What is density of a 36g object with a volume of 15

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Whitepunk [10]

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5 0

2 years ago

A 4.3 kg steel ball and 6.5 m cord of negligible mass make up a simple pendulum that can pivot without friction about the point

Alekssandra [29.7K]

Answer

given,

mass of steel ball, M = 4.3 kg

length of the chord, L = 6.5 m

mass of the block, m = 4.3 Kg

coefficient of friction, μ = 0.9

acceleration due to gravity, g = 9.81 m/s²

here the potential energy of the bob is converted into kinetic energy

$m g L = \dfrac{1}{2} mv^2$

$v= \sqrt{2gL}$

$v= \sqrt{2\times 9.8\times 6.5}$

v = 11.29 m/s

As the collision is elastic the velocity of the block is same as that of bob.

now,

work done by the friction force = kinetic energy of the block

$f . d = \dfrac{1}{2} mv^2$

$\mu m g. d = \dfrac{1}{2} mv^2$

$d=\dfrac{v^2}{2\mu g}$

$d=\dfrac{11.29^2}{2\times 0.9 \times 9.8}$

d = 7.23 m

the distance traveled by the block will be equal to 7.23 m.

8 0

3 years ago

If light has a speed of 122,000 mps in a transparent medium, what is the index of refraction of the medium? A. n = 1.52

castortr0y [4]

Answer:

$A.\hspace{3}n=1.52$

Explanation:

The refractive index of a medium is a measure to know how much the speed of light within the medium is reduced. It can be calculated with the next equation:

$n=\frac{c}{v}$ (1)

Where:

$c=Speed\hspace{3}of\hspace{3}light\hspace{3}in\hspace{3}vacuum\\n=Refractive\hspace{3}index\\$

$v=Velocity\hspace{3}of\hspace{3}light\hspace{3}in\hspace{3}the\hspace{3}medium$

The speed of light in the vacuum is approximately 300,000 km/s. In order to work with the same units let's do the proper conversion with the velocity of the medium:

$122,000\frac{mi}{s} *\frac{1.60934km}{1mi}=196339.48\frac{km}{s}$

Finally, replacing the data in (1):

$n=\frac{300,000}{196339.48} =1.527965746\approx1.52$

3 0

3 years ago

What factors determine the magnitude of the electric force between two particles ?

svp [43]

The electric force between the two particles are calculated through the equation,

F = kQ₁Q₂ / d²

where F is the force, k is a constant called Coulomb's law constant, Q₁ and Q₂ are the charges, and d is the distance. This equation is called the Coulomb's law.

It can be seen from the equation above that the electric forces between the objects are majorly affected by the substance's charges and distance.

The answer to this item is therefore letter A.

7 0

3 years ago

Read 2 more answers

A 9-μC positive point charge is located at the origin and a 6 μC positive point charge is located at x = 0.00 m, y = 1.0 m. Find

sukhopar [10]

Answer:

The coordinates of the point is (0,0.55).

Explanation:

Given that,

First charge $q_{1}=9\times10^{-6}\ C$ at origin