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lana66690 [7]
2 years ago
9

A 4 kg block is pushed up an incline that makes a 30° angle with the horizontal, as shown in the figure. Once the block is pushe

d a distance of d = 5.0 m up the incline, the block remains at rest.
What is the approximate change in the gravitational potential energy of the block-Earth system when the block is held at rest compared to its original location at the bottom of the incline?
A) OJ
B) 100 J
C) 100/33 J
D) 200 J
Physics
1 answer:
Sophie [7]2 years ago
8 0

Answer:

  B)  100 J

Explanation:

Assuming the distance given is measured along the incline, the vertical change in height is (5 m)(sin 30°) = 2.5 m. Then the change in potential energy is ...

  ∆PE = mg(∆h) = (4 kg)(10 m/s^2)(2.5 m) = 100 J

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An object of mass m = 5.0 kg hangs from a cord around a light pulley: The length of the cord between the oscillator and the pull
puteri [66]

Answer:

\mu=0.0049Kg/m

Explanation:

When a standing wave is formed with six loops means the normal mode of the wave is n=6, the frequency of the normal mode is given by the expression:

f_n=\frac{nv}{2L}

Where L is the length of the string and v the velocity of propagation. Use this expression to find the value of v.

f_6=\frac{6v}{2L}\\(150)=\frac{6v}{2(2)} \\150=\frac{3v}{2} \\3v=150(2)\\ v=\frac{300}{3} \\v=100m/s

The velocity of propagation is given by the expression:

v=\sqrt{\frac{T}{\mu }

Where \mu is the desirable variable of the problem, the linear mass density, and T is the tension of the cord. The tension is equal to the weight of the mass hanging from the cord:

T=W=mg=(5)(9.81)=49.05N

With the value of the tension and the velocity you can find the mass density:

v=\sqrt{\frac{T}{\mu}

v^2=\frac{T}{\mu}\\ \mu=\frac{T}{v^2} =\frac{49.05}{(100)^2} =\frac{49.05}{10000} =0.0049Kg/m

6 0
2 years ago
What's the Coulomb's law?
Ulleksa [173]

<span>
In layman's term: </span>like charges don't attract while opposite charges do<span>electrostatic forces between point A( which is charged) and point B (which is also charged) are proportional to the charge of point A and point B. </span><span>there is also something else about this  law that I don't quite remember.</span>

<span>___________________________________________________</span>

<span />Here is the formula:

<span>F = k x Q1 x Q2/d^<span>2</span></span>

<span>What the formula means:</span>

F=force between charges

Q1 and Q2= amount of charge

d=distance between these two charges

k= Coulombs constant (proportionally constant)

________________________________________________

I think that about covers it and hopefully this helped.

4 0
3 years ago
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A student observes a shadow move across the sun during a solar eclipse. Which list represents the position of Earth, the sun, an
Liono4ka [1.6K]
It goes sun moon earth the moon is blocking us from seeing the sun.
3 0
3 years ago
In the flow past a compression corner, the upstream Mach number and pressure are 3.5 and 1 atm, respectively. Downstream of the
yulyashka [42]

Answer:

\theta=23.7^{\circ}

Explanation:

The ratio of pressure 2 to 1 us 5.48/1= 5.48 rounded off as 5.5.

Referring to table A.2 of modern compressible flow then M_{\beta_1}=2.2

Also

M_{\beta_1}=M_1 sin \beta and making sin\beta the subject of the formula then

sin\beta=\frac {2.2}{3.5}\\\beta=38.94^{\circ}

Making reference to \theta-\beta-M diagram then

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4 0
3 years ago
Gibbons, small Asian apes, move by brachiation, swinging below a handhold to move forward to the next handhold. A 9.0 kg gibbon
LiRa [457]

Answer:

241.8 N.

Explanation:

The force on branch provides a reaction to the ape's weight force plus the centripetal force needed to keep the gibbon in a circular motion of radius 0.60 m.

Centripetal force = mv^2/r

F = mg + mv²/r

F = m(g + v²/r)

where,

m = mass

= 9 kg

g = acceleration due to gravity

= 9.8 m/s²

v = 3.2 m/s

r = 0.60 m

F = 9 * (9.8 + 3.2²/0.60)

= 241.8 N.

3 0
2 years ago
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