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2 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]2 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

Solve for the length of the inclined plane if the angle equals 19.45 degrees.

mel-nik [20]

The length of the inclined plane is approximately 12 ft

The situation forms a right angle triangle.

<h3>Right triangle</h3>

Right triangle have one of its angle as 90 degrees.

Therefore,

The length of the inclined plane is the hypotenuse of the triangle. The length of the inclined plane can be found using trigonometric ratios.

height = 4 ft

angle(∅) = 19.45°

sin 19.45 = 4 / h

h = 4 / 0.33298412235

h = 12.0125847796

h = 12 ft

Therefore, the length of the inclined plane is approximately 12 ft

learn more on inclined plane:brainly.com/question/14163589?referrer=searchResults

5 0

2 years ago

a ball is thrown inclined in to the air with a initial velocity of u. if it reaches the maximum height in 6 seconds , find the r

olga55 [171]

Answer:

11:1

Explanation:

At constant acceleration, an object's position is:

y = y₀ + v₀ t + ½ at²

Given y₀ = 0, v₀ = u, and a = -g:

y = u t − ½g t²

After 6 seconds, the ball reaches the maximum height (v = 0).

v = at + v₀

0 = (-g)(6) + u

u = 6g

Substituting:

y = 6g t − ½g t²

The displacement between t=0 and t=1 is:

Δy = [ 6g (1) − ½g (1)² ] − [ 6g (0) − ½g (0)² ]

Δy = 6g − ½g

Δy = 5½g

The displacement between t=6 and t=7 is:

Δy = [ 6g (7) − ½g (7)² ] − [ 6g (6) − ½g (6)² ]

Δy = (42g − 24½g) − (36g − 18g)

Δy = 17½g − 18g

Δy = -½g

So the ratio of the distances traveled is:

(5½g) / (½g)

11 / 1

The ratio is 11:1.

8 0

3 years ago

A spherical shell with a net charge of 3Q surrounds a point charge of -q at the center of the shell. The charges on the inner an

aleksley [76]

Answer:

1) The charge on the outer shell is +4·Q

2) The charge on the inner shell is +Q

Explanation:

1) The given parameters of the spherical shell are;

The net charge on the spherical shell = 3·Q

The point charge surrounded by the spherical shell = -Q

Let 'x' represent the charge on the outer shell, and let 'y', represent the charge on the inner shell, we have;

The net charge, 3·Q = -q + x

∴ x = 3·Q + Q = 4·Q

The charge on the outer shell, x = 4·Q

2) The net charge in the shell is zero, therefore, the charge on the inner shell, 'y', is given as follows;

-Q + y = 0

∴ y = +Q

The charge on the inner shell, y = +Q

5 0

3 years ago

An oil gusher shoots crude oil 25.0 m into the air through a pipe with a 0.100-m diameter. Neglecting air resistance but not the

Mariana [72]

Answer:3764.282 KPa

Explanation:

Given gusher shoots oil at h=25 m

i.e. the velocity of jet is

v=\sqrt{2gh}[/tex]

v=22.147 m/s

Now the pressure loss in pipe is given by hagen poiseuille equation

$\Delta P=\frac{32L\mu v}{D^2}$

$\Delta P=\frac{32\times 50\times 22.147\times 1}{10^{-2}}$

$\Delta P=3543.557 KPa$

For 25 m head in terms of Pressure

$\Delta P_2=\rho \times g\times h=220.725 KPa$

Total Pressure= $\Delta P+\Delta P_2$ =3543.557+220.725=3764.282 KPa

4 0

3 years ago

The flywheel of a steam engine runs with a constant angular velocity of 140 rev/min. When steam is shut off, the friction of the

ratelena [41]

Answer:

A) α = -1.228 rev/min²

B) 7980 revolutions

C) α_t = -8.57 x 10^(-4) m/s²

D) α = 21.5 m/s²

Explanation:

A) Using first equation of motion, we have;

ω = ω_o + αt

Where,

ω_o is initial angular velocity

α is angular acceleration

t is time the flywheel take to slow down to rest.

We are given, ω_o = 140 rev/min ; t = 1.9 hours = 1.9 x 60 seconds = 114 s ; ω = 0 rev/min

Thus,

0 = 140 + 114α

α = -140/114

α = -1.228 rev/min²

B) the number of revolutions would be given by the equation of motion;

S = (ω_o)t + (1/2)αt²

S = 140(114) - (1/2)(1.228)(114)²

S ≈ 7980 revolutions

C) we want to find tangential component of the velocity with r = 40cm = 0.4m

We will need to convert the angular acceleration to rad/s²

Thus,

α = -1.228 x (2π/60²) = - 0.0021433 rad/s²

Now, formula for tangential acceleration is;

α_t = α x r

α_t = - 0.0021433 x 0.4

α_t = -8.57 x 10^(-4) m/s²

D) we are told that the angular velocity is now 70 rev/min.

Let's convert it to rad/s;

ω = 70 x (2π/60) = 7.33 rad/s

So, radial angular acceleration is;

α_r = ω²r = 7.33² x 0.4

α_r = 21.49 m/s²

Thus, magnitude of total linear acceleration is;

α = √((α_t)² + (α_r)²)

α = √((-8.57 x 10^(-4))² + (21.49)²)

α = √461.82

α = 21.5 m/s²

4 0

3 years ago