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2 years ago

What is 1 and 2 I need it asap

Physics

1 answer:

ExtremeBDS [4]2 years ago

5 0

The correct answer for the first one is A) Breaking rocks into smaller pieces. The correct answer to the second one is A) It has more energy. Hope this helps.

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A 200kg bucket of cement<br>

ozzi

Answer:

Yes. A 200 kg bucket of cement = About 440.925 pounds of cements.

Explanation:

7 0

2 years ago

This is a problem about a child pushing a stack of two blocks along a horizontal floor. The masses of the blocks, and the coeffi

Rufina [12.5K]

Answer:

N = 23.4 N

Explanation:

After reading that long sentence, let's solve the question

The contact force is the so-called normal in this case we can find it by writing the translational equilibrium equation for the y axis

N - w₁ -w₂ =

N = m₁ g + m₂ g

N = g (m₁ + m₂)

let's calculate

N = 9.8 (0.760 + 1.630)

N = 23.4 N

This is the force of the support of the two blocks on the surface.

7 0

2 years ago

A guitar player tunes the fundamental frequency of a guitar string to 560 Hz. (a) What will be the fundamental frequency if she

lawyer [7]

Answer:

(a) if she increases the tension in the string is increased by 15%, the fundamental frequency will be increased to 740.6 Hz

(b) If she decrease the length of the the string by one-third the fundamental frequency will be increased to 840 Hz

Explanation:

(a) The fundamental, f₁, frequency is given as follows;

$f_1 = \dfrac{\sqrt{\dfrac{T}{\mu}} }{2 \cdot L}$

Where;

T = The tension in the string

μ = The linear density of the string

L = The length of the string

f₁ = The fundamental frequency = 560 Hz

If the tension in the string is increased by 15%, we will have;

$f_{(1 \, new)} = \dfrac{\sqrt{\dfrac{T\times 1.15}{\mu}} }{2 \cdot L} = 1.3225 \times \dfrac{\sqrt{\dfrac{T}{\mu}} }{2 \cdot L} = 1.3225 \times f_1$

$f_{(1 \, new)} = 1.3225 \times f_1 = (1 + 0.3225) \times f_1$

$f_{(1 \, new)} = 1.3225 \times f_1 =\dfrac{132.25}{100} \times 560 \ Hz = 740.6 \ Hz$

Therefore, if the tension in the string is increased by 15%, the fundamental frequency will be increased by a fraction of 0.3225 or 32.25% to 740.6 Hz

(b) When the string length is decreased by one-third, we have;

The new length of the string, $L_{new}$ = 2/3·L

The value of the fundamental frequency will then be given as follows;

$f_{(1 \, new)} = \dfrac{\sqrt{\dfrac{T}{\mu}} }{2 \times \dfrac{2 \times L}{3} } =\dfrac{3}{2} \times \dfrac{\sqrt{\dfrac{T}{\mu}} }{2 \cdot L} = \dfrac{3}{2} \times 560 \ Hz = 840 \ Hz$

When the string length is reduced by one-third, the fundamental frequency increases to one-half or 50% to 840 Hz.

6 0

3 years ago

Two mirrors are touching so they have an angle of 35.4 degrees with one another. A light ray is incident on the first at an angl

alexandr1967 [171]

Answer:

54.6°

Explanation:

From law of reflection i=r.

So, construct the reflected ray at 55.7°degrees from the normal and let it fall on the other mirror.

Now draw the second normal at the point of incidence and again measure the angle of incidence, and draw the angle of reflection.

If you consider triangle AOB, one angle is ∠AOB=90°

and ∠OAB is 54.6°

From angle sum property third angle ie ∠ABO=180°-90°-54.6°=35.4°

So, the second incident angle will be 54.6°

Hence, the second reflected angle will be 54.6 degrees.

8 0

3 years ago

to determine the mass of the central object, we must apply newton's version of kepler's third law, which requires knowing the or

mixer [17]

The approximate orbital period of this star is 13 years.

<h3>What is Kepler's third law?</h3>

The square of a planet's period of revolution around the sun in an elliptical orbit is directly proportional to the cube of its semi-major axis, states Kepler's law of periods.

T² ∝ a³

The time it takes for one rotation to complete depends on how closely the planet orbits the sun. With the use of the equations for Newton's theories of motion and gravitation, Kepler's third law assumes a more comprehensive shape:

P² = 4π² /[G(M₁+ M₂)] × a³

where M₁ and M₂ are the two circling objects' respective masses in solar masses.

Learn more about Kepler's third law here:

brainly.com/question/1608361

#SPJ1

6 0

1 year ago