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IRISSAK [1]
3 years ago
6

If the mass of a material is 104 grams and the volume of the material is 16 cm3, what would the density of the material be?

Physics
1 answer:
almond37 [142]3 years ago
5 0
<h3>Answer</h3>

Density of the material is  6.5 g/cm³

<h3>Explanation</h3>

Given in the question,

Mass of the material = 104g

Volume of the material = 16 cm³

Density is a quantitative expression of the amount of mass contained per unit volume and it's SI unit is kg/m³

Formula to calculate density

<h3> density = mass / volume</h3>

density = 104 / 16

            = 6.5 g/cm³

or

1 g/cm³ = 1000kg/m³

6.5 g/cm³ = 6500kg/m³ in SI units

Hence, density of the material is6.5 g/cm³

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A ball on a string makes 25.0 revolutions in 9.37 s, in a circle radius 0.450 m. What is it’s velocity?
Sauron [17]

The velocity of the ball is 12.5 m/s

Explanation:

The velocity of the ball is given by the ratio between the distance covered by the ball and the time taken:

v=\frac{d}{t}

First, we calculate the distance covered. We know that the radius of the circle is

r = 0.450 m

And the length of the circumference is

L=2\pi r = 2\pi(0.750)=4.7 m

The ball makes 25.0 revolutions, so a total distance of

d=(25.0)L=(25.0)(4.7)=117.5 m

In a time of

t = 9.37 s

So, its velocity is

v=\frac{117.5}{9.37}=12.5 m/s

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3 years ago
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Two spherical shells have a common center. A -2.1 10-6 C charge is spread uniformly over the inner shell, which has a radius of
julsineya [31]

Answer:

a) E_total = 6,525 10⁴ N /C ,field direction is radial outgoing

b)  E_total = 1.89 10⁶ N / C, field is incoming radial

c) E_total = 0

Explanetion:

For this exercise we can use that the charge in a spherical shell can be considered concentrated at its center and that the electric field inside the shell is zero, since Gauss's law is

                Ф = E .dA = q_{int} /ε₀

inside the spherical shell there are no charges

The electric field is a vector quantity, so we calculate the field created by each shell and add it vectorly.

We have two sphere shells with radii 0.050m and 0.15m respectively

a) point where you want to know the electric field d = 0.20 m

shell 1

the point is on the outside,d>ro,  therefore we can consider the charge to be concentrated in the center

            E₁ = k q₁ / d²

             

shell 2

the point is on the outside,d>ro

             E₂ = k q₂ / d²

the total camp is

              E_total = -E₁ + E₂

              E_total = k ( \frac{-q_1 + q_2}{d^2})

              E_total = 9 10⁹ (-2.1 10⁻⁶+ 5 10⁻⁶ / .2²

              E_total = 6,525 10⁵ N /C

The field direction is radial and outgoing ti the shells

b) the calculation point is d = 0.10m

shell 1

point outside the shell d> ro

                 E₁ = k q₁ / d²

shell 2

the point is inside the shell d <ro

Therefore, according to Gauss's law, since there are no charges in the interior, the electrioc field is zero

                E₂ = 0

               

                 E_total = E₁

                 E_total = k q₁ / d²

                 E_total = 9 10⁹ 2.1 10⁻⁶ / 0.1²

                 E_total = 1.89 10⁶ N / A

As the charge is negative, this field is incoming radial, that is, it is directed towards the shell 1

c) the point of interest d = 0.025 m

shell 1

point  is inside the shell d< ro

                 

as there are no charges inside

                     E₁ = 0

shell 2

point is inside the radius of the shell d <ro

                    E₂ = 0

the total field is

                    E_total = 0

3 0
2 years ago
How fast was a plane flying if it traveled 500 km in 30 min?
anyanavicka [17]

Answer:

16.7 km per min

Explanation:

divided 500 and 30

8 0
3 years ago
Read 2 more answers
Can anyone help me with 11 and 12 please
dexar [7]
The 1st one is basically B because it will stay in motion with the same speed and in the same direction unless acted on by an unbalanced force and the 2cd one is A because most of is transformed into thermos energy. hope this helps!
7 0
3 years ago
What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3
Pachacha [2.7K]

This question is incomplete, the complete question is;

Coulomb's law for the magnitude of the force F between two particles with charges Q and Q' separated by a distance d is

|FI = |QQ'I / d²

where K = 1/4π∈0, and

∈0 = 8.854 × 10⁻¹² C²/(N.m²) is the permittivity of free space.

Consider two point charges located on the x-axis:

one charge, q₁ = -18.5 nC, is located at

x₁ = -1.715m; the charge q₂ = 30.5 nC, is at the origin ( x₂=0 )

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q₃ = 51.0 nC placed between q₁ and q₂ at x₃ = -1.085 m ?

Answer: (Fnet3)x = -3.3287 × 10⁻⁵ N

Explanation:

Given that;

Q₁ = -18.5 nC       Q₃ = 51 nC        Q₂ = 30.5 nC

x₁ = - 1.715m         x₃ = - 1.085m     x₂ = 0

Now

x - component of Net force on charge Q₃ is

(Fnet3)x = -K|Q₁I|Q₃I / r₁3² - -K|Q₂I|Q₃I / r₂3²

(Fnet3)x = -(9×10⁹)(51×10⁻⁹) [ 18.5 / ((-1.085 + 1.715)²) + (30.5 / (-1.085)² ] × 10⁻⁹

(Fnet3)x = -3.3287 × 10⁻⁵ N

6 0
3 years ago
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