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mars1129 [50]
3 years ago
3

Two identical masses are connected to two different flywheels that are initially stationary. Flywheel A is larger and has more m

ass, but has hexagonal sections where material has been removed. The attached masses are released from rest and allowed to fall a height h.
Which of the following statements about their angular accelerations is true?

A. The angular acceleration of the two flywheels is different but it is impossible to tell which is greater.
B. The angular acceleration of flywheel A is greater The angular acceleration of flywheel B is greater.
C. Not enough information is provided to determine.
D. The angular accelerations of the two flywheels are equal.
Physics
2 answers:
Lesechka [4]3 years ago
8 0

Answer: C - not enough information is provided.

Explanation:

The flywheel consists of a heavy circular wheel fitted with an axle projecting on either side.The axle is mounted on ball bearings on two fixed supports.

When a torque is applied to body the angular acceleration α is given by the ratio of the torque and moment of inertia.

The angular acceleration depend not only on the torque τ but also on the moment of inertia I of the body about the given axis which is determined by all the below parameters:

 I = Moment of inertia of the flywheel assembly

             N = Number of rotation of the flywheel before it stopped

             m = mass of the rings

             n = Number of windings of the string on the axle

             g = Acceleration due to gravity of the environment.

             h = Height of the weight assembly from the ground.

             r = Radius of the axle.

All these parameters are not given. Therefore, enough information is not provided to determine the angular acceleration.

GalinKa [24]3 years ago
7 0

Answer:C

Explanation:

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0.8 mj of energy are used by a blender in 3 and a half minutes what is the power rating on the blender
solong [7]

Answer:

   Power rating on the blender = 3809.52 Watts

Explanation:

   We have expression for power equal to ratio of work and time,

   Energy used by blender = Work done by electricity = 0.8 MJ = 0.8*10^6J=800000J

   Time of using blender = 3.5 minutes = 210 seconds

   So power of blender = 800000/210 = 3809.52 Watts

   Power rating on the blender = 3809.52 Watts

5 0
4 years ago
Effciency of a lever is never 100% or more. why?Give reason​
Troyanec [42]

Answer:

Ideally, the work output of a lever should match the work input. However, because of resistance, the output power is nearly always be less than the input power. As a result, the efficiency would go below 100\%.  

Explanation:

In an ideal lever, the size of the input and output are inversely proportional to the distances between these two forces and the fulcrum. Let D_\text{in} and D_\text{out} denote these two distances, and let F_\text{in} and F_\text{out} denote the input and the output forces. If the lever is indeed idea, then:

F_\text{in} \cdot D_\text{in} = F_\text{out} \cdot D_\text{out}.

Rearrange to obtain:

\displaystyle F_\text{in} = F_\text{out} \cdot \frac{D_\text{out}}{D_\text{in}}

Class two levers are levers where the perpendicular distance between the fulcrum and the input is greater than that between the fulcrum and the output. For this ideal lever, that means D_\text{in} > D_\text{out}, such that F_\text{in} < F_\text{out}.

Despite F_\text{in} < F_\text{out}, the amount of work required will stay the same. Let s_\text{out} denote the required linear displacement for the output force. At a distance of D_\text{out} from the fulcrum, the angular displacement of the output force would be \displaystyle \frac{s_\text{out}}{D_\text{out}}. Let s_\text{in} denote the corresponding linear displacement required for the input force. Similarly, the angular displacement of the input force would be \displaystyle \frac{s_\text{in}}{D_\text{in}}. Because both the input and the output are on the same lever, their angular displacement should be the same:

\displaystyle \frac{s_\text{in}}{D_\text{in}} =\frac{s_\text{out}}{D_\text{out}}.

Rearrange to obtain:

\displaystyle s_\text{in}=s_\text{out} \cdot \frac{D_\text{in}}{D_\text{out}}.

While increasing D_\text{in} reduce the size of the input force F_\text{in}, doing so would also increase the linear distance of the input force s_\text{in}. In other words, F_\text{in} will have to move across a longer linear distance in order to move F_\text{out} by the same s_\text{out}.

The amount of work required depends on both the size of the force and the distance traveled. Let W_\text{in} and W_\text{out} denote the input and output work. For this ideal lever:

\begin{aligned}W_\text{in} &= F_\text{in} \cdot s_\text{in} \\ &= \left(F_\text{out} \cdot \frac{D_\text{out}}{D_\text{in}}\right) \cdot \left(s_\text{out} \cdot \frac{D_\text{in}}{D_\text{out}}\right) \\ &= F_\text{out} \cdot s_\text{out} = W_\text{out}\end{aligned}.

In other words, the work input of the ideal lever is equal to the work output.

The efficiency of a machine can be measured as the percentage of work input that is converted to useful output. For this ideal lever, that ratio would be 100\%- not anything higher than that.

On the other hand, non-ideal levers take in more work than they give out. The reason is that because of resistance, F_\text{in} would be larger than ideal:

\displaystyle F_\text{in} = F_\text{out} \cdot \frac{D_\text{out}}{D_\text{in}} + F(\text{resistance}).

As a result, in real (i.e., non-ideal) levers, the work input will exceed the useful work output. The efficiency will go below 100\%,

4 0
3 years ago
16. A 95kg fullback, running at 8.2m/s, collided in midair with a 128 kg defensive tackle moving in the opposite direction. Both
Daniel [21]

a) 779 kg m/s

The momentum of an object is given by:

p = mv

where

m is the mass of the object

v is its velocity

For the fullback before the collision,

m = 95 kg

v = 8.2 m/s

Therefore, his momentum was:

p=mv=(95)(8.2)=779 kg m/s

b) -779 kg m/s

After the collision, both the fullback and the tackle come to a stop: this means that their momentum after the collision is zero,

p' = 0

The initial momentum of the fullback was

p = 779 kg m/s

Therefore, his change in momentum is

\Delta p = p' -p =0-779  = -779 kg m/s

where the negative sign indicates that the direction is opposite to the initial direction of motion.

c) -779 kg m/s

Here we can apply the law of conservation of momentum. In fact, the total momentum before and after the collision must be conserved. So we can write:

p_f + p_t = p'

where

p_f is the initial momentum of the fullback

p_t is the initial momentum of the tackle

p' is the final combined momentum after the collision

We already know that

p_f = 779 kg m/s\\p' = 0

Therefore, we can find the tackle's original momentum:

p_t = p'-p_f = 0-(779) = -779 kg m/s

where the negative sign indicates that the direction is opposite to the initial direction of motion of the fullback.

e) -6.1 m/s

To find the velocity of the tackle, we can use again the equation of the momentum:

p = mv

where here we have

p=-779 kg m/s is the original momentum of the tackle

m = 128 kg is his mass

Solving the equation for v, we find the tackle's original velocity:

v=\frac{p}{m}=\frac{-779}{128}=-6.1 m/s

So, he was moving at 6.1 m/s in the direction opposite to the fullback.

4 0
4 years ago
A Fourier analysis of the instantaneous value of a waveform can be represented by:
Viefleur [7K]

Answer:

t = 1

Explanation:

you dont know the value of the variable

7 0
3 years ago
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What human body feature allows the respiratory system to take in oxygen
alexandr1967 [171]

The answer to your question is: the lungs.

The lungs allow the respiratory system to take in oxygen.


(Hope this helps!)

8 0
3 years ago
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