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suter
2 months ago
11

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

the loop in order to remain in contact with the loop
Physics
1 answer:
ValentinkaMS [3.4K]2 months ago
4 0
<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>
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A parachutist, after opening her parachute, finds herself gently floating downward, no longer gaining speed. She feels the upwar
Sav [3153]

As the parachutist is descending at a steady rate

we can conclude that

a = \frac{dv}{dt}

Acceleration indicates the change in velocity

given the constant velocity in this scenario

a = 0

Thus, in this situation, we find the acceleration to be zero

It’s understood from Newton's second law

F_{net} = ma

where a is equal to 0

F_{net} = 0

F_{net} = F_g - F_b

Here, the force due to gravity

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F_gHence, we can deduce

F_b

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1 month ago
When a vertical beam of light passes through a transparent medium, the rate at which its intensity I decreases is proportional t
Keith_Richards [3271]

Response:

The intensity of light 18 feet underwater is about 0.02%

Clarification:

Employing Lambert's law

Let dI / dt = kI, where k is a proportionality factor, I represents the intensity of incident light, and t indicates the thickness of the medium

Then dI / I = kdt

Taking logarithms,

ln(I) = kt + ln C

I = Ce^kt

At t=0, I=I(0) implies C=I(0)

I = I(0)e^kt

At t=3 & I=0.25I(0), we find 0.25=e^3k

Solving for k gives k = ln(0.25)/3

k = -1.386/3

k = -0.4621

I = I(0)e^(-0.4621t)

I(18) = I(0)e^(-0.4621*18)

I(18) = 0.00024413I(0)

The intensity of light 18 feet underwater is about 0.2%

3 0
2 months ago
A proton is accelerated from rest through a potential difference V0 and gains a speed v0. If it were accelerated instead through
inna [3103]
Answer: The resulting speed is \sqrt{2}v_{0}. Option (a) stands as the correct choice. Explanation: Given the context, the potential difference entails calculations linked to speed assessment. If instead accelerated through a different potential difference, the resulting speed will be computed accordingly.
6 0
1 month ago
A camera operator is filming a nature explorer in the Rocky Mountains. The explorer needs to swim across a river to his campsite
inna [3103]

Answer:

a. Angle= 28.82°

b. Approved. Although he might feel cold, he should be able to cross.

Explanation:

Velocity Vector

Velocity is a measure of how quickly something is moving in a specific direction. It is represented as a vector that has both magnitude and direction. If an object can only move in one direction, then speed can serve as the scalar equivalent of that velocity (only focusing on magnitude).

a.

The explorer aims to swim across a river to reach his campsite, as depicted in the image below. The river's velocity is vr and the explorer's swimming speed in still water is ve. If he were to swim straight towards the campsite, he would end up downstream due to the river's current. Therefore, he must swim at an angle that allows him to overcome the current while still moving towards his goal. This angle relative to the shore is what we need to determine. The explorer's speed can be broken down into its horizontal (vx) and vertical (vy) components. In order to counteract the river's flow:

v_{ey}=v_r

We can calculate the vertical component of the explorer's swimming speed as

v_{ey}=|v_e|cos\alpha

Thus

v_r=|v_e|cos\alpha

Finding the value of \alpha

\displaystyle cos\alpha=\frac{v_r}{|v_e|}

\displaystyle cos\alpha=\frac{0.665}{0.759}=0.876

Then the angle is given by

\alpha=28.82^o

b.

The component of the explorer's velocity that goes horizontally is

v_{ex}=0.759sin28.82^o

v_{ex}=0.366\ m/s

This represents the actual velocity directed towards the campsite

Considering that

\displaystyle v=\frac{x}{t}

To find t

\displaystyle t=\frac{x}{v}

Calculating the duration for the explorer to cross the river

\displaystyle t=\frac{29.3}{0.366}

t=80\ sec

As this time is under the hypothermia threshold (300 seconds), the conclusion is

Approved. Although he will feel cold, he should manage to cross successfully.

3 0
2 months ago
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