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DENIUS [597]
2 years ago
12

Which of the following is a strength training option?

Physics
2 answers:
Vilka [71]2 years ago
5 0

Answer:

D. All of the above

PLZ MARK ME AS BRAINLEIST ;)

polet [3.4K]2 years ago
3 0

Answer:

D All of the above

Explanation:

Check this site to understand why they all are strenght trainings. issaonline: weight machines vs free weights.

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Whenever a musician plays a guitar, they pluck one of the guitar strings to produce a standing wave in the string. Then pinch do
UkoKoshka [18]

Answer:

It's due to the distance from either ends of strings origin...

Explanation:

As we know that waves behave moving in a flow from one side to another side and this gives a prospective of motion. Suppose a wave is pinched from the near one end of a guitar then due to the distortion created by the point of tie of strings the wave super imposes and moves with a velocity v and produces a wave frequency f. as we the pinching go down to the center the wave stabilizes itself to a stationary origin right at the center and the frequency then changes accordingly as moving down on the string.

7 0
2 years ago
A plank of length L=2.200 m and mass M=4.00 kg is suspended horizontally by a thin cable at one end and to a pivot on a wall at
Elenna [48]

The tension in the cable is 23.2 N

<h3>What is the tension in the string?</h3>

The tension in the cable can be resolved into horizontal and vertical forces Tcosθ and Tsinθ respectively.

Tcosθ, is acting perpendicularly, Tcosθ = 0

Taking moments about the pivot:

Tsinθ * 2.2 = 4 * 9.8 * 0.7

Solving for θ;

θ = tan⁻¹(1.4/2.2) = 32.5°

T = 27.44/(sin 32.5 * 2.2)

T = 23.2 N

In conclusion, the tension in the cable is determined by taking moments about the pivot.

Learn more about moments of forces at: brainly.com/question/23826701

#SPJ1

3 0
2 years ago
Sugar is considered a crystalline solid because
Nina [5.8K]

Answer:

Im sure the awnser is option

A. arranged in a regular pattern.

8 0
2 years ago
A bug flying horizontally at 0.65 m/s collides and sticks to the end of a uniform stick hanging vertically. After the impact, th
irina [24]

The angular momentum is defined as,

L=I\omega

Acording to this text we know for conservation of angular momentum that

L_i=L_f

Where L_iis initial momentum

L_f is the final momentum

How there is a difference between the stick mass and the bug mass, we define that

Mass of the bug= m

Mass of the stick=10m

At the point 0 we have that,

L_i=mvl

Where l is the lenght of the stick which is also the perpendicular distance of the bug's velocity

vector from the point of reference (O), and ve is the velocity

At the end with the collition we have

L_f=(I_b+I_s)\omega

Substituting

L_f=(ml^2+\frac{10ml^2}{3})\omega

L_f=\frac{13}{10}ml^2w

m(0.65)l=\frac{13}{10}ml^2 \omega

\omega=\frac{1}{2l}

Applying conservative energy equation we have

\frac{1}{2}(I_b+I_s)\omega^2=mgh+10mgh'

\frac{1}{2}(ml^2+\frac{10ml^2}{3})(\frac{1}{2l})^2=mg(l-lcos\theta)+\frac{10}{2}mg(l-lcos\theta)

Replacing the values and solving

l=\frac{13}{0.54g}

Substituting

l=\frac{13}{0.54(9.8)}

l=2.45cm

7 0
3 years ago
An object is dropped from a platform 100 feet high. Ignoring wind resistance, what will its speed be when it reaches the ground?
Scorpion4ik [409]

Answer:

80 ft/s

Explanation:

Use III equation of motion

V^2 = U^2 + 2g h

Here, U = 0, g = 32 ft/s^2, h = 100 ft

V^2 = 0 + 2 × 32 ×100

V^2 = 6400

V = 80 ft/s

8 0
3 years ago
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