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3 years ago

When the IMA is compared to AMA in the real world, _____.

2 answers:

5 0

I believe the answer is A.) the value of the IMA is always larger than the AMA
(Not 100% on this, not my strongest topic, so I would double check)

Korvikt [17]3 years ago

4 0

AMA is actual mechanical advantage in where it would compare the force in as well as out in using actual measurements in the real life or with friction while in IMA which is the ideal mechanical advantage no friction then is needed. The answer then is a. the value of IMA is larger than AMA.

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Pendulum clocks are made to run at the correct rate by adjusting the pendulum’s length. Suppose you move from one city to anothe

levacccp [35]

Answer:

Obviously Lengthen... $T = 2\pi \sqrt{L/g}$ or $g = 4\pi ^{2} L/g$

Explanation:

As we can observe from the equation, time period of a simple pendulum depends upon the length directly. When the gravitational acceleration increases the time period of the pendulum decreases and vice versa. So, by increasing the length, the time period can be adjusted...

4 0

3 years ago

Two protons are aimed directly toward each other by a cyclotron accelerator with speeds of 1200 km/s , measured relative to the

Radda [10]

Answer:

The maximum electrical force is $2.512\times10^{-2}\ N$ .

Explanation:

Given that,

Speed of cyclotron = 1200 km/s

Initially the two protons are having kinetic energy given by

$\dfrac{1}{2}mv^2=\dfrac{1}{2}mv^2$

When they come to the closest distance the total kinetic energy is converts into potential energy given by

Using conservation of energy

$mv^2=\dfrac{kq^2}{r}$

$r=\dfrac{kq^2}{mv^2}$

Put the value into the formula

$r=\dfrac{8.99\times10^{9}\times(1.6\times10^{-19})^2}{1.67\times10^{-27}\times(1200\times10^{3})^2}$

$r=9.57\times10^{-14}\ m$

We need to calculate the maximum electrical force

Using formula of force

$F=\dfrac{kq^2}{r^2}$

$F=\dfrac{8.99\times10^{9}\times(1.6\times10^{-19})^2}{(9.57\times10^{-14})^2}$

$F=2.512\times10^{-2}\ N$

Hence, The maximum electrical force is $2.512\times10^{-2}\ N$ .

8 0

3 years ago

Suppose that a ball is released from the window of a train that is moving with constant velocity. The path of the ball, as obse

barxatty [35]

True, the path of the ball, as observed from the train window, will be a horizontal straight line.

An object projected from a certain height has a parabolic path when observed from a fixed point.

However, if the reference point is moving at the same velocity as the object, the path of the object's motion appears to be a straight line.

When the ball is released from the window of the train, it will move at the same constant velocity as the train, and the path of the ball's motion observed from the train window will be a straight line.

Thus, we can conclude that the given statement is true. The path of the ball, as observed from the train window, will be a horizontal straight line.

Learn more about path of motion of objects here: brainly.com/question/82610

7 0

2 years ago

7. Two people are pushing a 40.0kg table across the floor. Person 1 pushes with a force of 490N

artcher [175]

Answer:

20.4 $m/s^{2}$

Explanation:

To start doing this problem, first draw a free body diagram of the table. My teacher always tells us to do this, and I find that it is very helpful. I have attached a free body diagram to this answer- take a look at it.

First, let us see if Net force = MA. To do that, we need to determine whether the object is at equilibrium horizontally. For an object to be at equilibrium, it either needs to be moving at a constant velocity or not moving at all. Also, if an object is at equilibrium, there will not be any acceleration. But we know that there IS acceleration horizontally, so it cannot be in equilibrium. If it is not in equilibrium, we can use the formula ∑F= ma.

Let us determine the net force. Since the object is moving horizontally, we can ignore the weight and normal force, because they are vertical forces. The only horizontal forces we need to worry about are the applied force and force of friction.

Applied force = 1055 N (490 + 565)

Friction force= Unknown

To find the friction force, use the kinetic friction formula, Friction = μkN

μk is the coefficient, which the problem includes- it is 0.613.

N is the normal force, which we have to find.

*To find the normal force, we have to determine if the object is at equilibrium VERTICALLY. Since it has no acceleration vertically (it's not moving up/down), it is at equilibrium. Now, when an object is at equilibrium in one direction, it means that all the forces in that direction are equal. What are our vertical forces? Weight (mg) and Normal force (N). So it means that the Normal force is equal to the Weight.

Weight = mg = (40)(9.8) = 392 N

Normal force = 392 N

Now, plug it back into the formula (μkN): (0.613)(392) = 240.296 N

Friction = 240.296 N

Now that we know the friction, we can find the horizontal net force. Just subtract the friction force, 240.296 from the applied force, 1055 N

Horizontal Net Force: 814.704 N

Now that we know the net force, plug in the numbers for the formula

∑F= ma.

814.704 = (40.0)(a)

*Divide on both sides)

a = 20.3676 m/s^2

Round it to 3 significant figures, to get:

20.4 $m/s^{2}$

7 0

3 years ago

a car whose mass is 1000kg is traveling at a constant speed of 10m/s. Neglecting any friction how much force will the engine hav

AURORKA [14]

This next statement is a big deal. It should be up on a board, surrounded
by flashing red and yellow lights, and hung on the wall of every Science
classroom. Although we never see it in our daily lives, it's fundamental to
the workings of the universe, and it's also Newton's first law of motion:

Without friction, it doesn't take ANY force to keep a moving object
moving. Force is only required to change the object's speed, or to
change the direction in which it's moving.

The answer to the question is: On a level road, and neglecting any friction,
the engine doesn't have to supply ANY force to keep the car going at the
same speed.

7 0

3 years ago

Read 2 more answers