The volume of a piece of rock was 18.0 cm^3. The student measured the mass of the piece of rock as 48.6 g. Calculate the density of the rock in g/cm^3.

Answer: Frequency

Explanation: Frequency is term which best describes how often a person exercises.

Answer:

frequency

Explanation:

I took the. test

Answer:

[tex]n=1.56\times 10^{17}[/tex] electrons

Explanation:

Given that,

Total charge = 9 mC = 0.009 C

0.009 C of charge passes through a wire in 3.6 s.

Let q' is the charge that passes through it in 10 s.

So,

[tex]\dfrac{0.009 }{3.6}=\dfrac{q'}{10}\\\\q'=\dfrac{0.009 \times 10}{3.6}\\\\q'=0.025\ C[/tex]

We know that,

q = ne

Where

n is the number of electrons

So,

[tex]n=\dfrac{q}{e}\\\\n=\dfrac{0.025}{1.6\times 10^{-19}}\\\\n=1.56\times 10^{17}[/tex]

So, [tex]1.56\times 10^{17}[/tex]electrons must pass through the cross-sectional area.

Answer:

The right answer is "1010 V/m".

Explanation:

The given values are:

Intensity,

[tex]I=1350 \ W/m^2[/tex]

[tex]c = 3.00\times 108 \ m/s[/tex]

[tex]\mu_0= 4\pi\times 10^{-7} \ T.m/A[/tex]

Now,

The electric field's maximum value will be:

=  [tex]\sqrt{2\times u\times c\times I}[/tex]

On substituting the values in the above formula, we get

=  [tex]\sqrt{2\times 4\times \pi\times 10^{-7}\times 3\times 10^8\times 1350}[/tex]

=  [tex]\sqrt{32400\times 3.14\times 10^{-7}\times 10^8}[/tex]

=  [tex]1010 \ V/m[/tex]

The maximum value of the electric field associated with this radiation is :

Given data :

Intensity ( I ) = 1350 W/m²

C = 3.00 * 10⁸ m/s

The maximum value of the electric field can be calculated using the equation below

[tex]E_{max} = \sqrt{2*u*c*I}[/tex]  ----- ( 1 )

where :  I = 1350 W/m²,  μ =  4π * 10⁻⁷,  c = 3.00 * 10⁸

Insert values into equation ( 1 )

∴ [tex]E_{max}[/tex] = 1010 V/m.

Hence we can conclude that the maximum value of the electric field is 1010 V/m .

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Answer:

L = 0.1379 m = 13.79 cm

Explanation:

The Rayleigh criterion establishes that two objects are separated when the maximum of diffraction for slits coincides with the minimum of the other point, therefore the expression for the diffraction

            a sin θ = m λ

the first zero occurs when m = 1

let's use trigonometry to find the angle

          tan θ = y / L

as in these experiments the angles are very small

          tan θ = sin θ /cos θ = sin θ

          sin θ = y / L

we substitute

          a y /L = λ

In the case of circular aperture the system must be solved in polar coordinates, for which a numerical constant is introduced

           a y / L = 1.22 λ

           L = a y / 1.22 λ

We search the magnitudes to the SI system

           a = 1.04 cm = 1.04 10⁻² m

           y = 9.09 10⁻⁶ m

           λ = 562 10⁻⁹ m

let's calculate

           L = [tex]\frac{1.04 \ 10^{-2} \ 9.09 \ 10^{-6} }{1.22 \ 562 \ 10^{-9} }[/tex]

           L = 1.379 10⁻¹ m

           L = 0.1379 m = 13.79 cm

Answer: 5

Explanation: B is for sure 60°, c* cosB = 10*1/2 =5

Answer:

Power = 4000watts

Energy = 22.22Joules

Explanation:

Power = I²R

I is the current

R is the resistance

t is the time

Given the following

I = 20Amps

R = 10ohms

t = 3miuntes = 180secs

Substitute

P = 20²*10

P = 400*10

P = 4000Watts

Hence the amount of power used is 4000Watts

Energy used = Power/time

Energy used= 4000/180

Energy used = 22.22Joules

Answer:

n = 1.4

Explanation:

Given,

R1 = 18 cm, R2 = -18 cm

From lens makers formula

1/f = (n - 1)(1/18 + 1/18) = (n-1)/9

f = 9/(n-1)

Power, P = 1/f ( in m) = (n-1)/0.09

Now, this lens is in with conjunction with a concave mirror which then can be thought of as to be in conjunction with another thin lens

Power of concave mirror = P' = 1/f ( in m) = 2/R = 2/0.18 = 1/0.09

Net power of the combination = 2P + P' = 2(n-1)/0.09 + 1/0.09 = 1/0.05

n = 1.4

Answer:

[tex]\boxed {\boxed {\sf 2.40 \ meters}}[/tex]

Explanation:

Work is the product of force and distance, so the formula is:

[tex]W= F \times d[/tex]

The work is 6.00 Joules, but

The force is 2.50 Newtons.

The known values are:

Substitute the values into the formula.

[tex]6.00 \ N*m= 2.50 \ N *d[/tex]

We are solving for distance, so we must isolate the variable, d. It is being multiplied by 2.50 Newtons and the inverse of multiplication is division. Divide both sides by 2.50 N.

[tex]\frac {6.00 \ N*m}{2.50 \ N}=\frac{2.50 \ N * d}{2.50 \ N}[/tex]

[tex]\frac {6.00 \ N*m}{2.50 \ N}=d[/tex]

The units of Newtons cancel out.

[tex]\frac {6.00 m}{2.50 }=d[/tex]

[tex]2.40 \ m =d[/tex]

The force was applied to the rock over a distance of 2.40 meters and choice C is correct.

Answer:

A. The shorter the distance the greater the force, so now the force would have increased because they haven't moved away from each other.

B. The force would decrease because they are now further apart.

C. the force wouldn't be as great as before but the force would still be high because they are not far apart from each other.

I hope this helps you( ◜‿◝ )♡

Answer:

1) 392 joules

2) 625,000 joules

3) 156,800 joules

Explanation:

Gravitational potential energy = mgh. m = mass in kg, g = acceleration due to gravity and h = height in meters.

Kinetic energy = 1/2(m)(v)^2, m = mass in kg, v = velocity in meters per second

1) P.E = 2 x 9.8 x 20 = 392 joules

2) K.E = 1/2 x 2000 x (25)^2 = 625,000 joules

3) P.E = 80 x 9.8 x 200 = 156,800 joules

Answer:

[tex]1)\:392\:\text{J}\:(400\:\text{J with one significant figure)},\\2)\:625,000\:\text{J}\:(600,000\:\text{J with one significant figure)},\\3)\:156,800\:\text{J}\:(200,000\:\text{J with one significant figure)},[/tex]

Explanation:

1. The potential energy of an object is given by [tex]PE=mgh[/tex]. Substituting given values, we have:

[tex]PE=2\cdot 9.8\cdot 20=\boxed{392\:\text{J}}[/tex]

2. The kinetic energy of an object is given by [tex]KE=\frac{1}{2}mv^2[/tex]. Substituting given values, we have:

[tex]KE=\frac{1}{2}\cdot 2000\cdot 25^2 =\boxed{625,000\: \text{J}}[/tex]

3. 1. The potential energy of an object is given by [tex]PE=mgh[/tex]. Substituting given values, we have:

[tex]PE=80\cdot 9.8\cdot 200=\boxed{156,800\:\text{J}}[/tex]

Answer:

A/A₀ = 0.5106

Explanation:

To do this, we need to use several formulas and expressions. First, we need to know the period of time of the oscillator. This can be calculated using the following expression:

ω = 2π/T   (1)

But angular frequency (ω) can be also be calculated using:

ω = √(k/m)   (2)

Using (1) and (2), we can solve for the period T:

2π/T = √(k/m)    (mass in kg)

2π/T = √(130/0.360)

2π/T = √361.11

2π/T = 19.003

T = 2π/19.003 = 0.331 s

Now, the amplitude A at a time t, is:

A = x exp(-bt/2m)    (3)

At time 0, A = x. so A₀ = x

The problem states that we have 11 cycles respect to the initial amplitude, so expression (3) can be rewritten as:

A = x exp(-b(17t/2m))     using b as kg/s = 0.086 kg/s

Replacing the data we have:

A = x exp(-0.086(17*0.331)/2*0.36)

A = x exp(-0.086 * 7.815)

A = x exp(-0.6721)

A = 0.5106x    (4)

Now, doing the ratio with the innitial we have:

A / A₀ = 0.5106x / x

The ratio is:

Hope this helps

Answer:

150

Explanation:

P=F/A

10=15/A

10*15=15/A*15

150=A

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Answer:

P = 9622.9 Pa = 9.62 KPa

Explanation:

First, we will calculate the mass of all three liquids:

m = ρV

where,

m = mass of liquid

ρ = density of liquid

V = Volume of liquid

FOR LIQUID 1:

m₁ = (2.8 x 10³ kg/m³)(2 x 10⁻³ m³) = 5.6 kg

m₂ = (1 x 10³ kg/m³)(1.5 x 10⁻³ m³) = 1.5 kg

m₃ = (0.6 x 10³ kg/m³)(1 x 10⁻³ m³) = 0.6 kg

The total mass will be:

m = m₁ + m₂+ m₃ = 5.6 kg + 1.5 kg + 0.6 kg

m = 7.7 kg

Hence, the weight of the liquids will be:

W = mg = (7.7 kg)(9.81 m/s²) = 75.54 N

Now, we calculate the base area:

A = πr² = π(0.05 m)²

A = 7.85 x 10⁻³ m²

Now the pressure will be given as:

[tex]P = \frac{F}{A}\\\\P = \frac{75.54\ N}{7.85\ x\ 10^{-3}\ m^2}[/tex]

P = 9622.9 Pa = 9.62 KPa

Answer: If the magnification is positive, the image is upright compared to the object; if m is negative, the image is inverted compared to the object.

(I think, I was also stuck on this question for a bit)

The positive sign indicate about the image that

b) The image is on the same side of the mirror as the object.

d) The image is upright.

Magnification  is a quantification of comparing the size of the image with respect to the size of the object. It gives us information about the image in terms of how large or small is the image formed.

magnification   = height of image / height of object

since , height of object  is always positive as it is always kept upright hence , in order to make m positive ,  height of image  need to be positive

also magnification    = -(v/u )

v = image distance

u = object distance

u is always positive hence , in order to make m positive , v needs to be negative , which implies on the same side of the mirror as the object

correct option

b) The image is on the same side of the mirror as the object.

d) The image is upright.

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Answer:

Option C and D only

Explanation:

Option A is incorrect  because refractive index of a material is the ratio of speed of light in vacuum to the speed of light in a any given medium

Option B is correct as the speed of light in vacuum is always greater than the speed of light in any given medium.

Option C is correct

Option D is incorrect

Option E is incorrect because the denser the medium the more is the refractive index. Water is denser than air, hence it should have more refractive index as compared to that of air.

False

Explanation:

Called an asteroid for many years, Ceres is so much bigger and so different from its rocky neighbors that scientists classified it as a dwarf planet in 2006. Even though Ceres comprises 25 percent of the asteroid belt's total mass, tiny Pluto is still 14 times more massive.

Answer:

Y = 17.75 x 10⁻³ m = 17.75 mm

Explanation:

Using Young's Double Slit formula below:

[tex]Y = \frac{\lambda L}{d}[/tex]

where,

Y = fringe separation = ?

λ = wavelength = 568 nm = 5.68 x 10⁻⁷ m

L = distance between slits and screen = 3.5 m

d = slit separation = 0.112 mm = 1.12 x 10⁻⁴ m

Therefore,

[tex]Y = \frac{(5.68\ x\ 10^{-7}\ m)(3.5\ m)}{1.12\ x\ 10^{-4}\ m}[/tex]

Y = 17.75 x 10⁻³ m = 17.75 mm

Answer:

the energy when it reaches the ground is equal to the energy when the spring is compressed.

Explanation:

For this comparison let's use the conservation of energy theorem.

Starting point. Compressed spring

         Em₀ = K_e = ½ k x²

Final point. When the box hits the ground

         Em_f = K = ½ m v²

since friction is zero, energy is conserved

          Em₀ = Em_f

          1 / 2k x² = ½ m v²

          v = [tex]\sqrt{ \frac{k}{m} }[/tex]     x

Therefore, the energy when it reaches the ground is equal to the energy when the spring is compressed.

Based on the law of conservation of energy, the elastic potential energy of the system when the spring is fully compressed is equal to the kinetic energy of the system at the moment immediately before the box hits the ground.

The energy in a compressed spring is elastic potential energy given by the formula:

where

The kinetic energy of a body is the energy the body the has due to it's motion.

Kinetic energy, KE, is givenby the formula below:

From the law of conservation of energy, the total energy in a closed system is conserved.

Based on this law, all the energy in the compressed spring is converted to the kinetic energy of the box just before it reaches the ground.

Therefore, the elastic potential energy of the system when the spring is fully compressed is equal to the kinetic energy of the system at the moment immediately before the box hits the ground.

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Answer: C-  The intensity increases by a factor 9

Explanation: The intensity of a sound wave follows an inverse square law, that means that it is inversely proportional to the square of the distance: so  the new distance is the intensity will increase by a factor 9.

My sentence- I hope that helped!

Intensity of a sound wave decreases by a factor of 9, if the distance from the source is increased by a factor of 3. Hence option B is correct.

Intensity of Sound is inversely proportional to the square of the distance from the sound source. Since sound waves carry its energy though a two-dimensional or three-dimensional medium, the intensity of the sound wave decreases with increasing as second power of distance form the source.

Mathematically,

Intensity I ∝ 1/D²

If the distance from the source is increased by a factor of 3, Then

I ∝ 1/3² ∝ 1/9

Intensity ∝  1/9

Hence option B is correct.

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Answer:

40

Explanation:

Answer:

42.6083 mi/h

Explanation:

Given: A car travels 8km in 7 minutes.

To find: Find the speed of the car.

Formula: [tex]Speed = \frac{Distance}{Time}[/tex]

Solution: Since the formula for the speed of an object (which is the car) is speed = distance ÷ time, divide the distance (8km) by the time (7min)

Speed = 42.6083 miles per hour

Answer:

The answer is A

Explanation:

An insulator doesn't have to be a metal

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Answer:

a substance which does not readily allow the passage of heat or sound.

<3<3<3<3<3<3<3<3<3<3<3<3<3<3<3<3<3<3<3<3<3<3<3

The following that describes an electric insulator are :

C. A material that has a high resistance and prevents charges from  moving freely.    

Thus, the correct answer is C.

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Answer:

3,000 meters = 3 kilometers

Explanation:

Answer:

A) the magnetic field and the direction of motion must be perpendicular

For a current to be induced in a wire passing through a magnetic field, the magnetic field and the direction of motion must be perpendicular. The correct option is A.

The magnetic field is the region of space where another object experiences magnetic force and the current is induced in it.

According to the Fleming's right hand rule, the direction of motion, magnetic force and magnetic field are mutually perpendicular.

Thus, the correct option is A.

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Answer:

The wavelength will be "1.182 μm".

Explanation:

The given values are:

Wavelength

[tex]\lambda=591 \ nm[/tex]

or,

  [tex]=591\times 10^-9 \ m[/tex]

Distance,

[tex]d = 3.00 m[/tex]

[tex]n = 1[/tex]

Distance of fringe from center,

[tex]y = 4.84 \ mm[/tex]

We have to find the wavelength of first order dark fringe,

[tex]\lambda = ?[/tex]

As we know,

⇒ [tex]y_{bright} =\frac{1\times \lambda\times L}{d}[/tex]

On putting the given values in the formula, we get

   [tex]0.00484=\frac{1\times (591\times 10^{-9})\times 3}{d}[/tex]          

On applying the cross multiplication, we get

   [tex]\lambda = \frac{0.00484\times 000036632}{0.5\times 3}[/tex]

      [tex]=1182\times 10^{-9}[/tex]

or,

      [tex]=1.182 \ \mu m[/tex]    

Answer:

Definimos momento como el producto entre la masa y la velocidad

P = m*v

(tener en cuenta que la velocidad es un vector, por lo que el momento también será un vector)

Sabemos que el peso de la paca de heno es 175N, y el peso es masa por aceleración gravitatoria, entonces.

Peso = m*9.8m/s^2 = 175N

m = (175N)/(9.8m/s^2) = 17.9 kg

Ahora debemos calcular la velocidad de la paca justo antes de tocar el suelo.

Sabemos que la velocidad horizontal será la misma que tenía el avión, que es:

Vx = 36m/s

Mientras que para la velocidad vertical, usamos la conservación de la energía:

E = U + K

Apenas se suelta la caja, esta tiene velocidad cero, entonces su energía cinética será cero y la caja solo tendrá energía potencial (Si bien la caja tiene velocidad horizontal en este punto, por la superposición lineal podemos separar el problema en un caso horizontal y en un caso vertical, y en el caso vertical no hay velocidad inicial)

Entonces al principio solo hay energía potencial:

U = m*g*h

donde:

m = masa

g = aceleración gravitatoria

h = altura  

Sabemos que la altura inicial es 60m, entonces la energía potencial es:

U = 175N*60m = 10,500 N

Cuando la paca esta próxima a golpear el suelo, la altura h tiende a cero, por lo que la energía potencial se hace cero, y en este punto solo tendremos energía cinética, entonces:

10,500N = (m/2)*v^2

De acá podemos despejar la velocidad vertical justo antes de golpear el suelo.

√(10,500N*(2/ 17.9 kg)) = 34.25 m/s

La velocidad vertical es 34.25 m/s

Entonces el vector velocidad se podrá escribir como:

V = (36 m/s, -34.25 m/s)

Donde el signo menos en la velocidad vertical es porque la velocidad vertical es hacia abajo.

Reemplazando esto en la ecuación del momento obtenemos:

P = 17.9kg*(36 m/s, -34.25 m/s)  

P = (644.4 N, -613.075 N)