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Greeley [361]
3 years ago
9

What is the chemical name for a sand

Physics
1 answer:
vovikov84 [41]3 years ago
7 0
It is known as silicon dioxide or silica!

Hope this helps!
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Plzzz answer this question correctly
tekilochka [14]

Answer:

changing the direction in which a force is exerted

8 0
3 years ago
Air "breaks down" when the electric field strength reaches 3×106n/c, causing a spark. a parallel-plate capacitor is made from tw
aalyn [17]
Ok, I think this is right but I am not sure:
 Q = ϵ
 0AE
A= π π
 r^2

=(8.85x10^-12 C^2/Nm^2)
( π π (0.02m)^2)
(3x10^6 N/C) =3.3x10^-8 C = 33nC N = Q/e = (3.3x10^-8 C)/(1.60x10^-19 C/electron) = 2.1x10^11 electrons


4 0
3 years ago
is this true or false? an object with a mass of 9.0kg has an acceleration of 3m/s2. the force acting on it is 3N
ludmilkaskok [199]

The answer is "False". The force acting on the object is 27 N.

According to Newton's second law, when a force <em>F</em> acts on am object of mass <em>m</em>, it produces an acceleration <em>a</em>. The force is given by the expression,

F=ma

Thus, if the body has a mass of 9.0 kg and if it has an acceleration of 3 m/s², then, on substituting the values in the equation for force,

F=ma\\ =(9.0kg)(3m/s^2)\\ =27N

Thus, it can be seen that the force acting on the body is 27 N and not 3 N as is mentioned in the statement. Hence the statement is false.

5 0
3 years ago
Please awnser and show the ways​
topjm [15]

Answer:

Answers in solutions.

Explanation:

<u>Question 6:</u>

The density of gold is 19.3 g/cm³

The density of silver is 10.5 g/cm³

  • The density of the substance in Crown A;

Density = mass ÷ volume = \frac{1930}{100} = 19.3 g/cm³

Since the density of gold, given, is 19.3 g/cm³ and the density of the substance in Crown A has a density of 19.3 g/cm³ , then that substance must be gold.

  • The density of the substance in Crown B;

Density = mass ÷ volume = 1930 ÷ 184 = 10.48913043  g/cm³ ≈ 10.5 g/cm³  (answer rounded up to one decimal place)

Since the density of the substance in Crown B is approximately equal to 10.5 g/cm³ , then that substance is Silver.

  • The density of substance in Crown C;

Density = mass ÷ volume = 1930g ÷ 150cm³ = 12.86666667 ≈ 12.9 cm³ (answer rounded up to one decimal place)

<h3><u>The density of the mixture:</u></h3><h3 />

For 2 cm³ of the mixture, its mass equal 19.3 g + 10.5 g = 29.8 g

∴ for 1 cm³ of the mixture, its mass equal to \frac{29.8}{2} = 14.9 g

Hence the density of the mixture = 14.9 g/cm³ and is not equal to the density of the substance in Crown C.

* Crown C is not made up of a mixture of gold and silver.

<u>Question 7:</u>

<u />

  • An empty masuring cylinder has a mass of 500 g.
  • Water is poured into measuring cylinder until the liquid level is at the 100 cm³ mark.
  • The total mass is now 850 g

The mass of water that occupied the 100 cm³ space of the container = total mass - mass of the empty container = 850 g - 500 g = 350 g

Density of the liquid (water) poured into the container = mass ÷  volume = 350 g ÷ 100 cm³ = 3.5g/cm³

<u>Question 8:</u>

<u />

A tank filled with water has a volume of 0.02 m³

(a) 1 liter = 0.001 m³

How many liters? = 0.02 m³ ?

Cross multiplying gives:

\frac{0.02 * 1}{0.001} =  20 liters

(b) 1 m³ = 1,000,000 cm³

0.02 m³ = how many cm³ ?

Cross-multiplying gives;

\frac{0.02 * 1,000,000}{1} = 20,000 cm³

(c) 1 cm³ = 1 ml

∴ 0.02 m³ of the water = 20,000 cm³ = 20,000 ml

<u>Question 9:</u>

<u />

Caliper (a) measurement = 3.2 cm

Caliper (b) measurement = 3 cm

<u>Question 10:</u>

<u />

  • A stone is gently and completely immersed in a liquid of density 1.0 g/cm³
  • in a displacement can
  • The mass of liquid which overflow is 20 g

The mass of the liquid which overflow = mass of the stone = 20 g

1 gram of the liquid occupies 1 cm³ of space.

20 g of the liquid will occupy; \frac{20 * 1}{1} = 20 cm³

(a) Since the volume of the water displaced is equal to the volume of the stone.

∴ The volume of the stone = 20 cm³

(b) Mass = density ×  volume

Density of the stone = 5.0 g/cm³

Volume of the stone = 20 cm³

Mass of the stone = 5 g/cm³ × 20 cm³ = 100 g

7 0
3 years ago
On your first trip to Planet X you happen to take along a 180 g mass, a 40-cm-long spring, a meter stick, and a stopwatch. You'r
lys-0071 [83]

Answer:

g_x = 3.0 m / s^2

Explanation:

Given:

- Change in length of spring [email protected] = 22.6 cm

- Time taken for 11 oscillations t = 19.0 s

Find:

- The value of gravitational free fall g_x at plant X:

Solution:

- We will assume a simple harmonic motion of the mass for which Time is:

                                 T  = 2*pi*sqrt(k / m )    ...... 1

- Sum of forces in vertical direction @equilibrium is zero:

                                 F_net = k*x - m*g_x = 0

                                 (k / m) = (g_x / x)    .... 2

- substitute Eq 2 into Eq 1:

                                  2*pi / T = sqrt ( g_x / x )

                                   g_x = (2*pi / T )^2 * x

- Evaluate g_x:

                                  g_x = (2*pi / (19 / 11) )^2 * 0.226

                                  g_x = 3.0 m / s^2

                                 

                       

3 0
3 years ago
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