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alina1380 [7]
2 years ago
14

When the distance increases, the electrostatic force____; we call this relationship proportional

Physics
2 answers:
Ipatiy [6.2K]2 years ago
8 0

Answer:

See below

Explanation:

When the distance increases, the electrostatic force <u> DECREASES</u>; we call this relationship <u> INVERSELY</u>  proportional

Alik [6]2 years ago
5 0

Answer:

I think decreases inversely

Explanation:

the third

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A uniform rod is hung at one end and is partially submerged in water. If the density of the rod is 5/9 that of water, find the f
VashaNatasha [74]

Answer:

\frac{y}{L} = 0.66

Hence, the fraction of the length of the rod above water = \frac{y}{L} = 0.66

and fraction of the length of the rod submerged in water = 1 - \frac{y}{L} = 1 - 0.66 = 0.34  

Explanation:

Data given:

Density of the rod = 5/9 of the density of the water.

Let's denote density of Water with w

And density of rod with r

So,

r = 5/9 x w

Required:

Fraction of the length of the rod above water.

Let's denote total length of the rod with L

and length of the rod above with = y

Let's denote the density of rod = r

And density of water = w

So, the required is:

Fraction of the length of the rod above water = y/L

y/L = ?

In order to find this, we first need to find out the all type of forces acting upon the rod.

We know that, a body will come to equilibrium if the net torque acting upon a body is zero.

As, we know

F = ma

Density = m/v

m = Density x volume

Volume = Area x length = X ( L-y)

So, let's say X is the area of the cross section of the rod, so the forces acting upon it are:

F = mg

F = (Density x volume) x g

g = gravitational acceleration

F1 = X(L-y) x w x g (Force on the length of the rod submerged in water)

where,

X (L-y) = volume

w = density of water.

Another force acting upon it is:

F = mg

F2 =  X x L x r x g

Now, the torques acting upon the body:

T1 + T2 = 0

F1 ( y + (\frac{L-y}{2}) ) g sinФ - F2 x (\frac{L}{2}) x gsinФ = 0

plug in the  equations of F1 and F2 into the above equation and after simplification, we get:

(L^{2} - y^{2} ) . w = L^{2} . r

where, w is the density of water and r is the density of rod.

As we know that,

r = 5/9 x w

So,

(L^{2} - y^{2} ) . w = L^{2} . 5/9 x w

Hence,

(L^{2} - y^{2} ) = \frac{5L^{2} }{9}

\frac{L^{2} - y^{2}  }{L^{2} } = \frac{5}{9}

Taking L^{2} common and solving for \frac{y}{L}, we will get

\frac{y}{L} = 0.66

Hence, the fraction of the length of the rod above water = \frac{y}{L} = 0.66

and fraction of the length of the rod submerged in water = 1 - \frac{y}{L} = 1 - 0.66 = 0.34

8 0
3 years ago
In gym class you run 22 m horizontally, then climb a rope vertically for 6.2 m. What is the direction angle of your total displa
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Im 99% sure
tan^-1 (6.2/22)= 15.7º
with 2 significant figures 16º
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Do you get this ?????
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The first one is actually 10 times as big as the second one.

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A lamp draws a current of 0.1 A when it is connected to a 122-V source. (
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The power of the lamp would be calculated with the equation of ohm laws. P = U x I = 122V x 0.1A = 12.2W
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3 years ago
What does the term calorie refer to
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Hey there,
<span>It means- the amount of energy required to raise the temperature of a liter of water one degree centigrade at sea level
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Hope this helps :))

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6 0
3 years ago
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