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zalisa [80]
3 years ago
6

Let's first consider the dynamics behind the motion in this video. After the student releases the wheel by cutting the cord, wha

t forces are acting on the wheel
Physics
1 answer:
nydimaria [60]3 years ago
6 0

Answer:

Hello. You did not provide the video to which the question refers, however, given the context of your question, we can consider that the forces acting on the wheel are the weight of the wheel, the tension that this weight promotes on the rope and the normal force of the axle.

Explanation:

According to the context of the question, we can see that the wheel is being pulled by a rope. The weight of the wheel is the first force that acts on it, promoting resistance to the impulse that the rope promotes on the wheel. When this wheel is pulled by the rope, the weight of the wheel creates tension in the rope, which stretches and becomes tensioned in response to this force. However, this tension is also a force acting on the wheel, as is the normal force on the axis between these two elements.

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Unpolarized light with intensity S is incident on a series of polarizing sheets. The first sheet has its transmission axis orien
jeka94

Answer:

Explanation:

Given

Initial Intensity of light is S

when an un-polarized light is Passed through a Polarizer then its intensity reduced to half.

When it is passed through a second Polarizer with its transmission axis \theat =45^{\circ}

S_1=S_0\cos ^2\theta

here S_0=\frac{S}{2}

S_1=\frac{S}{2}\times \frac{1}{(\sqrt{2})^2}

S_1=\frac{S}{4}

When it is passed through third Polarizer with its axis 90^{\circ} to first but \theta =45^{\circ} to second thus S_2

S_2=S_0\cos ^2\theta

S_2=\frac{S}{4}\times \frac{1}{2}

S_2=\frac{S}{8}

When middle sheet is absent then Final Intensity will be zero                    

3 0
3 years ago
An amusement park ride consists of a car moving in a vertical circle on the end of a rigid boom of negligible mass. The combined
MrRa [10]

Incomplete question as the car's  speed is missing.I have assumed car's  speed as 6.0m/s.The complete question is here

An amusement park ride consists of a car moving in a vertical circle on the end of a rigid boom of negligible mass. The combined weight of the car and riders is 6.00 kN, and the radius of the circle is 15.0 m. At the top of the circle, (a) what is the force FB on the car from the boom (using the minus sign for downward direction) if the car's speed is v 6.0m/s

Answer:

F_{B}=-5755N

Explanation:

Set up force equation

∑F=ma

∑F=W+FB

\frac{mv^{2} }{R}=W+F_{B}\\  F_{B}=\frac{mv^{2} }{R}-W\\F_{B}=\frac{(W/g)v^{2} }{R}-W\\F_{B}=\frac{(6000N/9.8m/s^{2} )(6m/s)^{2} }{(15m)}-6000N\\F_{B}=-5755N

The minus sign for downward direction

6 0
3 years ago
A parallel circuit is constructed with three different branches. An ammeter is placed in each of the branches to measure the bra
neonofarm [45]

Answer:

Current in the battery will be 11.7 A

Explanation:

We have given that the there is parallel circuit with three different branches.

Current in each branch is I_1=2.8A, I_2=4.1A,I_2=4.8A

We have to find the current I in the battery

It is known that in parallel circuit current is sum of current in each branches

So current in the battery will be equal to

I=I_1+I_2+I_3

I=2.8+4.1+4.8=11.7A

Therefore current in the battery will be 11.7 A

8 0
3 years ago
Two technicians are discussing how a clutch disengages. Technician A says a gap on each side of the clutch disk facing when dise
victus00 [196]

Answer: A - a gap on each side of the clutch disk facing when disengaged

Explanation:

A clutch switch is used to ensure the clutch is disengaged

or Prevent the engine from starting unless the clutch pedal is depressed

When a clutch is disengageda gap will be on each side of the facing.

5 0
3 years ago
Read 2 more answers
How do you find average velocity (average) from acceleration) and time (t)?
Tasya [4]

Average velocity is defined as the ratio in change in position to change in time,

v[ave] = ∆x/∆t

which on its own doesn't have anything to do with acceleration.

<u>If acceleration is constant</u>, the average velocity is the literal average of the initial and final velocities,

v[ave] = (v[final] + v[initial]) / 2

If this constant acceleration has magnitude a, the final velocity can be expressed in terms of the initial velocity by

v[final] = v[initial] + a*t

and plugging this into the previous equation gives

v[ave] = (v[initial] + a*t + v[initial])/2

v[ave] = v[initial] + 1/2*a*t

If the body in consideration is <u>initially at rest</u>, then

v[ave] = 1/2*a*t

which might be the relation you're looking for. But bear in mind the conditions I've underlined.

<u>If acceleration is not constant and changes over time</u>, so that the acceleration is some function of time a(t), then you can determine the velocity function v(t) by using the fundamental theorem of calculus. You need to know a particular velocity for some time to completely characterize v(t), though. For example, if you're given the initial velocity v[initial] = v(0), then

\displaystyle v(t) = v(0) + \int_0^t a(u) \, du

or if you know any other velocity for some time t₀ > 0,

\displaystyle v(t) = v(t_0) + \int_{t_0}^t a(u) \, du

8 0
3 years ago
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