A hand dryer blows heated air downwards out of the exit duct at a velocity of 4 m/s. The temperature and density of the ambient air at the inlet are 15 C and 1.23 kg/m3, while at the outlet it has temperature 35 C and density 1.15 kg/m3 The blower power is 10.0 W and the heater power is 715 W. Consider the inlet to be at the large mass of ambient air which has negligible velocity.
a) What is the pressure at the outlet? 4 m/s, 35 C
b) You will be applying the energy equation. Why can you ignore any height differences in this situation?
c) If the specific heat of air C-1000 J/(kg K), where Δυ-C Δ T, find the change in internal energy per unit mass from the inlet to outlet.
d) Find the mass flow rate through the dryer.
e) What is the power loss in the system?
f) What is the loss in the system?
g) What is the head loss in the system?
h) What is the total loss coefficient of the system, referred to the outlet velocity?
i) If there were no heater, would the temperature of gas at the outlet be higher, the same, or lower than the inlet? Explain why.

Answer:

because they are found freely in nature uncombined so they are highly reactive with other elements

Answer:

[tex]1.62\times 10^{-8}\ \text{s}[/tex]

Explanation:

[tex]\epsilon_0[/tex] = Vacuum permittivity = [tex]8.854\times 10^{-12}\ \text{F/m}[/tex]

[tex]A[/tex] = Area = [tex]10\times 2\times 10^{-4}\ \text{m}^2[/tex]

[tex]d[/tex] = Distance between plates = 1 mm

[tex]V_c[/tex] = Changed voltage = 60 V

[tex]V[/tex] = Initial voltage = 100 V

[tex]R[/tex] = Resistance = [tex]1000\ \Omega[/tex]

Capacitance is given by

[tex]C=\dfrac{\epsilon_0A}{d}\\\Rightarrow C=\dfrac{8.854\times 10^{-12}\times 10\times 2\times 10^{-4}}{1\times 10^{-3}}\\\Rightarrow C=1.7708\times 10^{-11}\ \text{F}[/tex]

We have the relation

[tex]V_c=V(1-e^{-\dfrac{t}{CR}})\\\Rightarrow e^{-\dfrac{t}{CR}}=1-\dfrac{V_c}{V}\\\Rightarrow -\dfrac{t}{CR}=\ln (1-\dfrac{V_c}{V})\\\Rightarrow t=-CR\ln (1-\dfrac{V_c}{V})\\\Rightarrow t=-1.7708\times 10^{-11}\times 1000\ln(1-\dfrac{60}{100})\\\Rightarrow t=1.62\times 10^{-8}\ \text{s}[/tex]

The time taken for the potential difference to reach the required level is [tex]1.62\times 10^{-8}\ \text{s}[/tex].

Answer:

The force of gravity is not the same as being on the earth. when your on the earth there no gravitational pull its all up to the air

Explanation:

No explanation

Answer:

The photosphere is the visible "surface" of the sun. So your answer would be C.

Explanation: its right

Answer:

The second bulb will have thicker filament

Explanation:

Given;

First electric bulb: Power, P₁ = 100 W and Voltage, V₁ = 220 V

Second electric bulb: Power, P₂ = 200 W and Voltage, V₂ = 200 V

Resistivity of tungsten, ρ = 4.9 x 10⁻⁸ ohm. m

Resistance of the first bulb:

[tex]P = IV = \frac{V}{R} .V = \frac{V^2}{R} \\\\R = \frac{V^2}{P} \\\\R_1 = \frac{V_1^2}{P_1} = \frac{(220)^2}{100} = 484 \ ohms[/tex]

Resistance of the second bulb:

[tex]R_2 = \frac{V_2^2}{P_2} = \frac{(200)^2}{200} = 200 \ ohms[/tex]

Resistivity of the tungsten filament is given by the following equation;

[tex]\rho = \frac{RA}{L}[/tex]

where;

L is the length of the filament

R is resistance of each filament

A is area of each filament

[tex]A = \pi r^2[/tex]

where;

r is the thickness of each filament

[tex]\rho = \frac{R (\pi r^2)}{L} \\\\\frac{\rho L}{\pi} = Rr^2 \\\\Recall ,\ \frac{\rho L}{\pi} \ is \ constant \ for \ both \ filaments\\\\R_1r_1^2 = R_2r_2^2\\\\(\frac{r_1}{r_2} )^2 = \frac{R_2}{R_1} \\\\\frac{r_1}{r_2} = \sqrt{\frac{R_2}{R_1} } \\\\\frac{r_1}{r_2} = \sqrt{\frac{200}{484} } \\\\\frac{r_1}{r_2} = 0.64\\\\r_1 = 0.64 \ r_2\\\\r_2 = 1.56 \ r_1[/tex]

Therefore, the second bulb will have thicker filament

Answer:

20.0N

Becuase It's the largest

Answer:

20.0

Explanation:

It's the biggest number

Answer:

Approximately [tex]261\; \rm K[/tex], if this gas is an ideal gas, and that the quantity of this gas stayed constant during these changes.

Explanation:

Let [tex]P_1[/tex] and [tex]P_2[/tex] denote the pressure of this gas before and after the changes.

Let [tex]V_1[/tex] and [tex]V_2[/tex] denote the volume of this gas before and after the changes.

Let [tex]T_1[/tex] and [tex]T_2[/tex] denote the temperature (in degrees Kelvins) of this gas before and after the changes.

Let [tex]n_1[/tex] and [tex]n_2[/tex] denote the quantity (number of moles of gas particles) in this gas before and after the changes.

Assume that this gas is an ideal gas. By the ideal gas law, the ratios [tex]\displaystyle \frac{P_1 \cdot V_1}{n_1 \cdot T_1}[/tex] and [tex]\displaystyle \frac{P_2 \cdot V_2}{n_2 \cdot T_2}[/tex] should both be equal to the ideal gas constant, [tex]R[/tex].

In other words:

[tex]R = \displaystyle \frac{P_1 \cdot V_1}{n_1 \cdot T_1}[/tex].

[tex]R =\displaystyle \frac{P_2 \cdot V_2}{n_2 \cdot T_2}[/tex].

Combine the two equations (equate the right-hand side) to obtain:

[tex]\displaystyle \frac{P_1 \cdot V_1}{n_1 \cdot T_1} = \frac{P_2 \cdot V_2}{n_2 \cdot T_2}[/tex].

Rearrange this equation for an expression for [tex]T_2[/tex], the temperature of this gas after the changes:

[tex]\displaystyle T_2 = \frac{P_2}{P_1} \cdot \frac{V_2}{V_1} \cdot \frac{n_1}{n_2} \cdot T_1[/tex].

Assume that the container of this gas was sealed, such that the quantity of this gas stayed the same during these changes. Hence: [tex]n_2 = n_1[/tex], [tex](n_2 / n_1) = 1[/tex].

[tex]\begin{aligned} T_2 &= \frac{P_2}{P_1} \cdot \frac{V_2}{V_1} \cdot \frac{n_1}{n_2}\cdot T_1 \\[0.5em] &= \frac{1.67\; \rm atm}{1.12\; \rm atm} \times \frac{11.2\; \rm m^{3}}{15.7\; \rm m^{3}} \times 1 \times 245\; \rm K \\[0.5em] &\approx 261\; \rm K\end{aligned}[/tex].

Answer:

Answer: Oxygen

Explanation:

Oxygen would not be considered as a pollutant

Answer:

Hey there the answer is C. Oxygen

Explanation:

Oxygen is the chemical element with the symbol O and atomic number 8. It is a member of the chalcogen group in the periodic table, a highly reactive nonmetal, and an oxidizing agent that readily forms oxides with most elements as well as with other compounds. After hydrogen and helium, oxygen is the third-most abundant element in the universe by mass. At standard temperature and pressure, two atoms of the element bind to form dioxygen, a colorless and odorless diatomic gas with the formula O 2. Diatomic oxygen gas constitutes 20.95% of the Earth's atmosphere. Oxygen makes up almost half of the Earth's crust in the form of oxides. Hope this helps! Have a great day!

Answer:

The border is 8km above sea level.

Explanation:

We know that:

Density = 1.25 kg/m^3

Pressure = 10^5 N/m^2

g = 10m/s^2

Now, suppose that we have a virtual rectangle, such that its bases have an area of 1m^2 and the rectangle has a height equal to H.

This virtual figure has a volume V = 1m^2*H, and it is filled with air (which we know that has a density 1.25 kg/m^3)

Then the total mass inside that volume is:

M = (1.25 kg/m^3)*V = (1.25 kg/m^3)*(1m^2*H)

The weight of this mass is:

W = g*M = (10m/s^2)*(1.25 kg/m^3)*(1m^2*H)

And if we divide the weight in a given surface, let's say 1 m^2, we get the pressure per square meter, which we know is equal to  10^5 N/m^2

then:

P = 10^5 N/m^2 = (10m/s^2)*(1.25 kg/m^3)*(1m^2*H)*(1/m^2)

Whit this equation we can find the value of H.

10^5 N/m^2 = (10m/s^2)*(1.25 kg/m^3)*(1m^2*H)*(1/m^2)

10^5 N =  (10m/s^2)*(1.25 kg/m^3)*(1m^2*H)

(10^5 N)/(10 m/s^2) = (1.25 kg/m^3)*(1m^2*H)

(10^4 kg) = (1.25 kg/m^3)*(1m^2*H)

(10^4 kg)/( 1.25 kg/m^3) = 1m^2*H

8,000 m^3 = 1m^2*H

(8,000 m^3)/(1m^2) =H

8,000 m = H

And we want this answer in km, knowing that 1,000m = 1km

8,000m = 8km = H

The border is 8km above sea level.

Height of boundaries is 8.2 km

Normal density = 1.25 kg/m³

1 atm = 101325 N/m²

Height of boundaries

Pressure = Height × Density × Gravitational acceleration

101325 = Height × 1.25 × 9.8

101325 = Height × 12.25

Height of boundaries = 101325 / 12.25

Height of boundaries = 8271.42 m

Height of boundaries = 8.2 km

https://brainly.com/question/23358029

Answer:

question #1 is A

Question #2 is C

Explanation:

Answer:

1 W = 1 J / sec       Definition of watt is 1 joule / sec

So if a bulb uses 75 J / sec it must use

75 J/s * 60 sec / min = 4500 J/min    energy used by bulb

If bulb is 15% efficient then the light delivered is

P = 4500 J / min * .15 = 675 J / min

Answer:

Most likely (B)

Explanation:

B in the passage is the most representative out of all your choices and it has evidence from the passage  

Hope dis helps Jit!

Sorry i forgot to type C

B and C both measure mass while the others are calculations and are bias

The following experimental set-ups would allow a student to calculate the magnitude of the gravitational field strength using only the quantities measured:

A gravitational field is a model used in physics to explain the effects that a large thing has on the area surrounding it, exerting a force on smaller, less massive bodies.

When  a known mass from a spring scale is hung; by e; measuring the spring scale reading when the mass is at rest, the magnitude of the gravitational field strength ( reading/mass) can be calculated.

When  a heavy metal ball is dropped, by measuring e the drop height and the time it takes the ball to hit the ground, the magnitude of the gravitational field strength ( h = gt²/2) can be calculated. Hence, option (B) and option (D) is correct.

Learn more about  gravitational field here:

https://brainly.com/question/26690770

#SPJ2

Answer:

D:#2 African Plate

Explanation:

The African Plate is flanked by the Red sea rift and the minor African plate.

The Red sea rift is a small part of a greater line of rifts known as the Great African Rift Valley. The rift valley is making several small lakes all through Africa and it will eventually split up the African continent.

The Red sea lift is the divergent boundary between the African plate and the Arabian plate. It means that the two plates are moving apart or spreading apart.

Answer:

[tex]0.842\ \text{lb ft}[/tex]

[tex]0.1052\ \text{lb ft}[/tex]

Explanation:

d = Diameter of wheel = 6 in

r = Radius = 3 in = [tex]\dfrac{3}{12}=0.25\ \text{ft}[/tex]

t = Thickness = [tex]\dfrac{3}{4}=0.75\ \text{in}=\dfrac{0.75}{12}\ \text{ft}[/tex]

w = Specific weight = [tex]425\ \text{lb/ft}^3[/tex]

[tex]t_2[/tex] = Time taken to slow down = 35 s

[tex]t_1[/tex] = Time taken to reach operating speed = 5 s

[tex]\omega[/tex] = Angular velocity = [tex]3450\times \dfrac{2\pi}{60}\ \text{rad/s}[/tex]

Weight is given by

[tex]W=2\pi r^2tw\\\Rightarrow W=2\pi\times 0.25^2\times \dfrac{0.75}{12}\times 425\\\Rightarrow W=10.43\ \text{lbs}[/tex]

Mass is given by

[tex]m=\dfrac{W}{g}\\\Rightarrow m=\dfrac{10.43}{32}\\\Rightarrow m=0.326\ \text{lb}[/tex]

Moment of inertia is given by

[tex]I=\dfrac{mr^2}{2}\\\Rightarrow I=\dfrac{0.326\times 0.25^2}{2}\\\Rightarrow I=0.01019\ \text{lb ft}^2[/tex]

Angular acceleration while slowing down is given by

[tex]\alpha_f=\dfrac{\omega}{t_2}\\\Rightarrow \alpha_f=\dfrac{3450\times \dfrac{2\pi}{60}}{35}\\\Rightarrow \alpha_f=10.32\ \text{rad/s}^2[/tex]

Frictional moment is given

[tex]\tau_f=I\alpha_f\\\Rightarrow \tau_f=0.01019\times 10.32\\\Rightarrow \tau_f=0.1052\ \text{lb ft}[/tex]

Frictional moment is [tex]0.1052\ \text{lb ft}[/tex]

Angular acceleration while speeding up is given by

[tex]\alpha=\dfrac{\omega}{t_1}\\\Rightarrow \alpha=\dfrac{3450\times \dfrac{2\pi}{60}}{5}\\\Rightarrow \alpha=72.26\ \text{rad/s}^2[/tex]

Motor torque is given by

[tex]\tau_m=\tau_f+I\alpha\\\Rightarrow \tau_m=0.1052+0.01019\times 72.26\\\Rightarrow \tau_m=0.842\ \text{lb ft}[/tex]

Motor torque is [tex]0.842\ \text{lb ft}[/tex].

Answer:

10.1 m/s

Explanation:

By Newton's third law, the force on the squid and that due to the water expelled form an action reaction pair.

And by the law of conservation of momentum,

initial momentum of squid + expelled water = final momentum of squid + expelled water.

Now, the initial momentum of the system is zero.

So, 0 = final momentum of squid + expelled water

0 = MV + mv where M = mass of squid = 6.50kg, V = velocity of squid = 2.40m/s, m =mass of water in cavity = 1.55 kg and v = velocity of water expelled

So, MV + mv = 0

MV = -mv

v = -MV/m

= -6.50 kg × 2.40 m/s ÷ 1.55 kg

= -15.6 kgm/s ÷ 1.55 kg

= -10.1 m/s

So, speed must it expel this water to instantaneously achieve a speed of 2.40 m/s to escape the predator is 10.1 m/s

Answer:

239 rpm

Explanation: So the distance covered in one minute is 75,000 centimeters. The diameter of the wheel is 100 cm, so the radius is 50 cm, and the circumference is 100π cm. How many of these circumferences (or wheel revolutions) fit inside the 75,000 cm? In other words, if I were to peel this wheel's tread from the cart and lay it out flat, it would measure a distance of 100π cm. How many of these lengths fit into the entire distance covered in one minute? To find out how many of (this) fit into so many of (that), I must divide (that) by (this), so:

100πcm/rev

75,000cm/min

​750 min rev≈238.7324146RPM

Answer:

A black body radiator is an idealized body that absorbs all incoming electromagnetic radiation (thus the name of "black body").

A black body radiator is an object that has a lot of thermal energy, and it irradiates its thermal energy in the form of black body radiation (thermal radiation emitted by a black body).

a) Then, we could go to the trivial case of a mirror, a mirror is a poor blackbody radiator because a mirror reflects most of the incoming electromagnetic radiation, thus, a mirror is a really bad approximation for a black body, then a mirror is a poor black body radiator.

b) Any electromagnetic wave is a light wave (there exists "light" that we can not see). A black body radiator irradiates energy, and this radiation is in the form of electromagnetic waves, which are in essence, light waves.

Answer:

A black body radiator is an idealized body that absorbs all incoming electromagnetic radiation (thus the name of "black body").

A black body radiator is an object that has a lot of thermal energy, and it irradiates its thermal energy in the form of black body radiation (thermal radiation emitted by a black body).

a) Then, we could go to the trivial case of a mirror, a mirror is a poor blackbody radiator because a mirror reflects most of the incoming electromagnetic radiation, thus, a mirror is a really bad approximation for a black body, then a mirror is a poor black body radiator.

b) Any electromagnetic wave is a light wave (there exists "light" that we can not see). A black body radiator irradiates energy, and this radiation is in the form of electromagnetic waves, which are in essence, light waves.

Explanation:

Answer:

0.54 sec

Explanation:

Answer:

Explanation:

Speed = 14.2 m/s

Distance = 514 m

Time = Distance / Speed

Time = 514 / 14.2

Time = 36.19 seconds

Answer:

A. Increases

Explanation:

As altitude decreases, the amount of gas molecules in the air increases - the air becomes less dense. As altitude increases, the amount of gas molecules in the air decreases—the air becomes less dense.

Answer:

It decreases so it is B

Explanation:

As altitude rises, air pressure drops.

Answer:

B. 3.1 × 10^5 V

Explanation:

Answer:

B

Explanation:

e2021

Answer:

1. Wavelength = 3.2 m

2. Amplitude = 0.6 m

Explanation:

1. Determination of the wavelength.

The wavelength of a wave is defined as the distance between two successive crest. This implies that for every complete vibration, there is one wavelength.

From the diagram given above, we can see that the wave makes 2½ vibrations.

This means that there are 2½ equal wavelength of the wave. Therefore, the wavelength can be obtained as follow:

Length (L) = 8 m

Wavelength (λ) =?

2½ λ = L

5/2 λ = 8

5λ / 2 = 8

Cross multiply

5λ = 2 × 8

5λ = 16

Divide both side by 5

λ = 16 / 5

λ = 3.2 m

Therefore, wavelength of the wave is 3.2 m.

2. Determination of the amplitude.

The amplitude of a wave is defined as the maximum displacement of the wave from the origin.

From the diagram given above, the distance between the maximum and minimum displacement is given as 1.2 m. Thus, we can obtain the amplitude of wave as follow:

Distance between the maximum and minimum displacement (D) = 1.2

Amplitude (A) =?

A = ½D

A = ½ × 1.2

A = 0.6 m

Thus, the amplitude of the wave is 0.6 m

Answer:

a = 52s²

Explanation:

How to find acceleration

Acceleration (a) is the change in velocity (Δv) over the change in time (Δt), represented by the equation a = Δv/Δt. This allows you to measure how fast velocity changes in meters per second squared (m/s^2). Acceleration is also a vector quantity, so it includes both magnitude and direction.

Solve

We know initial velocity (u = 16), velocity (v = 120) and acceleration (a = ?)

We first need to solve the velocity equation for time (t):

v = u + at

v - u = at

(v - u)/a = t

Plugging in the known values we get,

t = (v - u)/a

t = (16 m/s - 120 m/s) -2/s2

t = -104 m/s / -2 m/s2

t = 52 s

Answer:

a)

the quarterback will be moving back at speed of 0.080625 m/s

b)

the distance moved horizontally by the quarterback is 0.0241875 m or 2.41875 cm

Explanation:

Given the data in the question;

a)

How fast will he be moving backward just after releasing the ball?

using conservation of momentum;

m₁v₁ = m₂v₂

v₂ = m₁v₁ / m₂

where m₁ is initial mass ( 0.43 kg )

m₂ is the final mass ( 80 kg )

v₁ is the initial velocity  ( 15 m/s )

v₂ is the final velocity

so we substitute

v₂ = ( 0.43 × 15 ) / 80

v₂ = 6.45 / 80

v₂ = 0.080625 m/s

Therefore, the quarterback will be moving back at speed of 0.080625 m/s

b) Suppose that the quarterback takes 0.30 to return to the ground after throwing the ball. How far d will he move horizontally, assuming his speed is constant?

we make use of the relation between time, distance and speed;

s = d/t

d = st

where s is the speed ( 0.080625 m/s )

t is time ( 0.30 s )

so we substitute

d = 0.080625 × 0.30

d = 0.0241875 m or 2.41875 cm

Therefore, the distance moved horizontally by the quarterback is 0.0241875 m or 2.41875 cm