The following items are connected in a circuit. in which one would you find a voltage gain?

Astronomers have discovered a new planet called "Xandar" beyond the orbit of Pluto (No, not really but I need a fake planet for

Burka [1]

Answer:

m = 1.82E+23 kg

Explanation:

G = universal gravitational constant = 6.67E-11 N·m²/kg²

r = radius of orbit = 72,600 km = 7.26E+07 m

C = circumference of orbit = 2πr = 4.56E+08 m

P = period of orbit = 12.9 d = 1,114,560 s

v = orbital velocity of satellite Jim = C/P = 409 m/s

m = mass of Xandar = to be determined

v = √(Gm/r)

v² = [√(Gm/r)]²

v² = Gm/r

rv² = Gm

rv²/G = m

m = rv²/G

mG = universal gravitational constant = 6.67E-11 N·m²/kg²

r = radius of orbit = 72,600 km = 7.26E+07 m

C = circumference of orbit = 2πr = 4.56E+08 m

P = period of orbit = 12.9 d = 1,114,560 s

v = orbital velocity of satellite Jim = C/P = 409 m/s

m = mass of Xandar = to be determined

v = √(Gm/r)

v² = [√(Gm/r)]²

v² = Gm/r

rv² = Gm

rv²/G = m

m = rv²/G

m = 1.82E+23 kg

3 0

2 years ago

The Bernoulli equation is valid for steady, inviscid, incompressible flows with a constant acceleration of gravity. Consider flo

irina1246 [14]

Answer:

$p+\frac{1}{2}ρV^{2}+ρg_{0}z-\frac{1}{2}ρcz^{2}=constant$

Explanation:

first write the newtons second law:

F $_{s}$ =δma $_{s}$

Applying bernoulli,s equation as follows:

∑ $δp+\frac{1}{2} ρδV^{2} +δγz=0\\$

Where, $δp$ is the pressure change across the streamline and $V$ is the fluid particle velocity

substitute $ρg$ for {tex]γ[/tex] and $g_{0}-cz$ for $g$

$dp+d(\frac{1}{2}V^{2}+ρ(g_{0}-cz)dz=0$

integrating the above equation using limits 1 and 2.

$\int\limits^2_1 \, dp +\int\limits^2_1 {(\frac{1}{2}ρV^{2} )} \, +ρ \int\limits^2_1 {(g_{0}-cz )} \,dz=0\\p_{1}^{2}+\frac{1}{2}ρ(V^{2})_{1}^{2}+ρg_{0}z_{1}^{2}-ρc(\frac{z^{2}}{2})_{1}^{2}=0\\p_{2}-p_{1}+\frac{1}{2}ρ(V^{2}_{2}-V^{2}_{1})+ρg_{0}(z_{2}-z_{1})-\frac{1}{2}ρc(z^{2}_{2}-z^{2}_{1})=0\\p+\frac{1}{2}ρV^{2}+ρg_{0}z-\frac{1}{2}ρcz^{2}=constant$

there the bernoulli equation for this flow is $p+\frac{1}{2}ρV^{2}+ρg_{0}z-\frac{1}{2}ρcz^{2}=constant$

note: $ρ$ =density(ρ) in some parts and change(δ) in other parts of this equation. it just doesn't show up as that in formular

4 0

2 years ago

A 4.00-kg box sits atop a 10.0-kg box on a horizontal table. The coefficient of kinetic friction between the two boxes and betwe

natta225 [31]

First, we have to calculate the normal forces on different surfaces.The normal force on the 4.00 kg, N1 = (4)(9.8) = 39.2 N. The normal force on the 10.0 kg, N2 = (14)(9.8) = 137.2 N. Looking at the 10.0 kg block, the static forces that counteract the pulling force equals the sum of the friction from the two surfaces. Fc = N1 * 0.80 + N2 * 0.80 = 141.12 N. Since the counter force is less than the pulling force, the blocks start to move and hence, kinetic frictions are considered.

Therefore, f1 = uk * N1 = (0.60)(39.2) = 23.52 N.

4 0

2 years ago

A student hears a police siren. What would change the frequency that the student hears? Check all that apply.

ser-zykov [4K]

<span>A student hears a police siren.

The arithmetic of the Doppler Effect shows that if the distance between
the source and observer is changing, then the observer hears a different
frequency compared to the frequency actually radiating from the source.

Thus the first four choices would cause the student to hear a different
frequency:

-- if the student walked toward the police car
-- if the student walked away from the police car
-- if the police car moved toward the student
-- if the police car moved away from the student

The last two choices wouldn't affect the frequency heard by the student,
since the perceived frequency of a sound doesn't depend on its intensity.

-- if the intensity of the siren increased
-- if the intensity of the siren decreased.</span>

4 0

2 years ago

Read 2 more answers

A flywheel of mass M is rotating about a vertical axis with angular velocity ω0. A second flywheel of mass M/5 is not rotating a

Contact [7]

Answer:

0.83 ω

Explanation:

mass of flywheel, m = M

initial angular velocity of the flywheel, ω = ωo

mass of another flywheel, m' = M/5

radius of both the flywheels = R

let the final angular velocity of the system is ω'

Moment of inertia of the first flywheel , I = 0.5 MR²

Moment of inertia of the second flywheel, I' = 0.5 x M/5 x R² = 0.1 MR²

use the conservation of angular momentum as no external torque is applied on the system.

I x ω = ( I + I') x ω'

0.5 x MR² x ωo = (0.5 MR² + 0.1 MR²) x ω'

0.5 x MR² x ωo = 0.6 MR² x ω'

ω' = 0.83 ω

Thus, the final angular velocity of the system of flywheels is 0.83 ω.

5 0

2 years ago