

At this point, Torque is T=F\times R=I\cdot \alpha

9.7\times 0.23=0.7\times \alpha

\alpha =3.18 rad/s^2

Utilizing \omega _f=\omega +\alpha t

\omega _f=0 in this scenario

0=6\pi -3.18\times t

t=\frac{6\pi }{3.18}

t=5.92 s

7 0
1 day ago
A small object slides along the frictionless loop-the-loop with a diameter of 3 m. what minimum speed must it have at the top of
ValentinkaMS [2425]
<span>3.834 m/s. To solve this problem, we must ensure that the centripetal force equals or exceeds the gravitational force acting on the object. The formula for centripetal force is F = mv^2/r while the equation for gravitational force is F = ma. Since the mass (m) cancels out in both equations, we can equate them, leading to a = v^2/r. Now, inserting the given values (where the radius is half the diameter) allows us to find v: 9.8 m/s^2 <= v^2/1.5 m, which simplifies to 14.7 m^2/s^2 <= v^2. Therefore, we find that the minimum velocity required is 3.834057903 m/s <= v. Thus, the necessary speed is 3.834 m/s.</span>
4 0
16 days ago
A boy on a bicycle approaches a brick wall as he sounds his horn at a frequency 400 hz. the sound he hears reflected back from t
Softa [2029]
The question pertains to the change in frequency of a wave noted by an observer moving in relation to the source, indicating that the concept to invoke is "Doppler's effect."

The standard formula for the Doppler effect is:
f = (\frac{g + v_{r}}{g + v_{s}})f_{o} -- (A)

Note that we don’t need to be concerned with the signs here, as all entities are moving toward each other. If something was moving away, a negative sign would apply, but that is not relevant to this scenario.

Where,
g = Speed of sound = 340m/s.
v_{r} = Velocity of the observer relative to the medium =?.
v_{s} = Velocity of the source in relation to the medium = 0 m/s.
f_{o} =  Frequency emitted from the source = 400 Hz.
f = Frequency recognized by the observer = 408 Hz.

Substituting the given values into equation (A) will yield:

408 = ( \frac{340 + v_{r}}{340 + 0})*400

\frac{408}{400} = \frac{340 + v_{r}}{340}

Solving the above will result in,
v_{r} = 6.8 m/s

The correct result = 6.8m/s



7 0
26 days ago
A person who weighs 625 N is riding a 98-N mountain bike. Suppose that the entire weight of the rider and bike is supported equa
Keith_Richards [2256]

Answer:

A = 4.76 x 10⁻⁴ m²

Explanation:

Given data:

Person's weight = 625 N

Bike's weight = 98 N

Pressure per tire = 7.60 x 10⁵ Pa

Find: Contact area per tire

Total system weight = 625 + 98 = 723 N

Let F represent the force supported by each tire

2F = 723 N

Therefore, F = 361.5 N

Using the formula F = P × A

A = \dfrac{F}{P}

A = \dfrac{361.5}{7.60 \times 10^5}

Contact area, A = 4.76 x 10⁻⁴ m²

7 0
1 month ago
Help asap please!! An aluminum block of mass 12.00 kg is heated from 20 C to 118 C. If the specific heat of aluminum is 913 J-1
Sav [2226]
Q = mCΔT, in which Q = energy required, m = mass of the block, C = specific heat, ΔT = temperature change.

Utilizing the values provided;

Q = 12*913*(118-20) = 1073688 J = 1073.688 kJ.

The correct option is B.
7 0
1 month ago
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