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choli [55]
3 years ago
14

How does the excretory system work with the circulatory system to help the body regulate itself?

Physics
1 answer:
Evgesh-ka [11]3 years ago
8 0

Answer:

The excretory system is a close partner with both the circulatory and endocrine system. The circulatory system connection is obvious. Blood that circulates through the body passes through one of the two kidneys. Urea, uric acid, and water are removed from the blood and most of the water is put back into the system.

Explanation:

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A total solar eclipse is a rare event. Although they occur somewhere on earth every 18 months on average, it is estimated that t
Setler79 [48]
Because the tip of the moon's shadow ... the area of "totality" ... is never more than a couple hundred miles across, It never covers a single place for more than 7 minutes, and can never stay on the Earth's surface for more than a few hours altogether during one eclipse.

If you're not inside that small area, you don't see a total eclipse.
3 0
3 years ago
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Merry-go-rounds are a common ride in park playgrounds. The ride is a horizontal disk that rotates about a vertical axis at their
Vera_Pavlovna [14]

Answer:

A = 2.36m/s

B = 3.71m/s²

C = 29.61m/s2

Explanation:

First, we convert the diameter of the ride from ft to m

10ft = 3m

Speed of the rider is the

v = circumference of the circle divided by time of rotation

v = [2π(D/2)]/T

v = [2π(3/2)]/4

v = 3π/4

v = 2.36m/s

Radial acceleration can also be found as a = v²/r

Where v = speed of the rider

r = radius of the ride

a = 2.36²/1.5

a = 3.71m/s²

If the time of revolution is halved, then radial acceleration is

A = 4π²R/T²

A = (4 * π² * 3)/2²

A = 118.44/4

A = 29.61m/s²

7 0
3 years ago
A 3 x 10^-6 C charge is 5m away from a -2x 10^-6 C charge A. Attractive because one is positive the other one is negative B. Det
Marizza181 [45]

Answer:

0.0021576N

Explanation:

F=(k)(q1q2/r^2)

F=(8.99×10^9)(3×10^-6)(2×10^-6)/(5^2)

F=0.0021576N

5 0
2 years ago
What do radio waves and microwaves have in common?
Tamiku [17]

Answer:

I Will say the Answer is A

Explanation:

5 0
3 years ago
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Determine the magnitude of the resultant force acting on a 1.5 −kg particle at the instant t=2 s, if the particle is moving alon
Phoenix [80]

Answer:

F = 63N

Explanation:

M= 1.5kg , t= 2s, r = (2t + 10)m and

Θ = (1.5t² - 6t).

magnitude of the resultant force acting on 1.5kg = ?

Force acting on the mass =

∑Fr =MAr

Fr = m(∇r² - rθ²) ..........equation (i)

∑Fθ = MAθ = M(d²θ/dr + 2dθ/dr) ......... equation (ii)

The horizontal path is defined as

r = (2t + 10)

dr/dt = 2, d²r/dt² = 0

Angle Θ is defined by

θ = (1.5t² - 6t)

dθ/dt = 3t, d²θ/dt² = 3

at t = 2

r = (2t + 10) = (2*(2) +10) = 14

but dr/dt = 2m/s and d²r/dt² = 0m/s

θ = (1.5(2)² - 6(2) ) = -6rads

dθ/dt =3(2) - 6 = 0rads

d²θ/dt = 3rad/s²

substituting equation i into equation ii,

Fr = M(d²r/dt² + rdθ/dt) = 1.5 (0-0)

∑F = m[rd²θ/dt² + 2dr/dt * dθ/dt]

∑F = 1.5(14*3+0) = 63N

F = √(Fr² +FΘ²) = √(0² + 63²) = 63N

7 0
3 years ago
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