The largest volume of water lost from the body for a sedentary person is through ___________

Three resistors of equal resistance are connected in series as shown. Compared to the voltage provided by the battery, voltage a

dsp73

Answer:

the voltage across each resistor is one third of the battery voltage

Explanation:

In a series circuit, the current is constant throughout the circuit, so the battery voltage is equal to the sum of the voltage drop in each part of the series circuit.

V = i (R₁ + R₂ + R₃)

in the exercise indicate that all resistance has the same value

R₀ = R₁ = R₂ = R₃

V = i 3 R₀

V = 3 V₀

V₀ = i R₀

V₀= V / 3

the voltage across each resistor is one third of the battery voltage

6 0

3 years ago

A particle initially located at the origin has an acceleration of = 2.00ĵ m/s2 and an initial velocity of i = 8.00î m/s. Find (a

sergejj [24]

a. The particle has position vector

$\vec r(t)=\left(\left(8.00\dfrac{\rm m}{\rm s}\right)t\right)\,\vec\imath+\left(\dfrac12\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)t^2\right)\,\vec\jmath$

$\vec r(t)=\left(8.00\dfrac{\rm m}{\rm s}\right)t\,\vec\imath+\left(1.00\dfrac{\rm m}{\mathrm s^2}\right)t^2\,\vec\jmath$

b. Its velocity vector is equal to the derivative of its position vector:

$\vec v(t)=\vec r'(t)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\vec\imath+\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)t\,\vec\jmath$

c. At $t=7.00\,\mathrm s$ , the particle has position

$\vec r(7.00\,\mathrm s)=\left(8.00\dfrac{\rm m}{\rm s}\right)(7.00\,\mathrm s)\,\vec\imath+\left(1.00\dfrac{\rm m}{\mathrm s^2}\right)(7.00\,\mathrm s)^2\,\vec\jmath$

$\vec r(7.00\,\mathrm s)=\left(56.0\,\vec\imath+49.0\,\vec\jmath\right)\,\mathrm m$

That is, it's 56.0 m to the right and 49.0 m up relative to the origin, a total distance of $\|\vec r(7.00\,\mathrm s)\|=\sqrt{(56.0\,\mathrm m)^2+(49.0\,\mathrm m)^2}=74.4\,\mathrm m$ away from the origin in a direction of $\theta=\tan^{-1}\dfrac{49.0\,\mathrm m}{56.0\,\mathrm m}=41.2^\circ$ relative to the positive $x$ axis.

d. The speed of the particle at $t=7.00\,\mathrm s$ is the magnitude of the velocity at this time:

$\vec v(7.00\,\mathrm s)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\vec\imath+\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)(7.00\,\mathrm s)\,\vec\jmath$

$\vec v(7.00\,\mathrm s)=\left(8.00\,\vec\imath+14.0\,\vec\jmath\right)\dfrac{\rm m}{\rm s}$

Then its speed at this time is

$\|\vec v(7.00\,\mathrm s)\|=\sqrt{\left(8.00\dfrac{\rm m}{\rm s}\right)^2+(14.0\dfrac{\rm m}{\rm s}\right)^2}=16.1\dfrac{\rm m}{\rm s}$

4 0

3 years ago

Why is people asking for segges this is so annoying how do i block them help mee

NikAS [45]

Answer:

woah

Explanation:

report

6 0

3 years ago

A domestic water heater holds 189 L of water at 608C, 1 atm. Determine the exergy of the hot water, in kJ. To what elevation, in

Gekata [30.6K]

A.

The energy of the hot water is 482630400 J

Using Q = mcΔT where Q = energy of hot water, m = mass of water = ρV where ρ = density of water = 1000 kg/m³ and V = volume of water = 189 L = 0.189 m³,

c = specific heat capacity of water = 4200 J/kg-°C and ΔT = temperature change of water = T₂ - T₁ where T₂ = final temperature of water = 608 °C. If we assume the water was initially at 0°C, T₁ = 0 °C. So, the temperature change ΔT = 608 °C - 0 °C = 608 °C

Substituting the values of the variables into the equation, we have

Q = mcΔT

Q = ρVcΔT

Q = 1000 kg/m³ × 0.189 m³ × 4200 J/kg-°C × 608 °C

Q = 482630400 J

So, the energy of the hot water is 482630400 J

B.

The elevation <u>the mass would have to be raised from zero elevation relative to the reference environment for its exergy to equal that of the hot water</u> is 49248 m.

Using the equation for gravitational potential energy ΔU = mgΔh where m = mass of object = 1000 kg, g = acceleration due to gravity = 9.8 m/s² and Δh = h - h' where h = required elevation and h' = zero level elevation = 0 m

Since the energy of the mass equal the energy of the hot water, ΔU = 482630400 J

So, ΔU = mgΔh

ΔU = mg(h - h')

making h subject of the formula, we have

h = h' + ΔU/mg

Substituting the values of the variables into the equation, we have

h = h' + ΔU/mg

h = 0 m + 482630400 J/(1000 kg × 9.8 m/s²)

h = 0 m + 482630400 J/(9800 kgm/s²)

h = 0 m + 49248 m

h = 49248 m

So, the elevation <u>the mass would have to be raised from zero elevation relative to the reference environment for its exergy to equal that of the hot water</u> is 49248 m.

Learn more about heat energy here:

brainly.com/question/11961649

5 0

3 years ago

Calculate how far ahead of the drop zone a pilot would have needed to drop humanitarian aid packages if the delivery occurred at

ozzi

Answer:

S = 11488.42 m

Explanation:

Given,

The speed of the jet, v = 0.74 Mach

= 253.82 m/s

The altitude of the jet, h = 10000 m

Since the jet is travelling horizontally, the vertical component of velocity Vy = 0

The equation for a projectile projected from a height h is given by,

S = Vx [Vy + √(Vy² +2gh)] / g

Since Vy = 0

S = Vx √(Vy² +2gh) / g

= 253.82 x √(2 x 9.8 x 10000) / 9.8

= 11488.42 m

Hence, the humanitarian aid package should be delivered ahead of distance, S = 11488.42 m

6 0

3 years ago

The largest volume of water lost from the body for a sedentary person is through ____________.