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3 years ago

I've got an energy and work problem. The premise of the problem is:

Physics

1 answer:

Alenkasestr [34]3 years ago

7 0

Refer to the diagram shown below.

μ = the coefficient of dynamic friction between the crate and the ramp.

1. The applied force of F acts over a distance, d.
The work done is F*d.

2. The component of the weight of the crate acting down the ramp is
mg sin(30) = 0.5mg.
The work done by this force is 0.5mgd.

3. The normal force is N = mgcos(30) = 0.866mg.
This force is perpendicular to the ramp, therefore the work done is zero.

4. The frictional force is μN = μmgcos(30) = 0.866μmg.
The work done by the frictional force is 0.866μmgd.

5. The total force acting on the crate up the ramp is
F - mgsin(30) - μmgcos(30) = F - mg(0.5 - 0.866μ)

6. The work done on the crate by the total force is
d*(F - 0.5mg - 0.866μmg)

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Starting from zero, an electric current is established in a circuit made of a battery of emf E, a resistor of resistance R and a

igor_vitrenko [27]

Answer:

time constant will decrease and steady state current will decrease on increasing the resistance

Explanation:

As we know that the EMF of cell is E which is used to connected across a resistor and an inductor.

So we will have

$E - iR - L\frac{di}{dt} = 0$

here we know that

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now here we have

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3 years ago

7. Two children of mass 20 kg and 30 kg sit balanced on a seesaw with the pivot point located at the center of the seesaw. If th

Ainat [17]

Answer:

Explanation:

Given

mass of children $m_1=20\ kg$

$m_2=30\ kg$

distance between two children $L=3\ m$

suppose small child is at a distance of x m from pivot point

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$2x=9-3x$

$5x=9$

$x=1.8\ m$

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3 years ago

An object in the shape of a thin ring has radius a and mass M. A uniform sphere with mass m and radius R is placed with its cent

madreJ [45]

Answer:

F = GMmx/[√(a² + x²)]³

Explanation:

The force dF on the mass element dm of the ring due to the sphere of mass, m at a distance L from the mass element is

dF = GmdM/L²

Since the ring is symmetrical, the vertical components of this force cancel out leaving the horizontal components to add.

So, the horizontal components add from two symmetrically opposite mass elements dM,

Thus, the horizontal component of the force is

dF' = dFcosФ where Ф is the angle between L and the x axis

dF' = GmdMcosФ/L²

L² = a² + x² where a = radius of ring and x = distance of axis of ring from sphere.

L = √(a² + x²)

cosФ = x/L

dF' = GmdMcosФ/L²

dF' = GmdMx/L³

dF' = GmdMx/[√(a² + x²)]³

Integrating both sides we have

∫dF' = ∫GmdMx/[√(a² + x²)]³

∫dF' = Gm∫dMx/[√(a² + x²)]³ ∫dM = M

F = GmMx/[√(a² + x²)]³

F = GMmx/[√(a² + x²)]³

So, the force due to the sphere of mass m is

F = GMmx/[√(a² + x²)]³

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3 years ago

The ability of water molecules to form hydrogen bonds with other water molecules and water's ability to dissolve substances that

Arlecino [84]

Answer:

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The partial negative charge on oxygen and partial positive charge on hydrogen enables them to make hydrogen bond and also makes it to dissolve the the other substances having partial charges.

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aksik [14]

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Hope this helps, please mark me brainliest

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3 years ago