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2 years ago

6

A stand shows a pendulum in 5 positions. A has the pendulum bob at a 45 degree angle to the left. C has the pendulum bob straigh

t down, and B is halfway between A and C. E is has the pendulum bob at a 45 degree angle to the right, and D is halfway between C and E.
Determine how potential and kinetic energy changes at each position of the pendulum as the ball swings from A to E.

Position A:

Position B:

Position C:

Position D:

Position E:

2 answers:

bazaltina [42]2 years ago

7 0

Answer:

position A : is all potential

position B : is losing potential and gaining kinetic energy

position c : is all kinetic

position D : is losing kinetic and gaining potential

position E : is all potential

kotegsom [21]2 years ago

6 0

Answer:

plz brainliest

Explanation:

position A : is all potential

position B : is losing potential and gaining kinetic energy

position c : is all kinetic

position D : is losing kinetic and gaining potential

position E : is all potential

thx have a great day

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During a compaction test in the lab a cylindrical mold with a diameter of 4in and a height of 4.58in was filled. The compacted s

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Answer:

part a : <em>The dry unit weight is 0.0616 </em> $lb/in^3$ <em />

part b : <em>The void ratio is 0.77</em>

part c : <em>Degree of Saturation is 0.43</em>

part d : <em>Additional water (in lb) needed to achieve 100% saturation in the soil sample is 0.72 lb</em>

Explanation:

Part a

Dry Unit Weight

The dry unit weight is given as

$\gamma_{d}=\frac{\gamma}{1+\frac{w}{100}}$

Here

$\gamma_d$ is the dry unit weight which is to be calculated
γ is the bulk unit weight given as

$\gamma =weight/Volume \\\gamma= 4 lb / \pi r^2 h\\\gamma= 4 lb / \pi (4/2)^2 \times 4.58\\\gamma= 4 lb / 57.55\\\gamma= 0.069 lb/in^3$

w is the moisture content in percentage, given as 12%

Substituting values

$\gamma_{d}=\frac{\gamma}{1+\frac{w}{100}}\\\gamma_{d}=\frac{0.069}{1+\frac{12}{100}} \\\gamma_{d}=\frac{0.069}{1.12}\\\gamma_{d}=0.0616 lb/in^3$

<em>The dry unit weight is 0.0616 </em> $lb/in^3$ <em />

Part b

Void Ratio

The void ratio is given as

$e=\frac{G_s \gamma_w}{\gamma_d} -1$

Here

e is the void ratio which is to be calculated

$\gamma_d$ is the dry unit weight which is calculated in part a
$\gamma_w$ is the water unit weight which is 62.4 $lb/ft^3$ or 0.04 $lb/in^3$

G is the specific gravity which is given as 2.72

Substituting values

$e=\frac{G_s \gamma_w}{\gamma_d} -1\\e=\frac{2.72 \times 0.04}{0.0616} -1\\e=1.766 -1\\e=0.766$

<em>The void ratio is 0.77</em>

Part c

Degree of Saturation

Degree of Saturation is given as

$S=\frac{G w}{e}$

Here

e is the void ratio which is calculated in part b

G is the specific gravity which is given as 2.72

w is the moisture content in percentage, given as 12% or 0.12 in fraction

Substituting values

$S=\frac{G w}{e}\\S=\frac{2.72 \times .12}{0.766}\\S=0.4261$

<em>Degree of Saturation is 0.43</em>

Part d

Additional Water needed

For this firstly the zero air unit weight with 100% Saturation is calculated and the value is further manipulated accordingly. Zero air unit weight is given as

$\gamma_{zav}=\frac{\gamma_w}{w+\frac{1}{G}}$

Here

$\gamma_{zav}$ is the zero air unit weight which is to be calculated

$\gamma_w$ is the water unit weight which is 62.4 $lb/ft^3$ or 0.04 $lb/in^3$

G is the specific gravity which is given as 2.72

w is the moisture content in percentage, given as 12% or 0.12 in fraction

$\gamma_{zav}=\frac{\gamma_w}{w+\frac{1}{G}}\\\gamma_{zav}=\frac{0.04}{0.12+\frac{1}{2.72}}\\\gamma_{zav}=\frac{0.04}{0.4876}\\\gamma_{zav}=0.08202 lb/in^3\\$

Now as the volume is known, the the overall weight is given as

$weight=\gamma_{zav} \times V\\weight=0.08202 \times 57.55\\weight=4.72 lb$

As weight of initial bulk is already given as 4 lb so additional water required is 0.72 lb.

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A plane initially traveling at 200 m/s due west experiences a 10 m/s head wind coming from the opposite direction. A). What will

Hitman42 [59]

Ok so it would be late and the relative velocity would be 190 m/s because 200 m/s - 10 m/s is 190 m/s. Hope this helps.

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2 years ago

A beam of microwaves with λ = 0.9 mm is incident upon a 9 cm slit. At a distance of 1.5 m from the slit, what is the approximate

liq [111]

Answer:

3 cm

Explanation:

According to the question,

$D=1.5 m$ .

$d=9 cm$ .

$\lambda =0.9 mm$ .

Now the approximate slit's image width is equal to width of central maxima.

And width of central maxima is twice the width from center to first maxima

So,

$y=2\frac{\lambda (D)}{d}$ .

Substitute all the variable in above equation.

$y=\frac{(2)0.9\times 10^{-2} m(1.5 m) }{0.09 m}$ .

$y=3 cm$ .

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Plug variables expressed in SI units in the kinematic equation given in article: a = -v0^2/(2sg). What value of g you get as exp

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Answer:

1) acceleration is increased by a factor of four 4X

2) the acceleration increases a factor of 2X

3) the correct answer of 400g

Explanation:

This is a kinematics exercise, where you use the velocity equation to obtain the acceleration, with the final velocity equal to zero.

v² = v₀² + 2 a x

0 = v₀² + 2 a x

a = - v₀² / 2 x

In the case of wanting to give the acceleration as a function of g, we can find the relationship between the two quantities

a / g = - v₀² / (2 x g)

Let's answer the different questions about this equation

1. The initial velocity is doubled, how much the acceleration is worth

a/g = - (2v₀) 2 / 2xg

a = 4 (-v₀² / 2xg) g

acceleration is increased by a factor of four 4X

2. if the stopping distance is reduced by 2, that is, x = x₀ / 2

we substitute

a/g = (- v₀² / 2g) 2/x

a =2 (-v₀² / 2x₀g) g

therefore the acceleration increases a factor of 2X

3. the initial velocity of the hockey player is v₀ = 20 m / s and the stopping distance is

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we calculate the acceleration

a / g = - 20² / (2 0.05)

a / g = - 4000 / g

a / g = - 4000 / 9.8 = 408

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2 years ago

A wheel rotates without friction about a stationary horizontal axis at the center of the wheel. A constant tangential force equa

love history [14]

Answer:

I = 16 kg*m²

Explanation:

Newton's second law for rotation

τ = I * α Formula (1)

where:

τ : It is the moment applied to the body. (Nxm)

I : it is the moment of inertia of the body with respect to the axis of rotation (kg*m²)

α : It is angular acceleration. (rad/s²)

Kinematics of the wheel

Equation of circular motion uniformly accelerated :

ωf = ω₀+ α*t Formula (2)

Where:

α : Angular acceleration (rad/s²)

ω₀ : Initial angular speed ( rad/s)

ωf : Final angular speed ( rad

t : time interval (rad)

Data

ω₀ = 0

ωf = 1.2 rad/s

t = 2 s

Angular acceleration of the wheel

We replace data in the formula (2):

ωf = ω₀+ α*t

1.2= 0+ α*(2)

α*(2) = 1.2

α = 1.2 / 2

α = 0.6 rad/s²

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τ = F *R

Where:

F = tangential force (N)

R = radio (m)

τ = 80 N *0.12 m

τ = 9.6 N *m

Rotational inertia of the wheel

We replace data in the formula (1):

τ = I * α

9.6 = I *(0.6 )

I = 9.6 / (0.6 )

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1 year ago