Which of the following statements correctly describe the various applications listed above?
a. None of these technologies use radio waves.
b. None of these technologies use lower-frequency microwaves.
c. None of these technologies use higher-frequency microwaves.
d. None of these technologies use infrared waves.
e. The radiation emitted by some wireless local area networks has the shortest wavelength of all the technologies listed above.
f. All these technologies emit waves with a wavelength in the range 0.10 to 10.0 cm.
g. All the technologies emit waves with a wavelength in the range 0.01 to 10.0 km.

Answer: 0.553

Explanation:

Answer:

Mass, m = 6.18 kg

Explanation:

Given the following data;

Frequency, F = 10 Hz

Spring constant, k = 250 N/m

We know that pie, π = 22/7

To find the mass, we would use the following formula;

F = 1/2π√(k/m)

Where;

F is the frequency of oscillation.

k is the spring constant.

m is the mass of the spring.

Substituting into the formula, we have;

10 = 1/2 * 22/7 * √250/m

10 = 22/14 * √250/m

Cross-multiplying, we have;

140 = 22 * √250/m

Dividing both sides by 22, we have;

140/22 = √250/m

6.36 = √250/m

Taking the square of both sides, we have;

6.36² = (√250/m)²

40.45 = 250/m

Cross-multiplying, we have;

40.45m = 250

Mass, m = 250/40.45

Mass, m = 6.18 kg

Answer:

A

Explanation:

because the speed divide by the frequency is equal to the wavelength(in meters)

5×10² m

Answer:

800 meters per hour

Explanation:

800 meters per hour is the average speed of an airplane that travels from New York to Los Angeles, a total distance of 4800 km, in 6.0 hours.

Answer:

u₂ = 3.7 m/s

Explanation:

Here, we use the law of conservation of momentum, as follows:

[tex]m_1u_1+m_2u_2=m_1v_1+m_2v_2\\[/tex]

where,

m₁ = mass of the car = 1250 kg

m₂ = mass of the truck = 2020 kg

u₁ = initial speed of the car before collision = 17.4 m/s

u₂ = initial speed of the tuck before collision = ?

v₁ = final speed of the car after collision = 6.7 m/s

v₂ = final speed of the truck after collision = 10.3 m/s

Therefore,

[tex](1250\ kg)(17.4\ m/s)+(2020\ kg)(u_2)=(1250\ kg)(6.7\ m/s)+(2020\ kg)(10.3\ m/s)\\\\(2020\ kg)(u_2) = 8375\ N.s + 20806\ N.s - 21750\ N.s\\\\u_2=\frac{7431\ N.s}{2020\ kg}[/tex]

u₂ = 3.7 m/s

Answer:

λ = 266 nm

Explanation:

In this case, we need to determine first the separation of the second fringe from the central maximum. We can determine that with the following expression:

d sinθ = mλ   (1)

However, as the slits are very narrow, we can assume that sin θ ≈ θ and so θ = x/l

Replacing this in (1) we have:

d(x/l) = mλ

and solving for x:

x = mλl / d    (2)

Where:

x: separation of the 2nd fringe from the central maximum

m: order of the fringe

λ: wavelength of light

l: distance of the screen

d: distance between the slits

All the units must be in meters (m), so we can convert the units first or during the resolution. In this case, we'll do it in the resolution. Replacing the given data, we have:

x = 2 * (520 nm * 1m/10⁻⁹ nm) * 1.6 m / (0.66 mm * 1m/1000 mm)

x = 0.00252 m or just 2.52 mm

With this value, we can compute or determine the separation of the 2nd order fringe of the unknown light in the central maximum:

x₂ = 2.52 - 1.23 = 1.29 mm or 0.00129 m

Now, using (2) we can solve for λ:

λ = dx / ml   (3)

Replacing we have:

λ = (0.00129 * 0.0066) / (2 * 1.6)

Hope this helps

Answer:

Total energy, T= 4704 Joules

Explanation:

Given the following data;

Mass = 8kg

Initial height, h1 = 100m

Final height, h2 = 40 m

We know that acceleration due to gravity is equal to 9.8 m/s².

To find the total energy, T;

T = mg(h1 - h2)

T = 8 * 9.8 * (100 - 40)

T = 78.4 * 60

Total energy, T= 4704 Joules

Answer:

F₁ = 499.61 N , this is the force that Bubba support

Explanation:

The trunk is in equilibrium with the two forces applied by man, let's use the equilibrium relation

let's set a reference frame at the extreme left and assume that the counterclockwise rotations are positive

Let's write the expression for the translational equilibrium

subscript 1 is for Bubba's mass and subscript 2 for his partner

               F₁ + F₂ -W = 0

                F₁ + F₂ = W

the expression for rotational equilibrium

               ∑ τ = 0

               F₁  2.2 + F₂  (6.2-0.9) - W  6.2/2 = 0

               2.2 F1 + 5.3 F2 = 3.1 W

let's write our system of equations

             F₁ + F₂ = W

             2.2 F₁ + 5.3 F₂ = 3.1 W

we solve for F₁ in the first equation and substitute in the second

            F₁ = W-F₂

            2.2 (W- F₂) + 5.3 F₂ = 3.1 W

            F₂ ( -2.2 +5.3) = W (3.1 - 2.2)

            F₂ = 704  0.9 / 3.1            

            F₂ = 204.39 N

This is the force that the partner supports

we look for F1

            F₁ = W-F₂

            F₁ = 704 - 204.39

             F₁ = 499.61 N

This is the force that Bubba support

Answer:

(a) 0.014 A

(b) 2 V

Explanation:

(a) Applying

V = IR'...................... Equation 1

Where V = Voltge, I = current, R = total resistance.

make I the subject of the equation

I = V/R'................... Equation 2

From the question,

Given: V = 12 V, R' = (122+232+500) ohms (The resistance are connected in series) = 854 ohms

Substitute these values into equation 2

I = 12/854

I = 0.014 A

(b)  Applying

V' = V(R₁)/(R₁+R₂+R₃)...................... Equation 3

Where V' = Voltage across the 100 ohms resistor.

From the question,

V = 12V, R₁ = 100 ohm, R₂ = 200 ohm, R₃ = 300 ohm.

Substitute these values into equation 3

V' = (12×100)/(100+200+300)

V' = 1200/600

V' = 2 V

Answer:

5.277 kg

Explanation:

Since the formula for work is W = F * d and we are given distance and work, the force on the grocery bag is 88 = F * 1.7 F = 88 / 1.7 = 51.765 N.

We also know that force follows the equation F = m * a. Since the constant gravitational acceleration on earth is 9.81 m / s^2, we can find the mass through 51.765 = m * 9.81 m = 51.765/9.81 = 5.277 kg

Answer:

I don't no the answer sorry

Farenheit is an interesting unit so the conversions get a bit weird.

To convert from Farenheit to Celcius you follow this formula:

T(°C) = (T(°F) - 32) × 5/9

It is also easier to convert to Celcius to Kelvin shown in this formula:

T(K) = T(°C) + 273.15

To Answer your question:

105° F = 40.5°C

-25° F = -31.7°C

105° F = 313.65K

-25° F = 241.45K

Answer:

[tex]0.01\ \text{A}[/tex]

[tex]0.0015\ \text{C}[/tex]

[tex]0.0608\ \text{s}[/tex]

Explanation:

[tex]V_0[/tex] = Voltage = 1.5 V

[tex]C[/tex] = Capacitance = [tex]1000\ \mu\text{F}[/tex]

[tex]R[/tex] = Resistance = [tex]150\ \Omega[/tex]

Current is given by

[tex]I=\dfrac{V_0}{R}\\\Rightarrow I=\dfrac{1.5}{150}\\\Rightarrow I=0.01\ \text{A}[/tex]

Current flowing in the resistor is [tex]0.01\ \text{A}[/tex].

Charge is given by

[tex]Q=CV\\\Rightarrow Q=1000\times 10^{-6}\times 1.5\\\Rightarrow Q=0.0015\ \text{C}[/tex]

The charge on the capacitor is [tex]0.0015\ \text{C}[/tex].

Voltage is given by

[tex]V=V_0e^{-\dfrac{t}{RC}}\\\Rightarrow t=-RC\ln\dfrac{V}{V_0}\\\Rightarrow t=-150\times 1000\times 10^{-6}\times\ln\dfrac{1}{1.5}\\\Rightarrow t=0.0608\ \text{s}[/tex]

Time taken to reach 1 V is [tex]0.0608\ \text{s}[/tex].

Answer:

c

Explanation:

units c mark me brainless k

Answer:

Explanation:

(a) Electron Emission (Beta- decay):

When an unstable nucleus decays by the emission of Beta- particle, its charge number ‘Z’ increases by 1 but, its mass number ‘A’ remains unchanged. The transformation is represented by the equation:  

[tex]_zX^A\ -------->\ _{Z+1}Y^A\ +\ _{-1}e^0[/tex]

It is called ‘Negative Beta Decay’. It is more common than alpha decay.

Example:

[tex]_6C^{14}\ -------->\ _{7}N^{14}\ +\ _{-1}e^0[/tex]

Note:  

There are no electrons in a nucleus so, with the emission of a particle, one of the neutrons is converted to a proton and an electron.

[tex]_0n^1\ --------->\ _1P^1\ +\ _{-1}e^0[/tex]

(b) Positron Emission (Beta+ decay):

When an unstable nucleus decays by the emission of the positron, its charge number ‘Z’ decreases by 1 but,  its mass number ‘A’ remains unchanged. The transformation is represented by the equation:

[tex]_zX^A\ -------->\ _{Z-1}Y^A\ +\ _{+1}e^0[/tex]

Examples:

[tex]_{15}P^{30}\ -------->\ _{14}Si^{30}\ +\ _{+1}e^0[/tex]

Note:  

Inside the nucleus, only a proton can be transformed into a neutron with the emission of a positron (anti-particle of electron)

[tex]_1P^1\ -------->\ _0n^1\ +\ _{+1}e^0[/tex]

Answer:

law of Action and Reaction     F = 2250 N

Explanation:

The tractor and the trailer are two bodies that interact, therefore, by the law of Action and Reaction, the force that one applies on the other is equal to the force that the second body (trailer) applies on the first (tractor), but with opposite direction

          F = 2250 N

directed from trailer to tractor

Answer:

A)    E / B = 2.99 10⁸  V/ mT,  B)     B = 2.50 10⁻⁶ T

Explanation:

A) the expressions for the energy densities are:

         u_E = ½ ε₀ E²

         u_B = ½ B² /μ₀

indicate that the two densities are equal

         ½ ε₀ E² = ½ B² /μ₀  

         E / B =  1 /[tex]\sqrt{\epsilon_o \ \mu_o }[/tex]

we calculate

         E / B = 1 / √( 8.85 10⁻¹² 4π 10⁻⁷

         E / B = 1 /√( 11.1212 10⁻¹⁸)

         E / B = 0.29986 10⁹9

         E / B = 2.99 10⁸  V/ mT

B) for this case E = 750 V / m, ask the magnetic field

         E / B =  1 /[tex]\sqrt{\epsilon_o \ \mu_o}[/tex]

         B = E [tex]\sqrt{\epsilon_o \ \mu_o}[/tex]

we calculate

        B = 750 √(8.85 10⁻¹² 4π 10⁻⁷

        B = 750 3.3348 10⁻⁹

        B = 2.50 10⁻⁶ T

Energy densities can be expressed as

        u_E = ½ ε₀ E²

        u_B = ½ B² /μ₀

When the two densities are equal we use the formula

        ½ ε₀ E² = ½ B² /μ₀  

             E / B =  1 / μ₀  

             E / B = 1 / √( 8.85 10⁻¹² 4π 10⁻⁷)

             E / B = 1 /√( 11.1212 10⁻¹⁸)

             E / B = 0.29986 10⁹

             E / B = 2.99 10⁸  V/ mT

      E / B =  1 / √ ξ₀ μ₀

       B = E √ ξ₀ μ₀

We substitute the values into the equation

        B = 750 √(8.85 10⁻¹² 4π 10⁻⁷)

        B = 750 3.3348 10⁻⁹

        B = 2.50 10⁻⁶ T

Read more on https://brainly.com/question/23360987

Explanation:

Given

mass of the rock is 10 kg

Force requires to hold the rock is equal to its weight

Weight is given by the product of mass and acceleration due to gravity

Weight on the earth surface

[tex]\Rightarrow W_e=10\times 9.8\\\Rightarrow W_e=98\ N[/tex]

Weight on the moon surface

[tex]\Rightarrow W_m=1.6\times 10\\\Rightarrow W_m=16\ N[/tex]

So, the force holding the rock on earth is approximately 6 times the force on the moon.

Answer:

No

Explanation:

because there is no pressure

Answer:

The mass of the man is 71 kg

Explanation:

Given;

kinetic energy of the man, K.E = 887.5 J

velocity of the man, v = 5 m/s

The mass of the man is calculated as follows;

K.E = ¹/₂mv²

where;

m is the mass of the man

2K.E = mv²

m = 2K.E / v²

m = (2 x 887.5) / (5)²

m = 71 kg

Therefore, the mass of the man is 71 kg

Answer:

The difference between the positions of the two events as measured in = 3.53 *10^8 m/s

Explanation:

As we know -

[tex]\Delta x = -\gamma \mu\Delta t[/tex]

Here,

[tex]\Delta x[/tex] is the difference between the positions of the two events as measured in S^

[tex]\gamma[/tex] [tex]= \frac{1}{\sqrt{1-\frac{\mu^2}{c^2} } }[/tex]

And

[tex]\mu[/tex] = 0.547 c

Substituting the given values in above equation, we get -

[tex]\Delta x = (0.547 c)*\frac{1}{\sqrt{1-\frac{\mu^2}{c^2} } }*2.15\\\Delta x = (0.547 c)*\frac{1}{\sqrt{1-\frac{(0.547 c)^2}{c^2} } }*2.15\\\Delta x = (0.547 *3*10^8)*\frac{1}{\sqrt{(1-\(0.547 )^2 } }*2.15\\\Delta x = 3.53 *10^8[/tex]meter per second

Answer:

1568 J

Explanation:

From the question given above, the following data were obtained:

Mass (m) = 80 Kg

Distance (d) = 5 m

Coefficient of kinetic friction (μ) = 0.4

Workdone (Wd) =?

Next, we shall determine the normal reaction. This can be obtained as follow:

Mass (m) = 80 Kg

Acceleration due to gravity (g) = 9.8 m/s²

Normal reaction (N) =?

N = mg

N = 80 × 9.8

N = 784 N

Next, we shall determine force of friction. This can be obtained as follow:

Coefficient of kinetic friction (μ) = 0.4

Normal reaction (N) = 784 N

Force of friction (F) =?

F = μN

F = 0.4 × 784

F = 313.6 N

Finally, we shall determine the work done. This can be obtained as follow:

Distance (d) = 5 m

Force of friction (F) = 313.6 N

Workdone (Wd) =?

Wd = F × d

Wd = 313.6 × 5

Wd = 1568 J

Thus, the workdone is 1568 J

Answer:

I think its d matter accreted from a companion star unstable ignites on the surface of a white dwarf.

Answer:

True?

Explanation:

Genotype and phenotype are two fundamental terms in the science of genetics. ... An organism's genotype is the set of genes in its DNA responsible for a particular trait. An organism's phenotype is the physical expression of those genes.

The term "phenotype" refers to the observable physical properties of an organism; these include the organism's appearance, development, and behavior. An organism's phenotype is determined by its genotype, which is the set of genes the organism carries, as well as by environmental influences upon these genes.

This is why I think it's true. You don't have to read the rest,but that's why I know.

Answer:

The answer is true

Explanation:

I dont know but trust is probably true.

Answer:

6.69 kg

Explanation:

Given that:

The mass of the athlete = 89.2 kg

Height = 1.85 m

Resting HR = 47 bpm

BMI = ~26

The essence of the Wingate anaerobic test is to evaluate and access how much energy is released by phosphagen breakdown and anaerobic glycolysis. By applying a fixed amount of weight to that same cycle's weight tray, tension is added to the flywheel. The flywheel resistance is usually set at 0.075 kg per kilogram of body weight.

For athlete sprinters, the resistance is usually set at 1.0 kg which is then multiplied by the body mass up to 1.3 kg

Hence, the amount of the resistance applied for the athlete  = (0.075 × 89.2)kg

= 6.69 kg

Answer:

v = 4.233  m/s

Explanation:

By applying the rate of boiling from [tex]Q= mL_v[/tex];

the rate of the boiling can be described as:

[tex]\mathcal{P} = \dfrac{Q}{\Delta t} \\ \\ \mathcal{P} = \dfrac{mL_v}{\Delta t}[/tex]

The mode of the steam (water vapor) as an ideal gas can be illustrated by formula:

[tex]P_oV_o = nRT[/tex] --- (1)

where;

n = number of moles;

[tex]n = \dfrac{mass (m)}{Molar mass (M)}[/tex]

Then; equation (1) can be rewritten as:

[tex]P_oV_o = (\dfrac{m}{M}) RT \\ \\ \dfrac{P_oV}{\Delta T} = \dfrac{m}{\Delta t} ( \dfrac{RT}{M})[/tex]

∴

[tex]\dfrac{m}{\Delta t} = \dfrac{\mathcal{P}}{L_v}[/tex]

Then:

[tex]P_o \times A \times v= \dfrac{\mathcal{P}}{L_v}\Big ( \dfrac{RT}{M }\Big)[/tex]

making (v) the subject of the formula:

[tex]v= \Big ( \dfrac{\mathcal{P} RT}{M\times L_v \times P_o \times A }\Big)[/tex]

Given that:

[tex]\mathcal{P}[/tex] = 0.90  kW = 900 W

R(rate constant) = 8.314 J/mol.K

Temperature at 100° C = 373K

molar mass= 18.015 g/mol ≅ 0.0180 kg/mol

Latent heat of vaporisation [tex]L_v[/tex] = 2.26 × 10⁶ J/kg

Atmospheric pressure [tex]P_o = 1.013 \times 10^6 \ N/m^2[/tex]

Cross sectional area A =1.60 cm² = 1.60 × 10⁻⁴ m²

[tex]v= \Big ( \dfrac{900 W (8.314 \ J/mol.K)(373)}{0.0180 \ kg/mol) (2.26 \times 10^6 \ J/kg) (1.013 \times 10^5 \ N/m^2)(1.60 \times 10^{-4} \ m^2)}\Big)[/tex]

v = 4.233  m/s

Answer:

A stormy ocean, having the most kinetic energy means it moves more and faster