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luda_lava [24]
2 years ago
7

a rock with a mass of 10.0 kg is balanced on top of a large boulder. describe the forces acting on the rock, and use the concept

of forces to explain why it stays on top the boulder.
Physics
1 answer:
denis-greek [22]2 years ago
7 0

Explanation:

The forces acting on the rock include Normal Force, Gravitational force & Friction force

It's possible for it to stay on the boulder because the normal force balances it's weight. also because static friction acts on the boulder up to it's limiting friction even if it were on an attempt to move as a result of air resistance. gravitational forces act upon it by mainly affecting it it's weight. as altitude increases, it's weight decreases since gravity varies from a height to another.

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3 years ago
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What is one property of iodine
mamaluj [8]

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Explanation:

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Which component is used to change alternating current to direct current
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A charge 3q is at the origin, and a charge −2q is on the positive x axis at x=a. part a where on the x-axis would you place a th
Mila [183]

Answer:

x = a + \frac{a\sqrt2}{\sqrt3 - \sqrt2}

Explanation:

Position of charge 3q is x = 0

position of charge -2q is x = a

so here we know that

when two charges are of opposite nature then the electric field will be zero on the line joining the two charges at the position near to smaller magnitude charge

So here the electric field will be zero if the field due to 3q is counterbalanced by field due to -2q

so here we can say

\frac{k(3q)}{(r+a)^2} + \frac{k(-2q)}{r^2} = 0

\frac{3}{(r + a)^2} = \frac{2}{r^2}

\frac{r+a}{r} = \sqrt{\frac{3}{2}}

1 + \frac{a}{r} =\sqrt{\frac{3}{2}}

\frac{a}{r} = \frac{\sqrt3 - \sqrt2}{\sqrt2}

so we will have

r = \frac{a\sqrt2}{\sqrt3 - \sqrt2}

so the x coordinate of this position is given as

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6 0
3 years ago
A nonconducting container filled with 25 kg of water at 23°C is fitted with a stirrer, which is made to turn by gravity acting o
Paul [167]

Explanation:

Given that,

Weight of water = 25 kg

Temperature = 23°C

Weight of mass = 32 kg

Distance = 5 m

(a). We need to calculate the amount of work done on the water

Using formula of work done

W=mgh

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(b). We need to calculate the internal-energy change of the water

Using formula of internal energy

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The  change in internal energy is 1568 J.

(c). We need to calculate the final temperature of the water

Using formula of the change internal energy

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\Delta U=mc_{p}(T_{2}-T_{1})

T_{2}=T_{1}+\dfrac{\Delta U}{mc_{p}}

T_{2}=23+\dfrac{1568}{25\times4.18\times10^{3}}

T_{2}=23.01^{\circ}\ C

The final temperature of the water is 23.01°C.

(d). The amount of heat removed from the water to return it to it initial temperature is the change in internal energy.

The amount of heat is 1568 J.

Hence, This is the required solution.

6 0
3 years ago
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