solve this with figure.help me ......​

Answer:

The marginal revenue is positive when Roland sells 144 driers per week ( D

Explanation:

Elasticity of demand = -0.463

P = $127

116 hair dryers sold per week ( q )

demand function is differentiable

prove  the true statement below

Revenue = price per unit * quantity

determine the Marginal revenue = dr / dq

∴ MR = P + q dp/dq

price elasticity of demand E(p) = - dq / dp * p/q

where : E ( p ) = - 0.463  , q = 116 , p = $127

q * dp/dq = P / 0.463

∴ MR = P * 1.463 / 0.463

         = 127 * 3.16 = $401.32

since the Marginal revenue > 0 HENCE The marginal revenue will be  positive when Roland sells 144 driers per week.

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Force is defined as the rate of change of momentum. For an unchanging mass, this is equivalent to mass x acceleration. So, 1 N = 1 kg m s-2, or 1 kg m/s2.

Answer:

I don't know this, sorry

Answer:

All electromagnetic radiation, of which radio waves and X-rays are examples, travels at the speed c in a vacuum. The only difference between the two is that the frequency of X-rays is very much higher than radio waves

Answer:

medium

Explanation:

I am not really sure

Answer:

y₀ = 10.625 m

Explanation:

For this exercise we will use the kinematic relations, where the upward direction is positive.

         y = y₀ + v₀ t - ½ g t²

in the exercise they indicate the initial velocity v₀ = 8 m / s.

when the rock reaches the ground its height is zero

         0 = y₀ + v₀ t - ½ g t²

        y₀i = -v₀ t + ½ g t²

let's calculate

         y₀ = - 8  2.5 + ½  9.8  2.5²

         y₀ = 10.625 m

Answer:

12.4m

Explanation:

Given

Length of string = 6.2m

L = λ/2 (for natural frequency (first harmonic)

λ = 2L

λ  = 2(6.2)

λ = 12.4m

Hence the wavelength of the produced standing wave is 12.4m

Answer:

Cumulus, Stratus, and Cirrus. There are three main cloud types.

Explanation:

hopes it help^_^

Answer:

I study physics too, want me to study with you?

Answer:

Wind is a renewable energy source. Overall, using wind to produce energy has fewer effects on the environment than many other energy sources. Wind turbines do not release emissions that can pollute the air or water (with rare exceptions), and they do not require water for cooling.

Explanation:

Answer:

When mass increases, kinetic energy also increases. Changing mass doesn’t affect his speed on a given ramp.

Explanation:

Answer:

When mass increases, kinetic energy also increases. Changing mass doesn’t affect his speed on a given ramp.

Explanation:

pluto

Answer:

25.2 ml

Explanation:

When the aluminium block is inserted in the container, the overall amount of only the water in the beaker can equal V o.

The weight of the water expelled by the plastic container should be equal to the weight of the aluminium block, according to the buoyancy balance relation.

i.e.

[tex]\rho_w V_wg = m_{Al}g = \rho_{Al}V_{Al} g \\ \\ V_w = \dfrac{\rho_{Al}V_{AL}}{\rho_{w}}[/tex]

When the aluminium block is inserted into the plastic container, the initial volume of water = 60 ml

[tex]V_i = V_o + V_w[/tex]

[tex]V_i = V_o + \dfrac{\rho_{Al}V_{Al}}{\rho_{w}}---(1)[/tex]

When the aluminium block is placed outside the container, the volume of the water

[tex]V_f = V_o +V_{Al} ---(2)[/tex]

By subtracting equation (1) and (2)

[tex]V_i -V_f = V_o + \dfrac{\rho_{Al} V_{Al}}{\rho_w}- ( V_o + V_{Al}}) \\ \\ =\dfrac{\rho _{Al}V_{Al}}{\rho_w}-V_{Al} \\ \\ = V_{Al} \Big( \dfrac{\rho_{Al}}{\rho_{w}}-1 \Big)[/tex]

since;

[tex]m_{Al} = 40 g[/tex]

[tex]V _{Al} = \dfrac{40 \ g}{2.7 \ g/cm^3} \\ \\ V_{Al} = 14.815 \ cm^3[/tex]

Similarly;

[tex]\dfrac{\rho_{Al}}{\rho_{w}}= \dfrac{2.7 }{1.0}[/tex]

= 2.7

[tex]V_i -V_f =14.815\Big( 2.7-1 \Big) \\ \\ V_i -V_f = 25.1855 \ ml \\ \\ = \mathbf{25.2 \ ml}[/tex]

Answer:

[tex]8.16\ \text{m/s}[/tex]

Explanation:

[tex]d_1[/tex] = Initial diameter = 4 cm

[tex]v_1[/tex] = Initial velocity = 1 m/s

[tex]d_2[/tex] = Final diameter = 7.8 m

[tex]v_2[/tex] = Final velocity

[tex]A[/tex] = Area = [tex]\pi\dfrac{d^2}{4}[/tex]

From the continuity equation we get

[tex]A_1v_1=A_2v_2\\\Rightarrow \pi\dfrac{d_1^2}{4}v_1=\pi\dfrac{d_2^2}{4}v_2\\\Rightarrow v_2=\dfrac{d_1^2}{d_2^2}v_1\\\Rightarrow v_2=\dfrac{4^2}{1.4^2}\times 1\\\Rightarrow v_2=8.16\ \text{m/s}[/tex]

The speed of water at the second floor is [tex]8.16\ \text{m/s}[/tex].

An action which could increase the pair's angular velocity is the figure skater on the left pointing his right arm down instead of up.

Angular velocity simply refers to the rate of change of angular displacement of a physical object (body) with respect to time.

This ultimately implies that, angular velocity is a measure of how fast and quickly a physical object (body) rotates with respect to another point or how its angular displacement (position) changes with respect to time.

In this scenario, an action which could increase the pair's angular velocity is when the figure skater on the left points his right arm down instead of up.

Read more on angular velocity here: https://brainly.com/question/6860269

#SPJ1

the skater on the left pulls the skater on the right closer to him

Explanation:

apex bliches

Answer:

The angular speed of the ball in radians per second is 5.55 rad/s.

Explanation:

Given;

mass of the ball, m = 1.8 kg

number of the ball's rotation per minute, n = 53 RPM

The angular speed of the ball in radians per second is calculated as follows;

[tex]\omega = 53\frac{rev}{\min} \times \frac{2\pi \ rad}{1 \ rev} \times \frac{1\min}{60 \ s } \\\\\omega = 5.55 \ rad/s[/tex]

Therefore, the angular speed of the ball in radians per second is 5.55 rad/s.

Answer:

[tex]Q_net=534.67\frac{w}{m}[/tex]

Explanation:

From the question we are told that:

Steam line diameter [tex]D=89mm \approx 0.089m[/tex]

Surface emissivity [tex]\mu=0.8[/tex]

Steam temp [tex]T_s=200\textdegree C \approx 200+273=473k[/tex]

Surrounding temp [tex]T_a=20 \textdegree C \aprrox 20+273= 293k[/tex]

Generally the equation for heat loss per unit length due to radiation [tex]Q_{net}[/tex] is mathematically given by

[tex]Q_net=\sigma*\mu>(\pi *d)*(T_s^4-T_a^4)[/tex]

[tex]Q_net=5.6*10^8*0.8*(\pi *0.089)*(473^4-293^4)[/tex]

[tex]Q_net=534.67\frac{w}{m}[/tex]

Answer:

F1 = K Q1 Q2 / R1^2

F2 = K Q1 / 2 * Q2 / (2 R1)^2

F2 / F1 = 1/2 / 4 = 1/8

The new force is 5N   (1/2 due to charge and 1/4 due to distance)

Answer:

C

Explanation:

The said wave takes the shortest time to move/get transmitted

Answer:

80

Explanation:

800/10

=80

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Answer:

the difference in path lengths from each of the slits to the location of the center of a fifth-order bright fringe on the screen is 28 × 10⁻⁷ m

Explanation:

Given the data in the question;

slit separation d = 0.165 mm = 0.165 × 10⁻³ m

wavelength λ = 560 nm = 560 × 10⁻⁹ m

distance between the screen and slits D = 4.05 m

now,

for fifth-order bright fringe path difference = mλ

where m is 5

so, the difference in path lengths from each of the slits will be;

Δr = mλ

we substitute

Δr = 5( 560 × 10⁻⁹ m )

Δr = 28 × 10⁻⁷ m

Therefore, the difference in path lengths from each of the slits to the location of the center of a fifth-order bright fringe on the screen is 28 × 10⁻⁷ m

Answer:

Centripetal acceleration = 3.125 m/s²

Explanation:

Given the following data;

Radius, r = 50 m

Velocity, V = 12.5 m/s

To find the centripetal acceleration;

Centripetal acceleration = Velocity²/radius

Centripetal acceleration = 12.5²/50

Centripetal acceleration = 156.25/50

Centripetal acceleration = 3.125 m/s²

Therefore, the centripetal acceleration of the greyhound running on a circular track is 3.125 meters per seconds square.

Answer:

- the power rating of the resistance heater is 24139.5 W

- the inner surface temperature of the pipe at the exit is 96.34°C

Explanation:

Given the data in the question;

Flow rate of water in the tube V" = 5L/min = 8.333 × 10⁻⁵ m³/s        

The water is to be heated from 10°C to 80°C;

so Average or mean temperature [tex]T_{avg[/tex] will be;

[tex]T_{avg[/tex] = (T₁ + T₂) / 2 = (10 + 80) / 2 = 90/2 = 45°C

Now, from the Table " Properties of Water " at average temperature;

at [tex]T_{avg[/tex] = 45°C

density p = 990.1 kg/m³

specific heat [tex]C_p[/tex] = 4180 J/kg-k

thermal conductivity k = 0.637 W/m-°C

Now, we determine the mass flow;

m" = pV"

we substitute

m" = 990.1 × 8.333 × 10⁻⁵

m" = 0.08250 kg/s

we know that the power rating of the resistance heater is equal to the heat transfer rate to the water;

Q' = m"[tex]C_p[/tex]( T₂ - T₁ )

we substitute

Q' = (0.08250 × 4180 ) ( 80 - 10 )

Q' =  344.85 × 70

Q' = 24139.5 W

Hence, the power rating of the resistance heater is 24139.5 W

Next, we determine the average velocity of water in the tube;

[tex]V_{avg[/tex] = V" / [tex]A_c[/tex]

[tex]V_{avg[/tex] = V" / ( [tex]\frac{1}{4}[/tex]πD² )

given that;  flows through a 2-cm-internal-diameter; D = 0.02 m

we substitute

[tex]V_{avg[/tex] = (8.333 × 10⁻⁵) / ( [tex]\frac{1}{4}[/tex]π × (0.02)² )

[tex]V_{avg[/tex] = (8.333 × 10⁻⁵) / ( 3.14159 × 10⁻⁴ )

[tex]V_{avg[/tex] =  0.265 m/s

Also, from table " saturated water property table "

At 45°C

viscosity μ = 0.596 × 10⁻³ kg/m-s

Prandtl number Pr = 3.91

Now, we determine the kinematic viscosity

v = μ / p

we substitute

v = ( 0.596 × 10⁻³ ) / 990.1

v = 6.01959 × 10⁻⁷ m²/s

so, Reynolds number in the flow region will be;

Re = ([tex]V_{avg[/tex] × D) / v

we substitute

Re = ( 0.265 × 0.02) / (6.01959 × 10⁻⁷)

Re = 8804.586

we can see that our Reynolds number (  8804.586 ) more than 2300 and less than 10,000.

Hydraulic and thermal entry length are equal in this flow region,

such that;

[tex]L_h[/tex] = [tex]L_t[/tex]

⇒ 10 × D = 10 × 0.02 = 0.2 m

we can see that the entry length ( 0.2 m ) is smaller than the given length ( 13 m ) in the question; the flow is a turbulent flow.

So we the Nuddelt number

Nu = [tex]0.023Re^{0.8} Pr^{0.4[/tex]

Nu = 0.023 × [tex]8804.586^{0.8[/tex] × [tex]3.91^{0.4[/tex]

Nu = 56.8

Hence, the heat transfer coefficient h will be;

h = [tex]\frac{k}{D}[/tex] × Nu

we substitute

h = [tex]\frac{0.637}{0.02}[/tex] × 56.8

h = 31.85 × 56.8

h = 1809.1 W/m²-°C

Now, area of the heat transfer will be

A[tex]_s[/tex] = πDL

we substitute

A[tex]_s[/tex] = π × 0.02 × 13

A[tex]_s[/tex] = 0.8168 m²

Finally we determine the inner temperature of the pipe at exit. using the relation;

Q' = hA[tex]_s[/tex]( T₃ - T₂ )

we substitute

24139.5 = 1809.1 × 0.8168( T₃ - 10 )

24139.5 = 1477.67288( T₃ - 80 )

24139.5 = 1477.67288T₃ - 118213.8304

24139.5 + 118213.8304 = 1477.67288T₃

1477.67288T₃ = 142353.3304

T₃ = 142353.3304 / 1477.67288T

T₃ = 96.34°C

Therefore, the inner surface temperature of the pipe at the exit is 96.34°C

Answer:

a

Explanation:

a

Answer:

1

Explanation:

Answer:

True

Explanation:

Answer:

your answer is true hope this helps

Answer:

Use some Glue.. . . Aluminium is not magnetic, so your magnet can't sticl to aluminium without using Glue. You cannot get a magnet to stick to aluminum, no matter how hard you try. Aluminum is diamagnetic, which means it repels a magnetic field.

Explanation:

Hope It Help

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