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babymother [125]
3 years ago
13

Grains of fine california beach sand are approximately spheres with an average radius of 50 μm and are made of silicon dioxide,

which has a density of 2.6 × 103 kg/m3. what mass of sand grains would have a total surface area (the total area of all the individual spheres) equal to the surface area of a cube 1.1 m on an edge?
Physics
1 answer:
kykrilka [37]3 years ago
3 0

The average radius(r) of each grain is r = 50 nanometers = 50*10^-6 meters

Since it is spherical, so

 Volume=(4/3)*pi*r^3

V= (4/3)*pi*(50*10^-6)^3

V=5.23599*10^-13 m^3

 

We are given the Density(ρ) =2600kg/m^3

 

We know that:

Density(p) = mass(m)/volume(V)

m = ρV

 

So the mass of a single grain is:

m = 5.23599*10^-13 * 2600 = 1.361357*10^-9 kg

 

 

The surface area of a grain is:

a = 4*pi*r^2

a = 4*pi*(50*10^-6)^2

a = 3.14*10^-8 m^2

 

 

Since we know the surface area and mass of a grain, the conversion factor is:

1.361357*10^-9 kg / 3.14*10^-8 m^2

 

Find the Surface area of the cube:

cube = 6a^2

cube = 6*1.1^2  = 7.26m^2

 

multiply this by the converions ratio to get:

total mass of sand grains = (7.26 m^2 * 1.361357*10^-9 kg) / (3.14*10^-8 m^2)

total mass of sand grains = 0.3148 kg = 314.80 g

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5 0
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an electrically charged particle with mass m, velocity v, and charge q moves in a circle in magnetic field b. what is the formul
kolbaska11 [484]

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r=\frac {mv}{qb}

Explanation:

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\frac {mv^{2}}{r}=qvb\\\frac {mv}{r}=qb\\r=\frac {mv}{qb}

5 0
3 years ago
The ability of your joints and muscles to move in their full range of motion is called
Studentka2010 [4]

Answer:

Dynamic flexibility

Explanation:

Dynamic flexibility can be generally defined as the ability of the body muscles and joints to move in full range of motion. High flexibility in these joints and muscles leads to the decreasing pain and injury in different parts of the body.

Proper warm up exercises are needed to be carried out that involves both the combination of controlling movements and stretching of the body, and this directly enhances the dynamic flexibility of the body.

The athletes and sports persons possesses a good dynamic flexibility of their body as they carry our different types of body exercises.

6 0
3 years ago
You look down an old well, cannot see the bottom, and mutter to yourself "Oh well!". In order to estimate the depth of the well,
telo118 [61]

Answer:

The best estimate of the depth of the well is 2.3 sec.

Explanation:

Given that,

Record time,

t_{1}=2.19\ sec

t_{2}=2.30\ sec

t_{3}=2.26\ sec

t_{4}=2.29\ sec

t_{5}=2.27\ sec

We need to find the best estimate of the depth of the well

According to record time,

We can write of the record time

t_{1}=2.19\approx 2.2\ sec

t_{2}=2.30\approx 2.3\ sec

t_{3}=2.26\approx 2.3\ sec

t_{4}=2.29\approx 2.3\ sec

t_{5}=2.27\approx 2.3\ sec

Here, all time is nearest 2.3 sec.

So, we can say that the best estimate of the depth of the well is 2.3 sec.

Hence, The best estimate of the depth of the well is 2.3 sec.

6 0
3 years ago
1pt A cannon fires a 5-kg ball horizontally from a
Klio2033 [76]

Answer: Both cannonballs will hit the ground at the same time.

Explanation:

Suppose that a given object is on the air. The only force acting on the object (if we ignore air friction and such) will be the gravitational force.

then the acceleration equation is only on the vertical axis, and can be written as:

a(t) = -(9.8 m/s^2)

Now, to get the vertical velocity equation, we need to integrate over time.

v(t) = -(9.8 m/s^2)*t + v0

Where v0 is the initial velocity of the object in the vertical axis.

if the object is dropped (or it only has initial velocity on the horizontal axis) then v0 = 0m/s

and:

v(t) = -(9.8 m/s^2)*t

Now, if two objects are initially at the same height (both cannonballs start 1 m above the ground)

And both objects have the same vertical velocity, we can conclude that both objects will hit the ground at the same time.

You can notice that the fact that one ball is fired horizontally and the other is only dropped does not affect this, because we only analyze the vertical problem, not the horizontal one. (This is something useful to remember, we can separate the vertical and horizontal movement in these type of problems)

7 0
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