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Alenkasestr [34]
4 years ago
10

ANSWER QUESTIONS WITH CORRECT ANSWERS FOR BRAINLIST! ANSWER ASAP!!

Physics
1 answer:
devlian [24]4 years ago
7 0

Answer:

So, this is not the answer because I suck like that (hehe >:3) but here's some helps.

Explanation:

1. The magnitude of the net force acting on an object is equal to the mass of the object multiplied by the acceleration of the object as shown in the formula below.

2. If the net force acting on an object is zero, then the object is not accelerating, and is in a state known as equilibrium.

3. This will be were you use the info I gave to figure it out. I hope this wasn't completely unhelpful.

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An automobile engine can produce 153 N · m of torque. Calculate the angular acceleration (in rad/s^2) produced if 85.2% of this
galina1969 [7]

Answer:

46.2 rad/s2

Explanation:

Angular acceleration works very similar to linear acceleration, it follows this equation:

\gamma = \frac{Mt}{J}

Where:

γ: angular acceleration

Mt: torque

J: moment of inertia of the load from its turning axis

Since we have the torque we just need the moment of inertia. We have to add together the moments of the drive shaft, tires, wheel walls and wheels.

The wheels act like disks. For disks the moment of inertia is:

J = \frac{1}{2} * m * r^2

Jwheel = \frac{1}{2} = 15 * 0.18^2 = 0.243 kg*m^2

The wheel walls act like annular rings, for these the moment of inertia is:

J = \frac{1}{2} * m * (re^2 - ri^2)

Jwall = \frac{1}{2} * 2 * (0.32^2 - 0.18^2) = 0.07 kg * m^2

The tread acts like a hoop, as in mass concentrated into a circunference, for these:

J = m * r^2

Jtread = 10 * 0.33^2 = 1.09 kg*m^2

The axle acts like a rod, which is the same as the disk:

Jaxle = \frac{1}{2} * 14.1 * 0.02^2 = 0.0028 kg*m^2

The drive shaft acts like a rod too:

Jshaft = \frac{1}{2} * 31.7 * 0.032^2 = 0.016 kg*m^2

SO, the total moment of inertia is:

J = 2*Jwheel + 2*Jwall + 2*Jtread + Jaxle + Jshaft

J = 2*0.243 + 2*0.07 + 2*1.09 + 0.0028 + 0.016 = 2.82 kg*m2

Finally the angular acceleration is:

\gamma = \frac{0.852 * 153}{2.82} = 46.2 \frac{rad}{s^2}

4 0
3 years ago
You have designed a machine that requires 1000 J of work from a motor for every 800 J of useful work done by the machine. What i
valentinak56 [21]

Answer:

80%

Explanation:

efficiency =   \frac{useful \: work \: done}{total \: energy \: input}

800 / 1000 = 0.8

Efficiency = 0.8 *100 = 80%

4 0
2 years ago
If the number of troughs that pass a point in a given time increases, what has increased?
Lerok [7]
Wave length i think                                               hope it helps
8 0
3 years ago
Read 2 more answers
_____ have a nearly circular orbit.
bezimeni [28]
B. Asteroids have a nearly circular orbit.
4 0
3 years ago
a man drags a 8.10 kg bag of mulch at a constant speed, applying a 29.5 N at 38°. what is the coefficient of friction?​
lesantik [10]

Answer:

The coefficient of friction is 0.38.

Explanation:

The free body diagram is drawn below.

Let f be frictional force acting in the backward direction as shown. Let the coefficient of friction be \mu. Let N be the normal reaction force acting on the bag.

Given:

Mass of the bag is, m=8.10\textrm{ kg}

Force acting at \theta = 38° is F= 29.5\textrm{ N}

Acceleration due to gravity is, g=9.8\textrm{ }m/s^{2}

The force F can be resolved into its components as F_{x}=F \cos \theta and F_{y}=F \sin \theta

Therefore,

F_{x}=29.5\cos(38)=23.25\textrm{ N}\\F_{y}=29.5\sin(38)=18.16\textrm{ N}

Now, as there is no acceleration in vertical direction, therefore,

Sum of upward forces = Sum of downward forces

N+F_{y}=mg\\N=mg-F_{y}=8.10\times 9.8-18.16\\N=79.38-18.16=61.22\textrm{ N}

Now, as the bag is moving at a constant speed, so acceleration in the horizontal direction is also zero as acceleration is the rate of change of velocity.

Therefore, backward force = forward force.

f=F_{x}\\f=23.25\textrm{ N}

Now, frictional force is given as:

f=\mu N\\\mu = \frac{f}{N}=\frac{23.25}{61.22}=0.38

Therefore, the coefficient of friction is 0.38.

8 0
3 years ago
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