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3 years ago

What's needed to know

2 answers:

8 0

You're a little late. But if you want some short, quick rules, then these are
a couple that I would take in with me (stored only in my brain, of course):

-- If something is not accelerating or moving at all, then all the forces on it
must add up to zero. That could even mean a hanging rope.

-- In a vertical rope, the tension in it is the same everywhere in the rope.
The tension is the weight of whatever is hanging from the bottom.

That's really all I'm sure of, based on your hazy, fuzzy description of
what you've been doing in class. I don't want to get into things that
you might not have learned yet, and confuse you.

Orlov [11]3 years ago

5 0

You're absolute best bet is to read and re-read the entire chapter one of the book. This is for an absolute best strategy. On the other hand, there are usually key words and terms that are highlighted that most teachers use as a frame for test taking. Studying those highlighted words would be the next best bet. If there are no highlighted words, tough luck. You need to figure out what is important. Which is part of the learning process, and also why I won't directly answer your question. I care more about you learning than you getting a good grade.

You might be interested in

Which of the following statements about iron filings placed upon glass resting on top of a bar magnet is false?

marin [14]

Statements A, B, and C are true.

Statement <em> D is false</em>.

7 0

4 years ago

A 26.0 kg beam is attached to a wall with a hinge while its far end is supported by a cable such that the beam is horizontal. If

Nuetrik [128]

Answer:

Force exerted by the hinge on the beam = 109.24N

Explanation:

Weight = mg = 26 x 9.81 = 255.06 N

Vertical component = T sin θ

Horizontal component = Tcos θ

Now, there are 3 vertical forces acting on the beam. These are;

- The downward force which is the weight of the beam.

- The vertical components of the tension in the cable.

-The force that hinge exerts on the beam are the upward forces.

Hence, for the beam to remain horizontal, the sum of the upward forces must be equal to the weight of the beam.

For us to determine the vertical component of the tension in the cable, we will do a torque problem. Let the pivot point be at the hinge. Let’s assume that the length of the beam is L. The vertical component of the tension in the cable will produce clockwise torque while the weight of the beam will produce counter clockwise torque.

Tbus;

Clockwise torque = TL sin 61

Since the center of mass of beam is at the middle of the beam, the distance from the hinge to the weight of the beam is L/2.

Counter clockwise torque = WL/2

Thus;

TL sin 61 = WL/2

L will cancel out.

T sin 61 = 255.06/2

T x 0.8746 = 127.53

T = 127.53/0.8746 = 145.82 N

Now, the equation to determine the vertical component of the force that the hinge exerts on the beam is given as;

T + F = W

Thus;

145.82 + F = 255.06

F = 255.06 - 145.82 = 109.24 N

8 0

3 years ago

Earth's atmosphere is made up of mostly nitrogen (78%) and oxygen (21%); however, there are other gases in smaller amounts, like

pishuonlain [190]

Answer:

the temperature would increase globally because carbon dioxde contributes to global warming increase the temperature of the earth

Explanation:

i hope this will help you

5 0

3 years ago

Read 2 more answers

Which is an example of a mixture made of parts that can easily be separated?

Gemiola [76]

Answer

sand and water is an example of a mixture made of parts that can easily be separated

Explatnation

mixture is a material which is formed by mixing two or more different types of materials. No chemical reaction occurs.During combination of two or more type of material to form a mixture.

Example of mixture are

oil and water

sand and water

8 0

3 years ago

A light platform is suspended from the ceiling by a spring. A student with a mass of 90 kg climbs onto the platform. When it sto

Ilya [14]

Refer to the diagram shown.

When the student climbs onto the platform, the spring stretches by 0.82 m to reach the equilibrium position.
The mass of the student is m = 90 kg, so his weight is
mg = (90 kg)*(9.8 m/s²) = 882 N

By definition, the spring constant is
k = (882 N)/(0.82 m) = 1075.6 N/m

When the spring is stretched by x from the equilibrium position, the restoring force is
F = - k*x.

If damping is ignored, the equation of motion is
F = m * acceleration
or
$m \frac{d^{2}x}{dt^{2}} = -kx \\ \frac{d^{2}x}{dt^{2}} + \frac{k}{m} x = 0$

Define ω² = k/m = 11.751 => ω = 3.457.
Then the solution of the ODE is
x(t) = c₁ cos(ωt) + c₂ sin(ωt)

x'(t) = -c₁ω sin(ωwt) + c₂ω cos(ωt)
When t=0, x' =0, therefore c₂ = 0

The solution is of the form
x(t) = c₁ cos(ωt)
When t = 0, x = 0.32 m. Therefore c₁ = 0.32

The motion is
x(t) = 0.32 cos(3.457t)
The single amplitude is 0.32 m, and the double amplitude is 0.64 m.

Answer:
0.32 m (single amplitude), or
0.64 m (double amplitude)

6 0

3 years ago