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almond37 [142]
3 years ago
7

A small block of mass m on a horizontal frictionless surface is attached to a horizontal spring that has force constant k. The b

lock is pushed against the spring, compressing the spring a distance d. The block is released, and it moves back and forth on the end of the spring. During its motion, what is the maximum speed of the block?
Physics
1 answer:
Rashid [163]3 years ago
5 0

Answer:

v_{max} = |d|\cdot \sqrt{\frac{k}{m} }

Explanation:

The maximum speed of the block occurs when spring has no deformation, that is, there is no elastic potential energy, which can be remarked from appropriate application of the Principle of Energy Conservation:

\frac{1}{2}\cdot k \cdot d^{2} = \frac{1}{2}\cdot m \cdot v^{2}

k \cdot d^{2} = m\cdot v^{2}

v_{max} = |d|\cdot \sqrt{\frac{k}{m} }

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Find the west component of 45 m 19º S of W
BaLLatris [955]

Answer:

The west component of the given vector is - 42.548 meters.

Explanation:

We need to translate the sentence into a vectoral expression in rectangular form, which is defined as:

(x, y) = (r_{x}, r_{y})

Where:

r_{x} - Horizontal component of vector distance, measured in meters.

r_{y} - Vertical component of vector distance, measured in meters.

Let suppose that east and north have positive signs, then we get the following expression:

(x, y) = (-45\cdot \cos 19^{\circ}, -45\cdot \sin 19^{\circ})\,[m]

(x, y) = (-42.548,-14.651)\,[m]

The west component corresponds to the first component of the ordered pair. That is to say:

x = -42.548\,m

The west component of the given vector is - 42.548 meters.

8 0
4 years ago
WHATS THE ANSWER HELP
cluponka [151]

Explanation:

I believe part of the question is missing. can you please check on it?

is there a part where the mass of the object is mentioned?

6 0
3 years ago
Read 2 more answers
How much force is required to produce a torque of 100 Nm if the force is applied at an angle of 40 at a distance of 25 cm from t
Fudgin [204]

Answer:

Force, F = 622.28 N

Explanation:

It is given that,

Torque, \tau=100\ N-m

Angle between force and displacement, \theta=40^{\circ}

Distance, d = 25 cm = 0.25 m

We need to find the force required to produce the torque. Torque produced by an object is given by :

\tau=Fd\sin\theta

F is force required to produce torque

F=\dfrac{\tau}{d\sin\theta}\\\\F=\dfrac{100}{0.25\times \sin(40)}\\\\F=622.28\ N

So, applied force is 622.28 N.

7 0
3 years ago
A ship's anchor weighs 5000N. It's cable passes over a roller of negligible mass and is wound around a hollow cylindrical drum o
deff fn [24]
Hi! Great first step would be to understand the scenario (in my opinion). So two great ways would be to draw a picture or rephrase it. If something else works, do that! You just need to "see" the situation so that you can take some away from it.

Then I think a good next step is to conceptualize everything. Put everything into a context like a physics book would. The anchor is pulled 5000N downward - that's weight. The roller will act like a pulley, and we can ignore it's properties except that it's part of a pulley system (we can ignore stuff because it has "negligible" mass and no other details are given). And then we have the hollow cylindrical drum with one radius measurement given; so we can think of this as a made-up shape with mass - a cylindrical soda can without a top or bottom (but no thickness) and a 380kg mass. The anchor is drops 16m. It hints at energy. The energy that the drum gets is all do to this anchor pulling on the rope (which is really just a means of transferring force, since we neglect its mass and get no details).

Feel free to pause here to make sure you can get the scenario in your head.

So, we want to know something about the barrel as it's rolling. The rotation rate. How many turns per some time. But don't worry yet, we can find a way to work that in. Since the rope pulls and spins the drum, the drum is spun, and gets energy. One way to find the kinetic energy of the spinning drum uses the radius, mass, and rate of rotation. More on that soon.

And how does having some equation with the drum's kinetic energy, radius, mass, and rate of rotation help? Well, we can find all of those except our rate of rotation and solve for the rate of rotation. The energy is the only mystery, but that all comes from the dropping anchor. Can we find that energy? Yeah, there's a way to find the energy that gravity gives our anchor based on it's the force and how far that force moves it.

So, first for the anchor. Linear work is simple:  W=F d
So you have your force and distance we associate with the anchor, so you have your work. We'll call that "W_1" when we need it.

Next the drum's situation. Thanks to http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html, we have the equation for kinetic energy.
Generally, we have <em></em>KE=\frac12I\omega^2, and we need the "I," which deals with rotational inertia. That is pretty much how hard it is to rotate the drum based only on the idea that your getting the mass to move (acceleration). That site refers to our hollow drum as a "hoop," and gives says that we can consider the rotational inertia to be I=MR^2. Now that we know the rotational inertia, we can use good old mathematical substitution to get the kinetic energy to look like
KE=\frac12MR^2\omega^2
And we can rearrange that to get
\omega=\sqrt{\frac{2KE}{MR^2}}=\sqrt{\frac{2KE}{M}}\cdot\frac1R

Since the energy change from the anchor's fall is the energy change of the drum, this KE is the "W_1" from before. So
\omega=\sqrt{\frac{2W_1}{M}}\cdot\frac1R=\sqrt{\frac{2\left(F d\right)}{M}}\cdot\frac1R

Now everything's set up. It's a matter of checking my work, carefully using a calculator, and making sure the answer makes sense (ie. this should be a lot of energy - much more than 1 Joule). Also, follow up by making sure you can do it again, alone. And feel free to ask or lookup questions you need along the way if there are missing pieces in your understanding.

Good luck! :)
5 0
4 years ago
This refers to the bending of a wave as it crosses a boundary between two media at an angle
Evgesh-ka [11]
The bending of a wave as it crosses a boundary between 2 mediums at an angle is called refraction.
3 0
3 years ago
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