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spayn [35]
3 years ago
5

Which organ system do blood cells and blood vessels belong to

Physics
2 answers:
Neporo4naja [7]3 years ago
7 0

Answer:

The cardiovascular system

Explanation:

AfilCa [17]3 years ago
6 0
The cardiovascular system
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What is the equivalent resistance of the circuit?
Sonbull [250]

Answer:

80Ω

Explanation:

Resistance in a series circuit (when one resistor is after another in the same line) is calculated with the following equation

Total resistance = R1 + R2 + R3...Rn

Substitute in your values and solve for total resistance

Total resistance = 60 + 20 = 80Ω

8 0
3 years ago
Read 2 more answers
a solid metal sphere of radius 3.00m carries a total charge of -5.50. what is the magnitude of the electric field at a distance
aivan3 [116]

Answer:

(a) Electric field at 0.250 m is zero.

(b)  Electric field at 2.90 m is zero.

(c) Electric field at 3.10 m is - 5.15 x 10³ V/m.

(d) Electric field at 8.00 m is - 0.77 x 10³ V/m.

Explanation:

Let Q and R are the charge and radius of the solid metal sphere. The solid metal sphere behave as conductor, so total charge Q is on the surface of the sphere.

Electric field inside and outside the metal sphere is :

E = 0 for r ≤ R ( inside )

  = \frac{KQ}{r^{2} } for r > R ( outside )

Here K is electric constant and r is the distance from the center of the metal sphere.

(a) Electric field at 0.250 m is zero as r < R i.e. 0.250 m < 3 m from the above equation.

(b)  Electric field at 2.90 m is zero as r < R i.e. 2.90 m < 3 m from the above equation.

(c) Electric field at 3.10 m is given by the relation as r > R :

E = \frac{KQ}{r^{2} }

Substitute 9 x 10⁹ N m²/C² for K, -5.50 μC for Q and 3.10 m for r in the above equation.

E = - \frac{9\times10^{9}\times5.50\times10^{-6}  }{3.10^{2} }

E = - 5.15 x 10³ V/m

(d) Electric field at 8.00 m is given by the relation as r > R :

E = \frac{KQ}{r^{2} }

Substitute 9 x 10⁹ N m²/C² for K, -5.50 μC for Q and 8.00 m for r in the above equation.

E = - \frac{9\times10^{9}\times5.50\times10^{-6}  }{8^{2} }

E = - 0.77 x 10³ V/m

8 0
2 years ago
Which statement best explains the movement of electric current from the clouds to the ground during a lightning storm? .
olga2289 [7]
"The <span>ground is positively charged and the clouds are negatively charged " is the statement among the statements given in the question that </span><span>best explains the movement of electric current from the clouds to the ground during a lightning storm. The correct option among all the options that are given in the question is the third option or option "C". </span>
3 0
2 years ago
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Suppose that a charged particle of diameter 1.00 micrometer moves with constant speed in an electric field of magnitude 1.00×105
Dovator [93]
It's a bit of a trick question, had the same one on my homework. You're given an electric field strength (1*10^5 N/C for mine), a drag force (7.25*10^-11 N) and the critical info is that it's moving with constant velocity(the particle is in equilibrium/not accelerating). 
<span>All you need is F=(K*Q1*Q2)/r^2 </span>
<span>Just set F=the drag force and the electric field strength is (K*Q2)/r^2, plugging those values in gives you </span>
<span>(7.25*10^-11 N) = (1*10^5 N/C)*Q1 ---> Q1 = 7.25*10^-16 C </span>
3 0
3 years ago
Read 2 more answers
3 a There is a thin layer of water between the blade and the ice. Suggest how this affects friction .​
Gelneren [198K]

Answer:

The water acts like a lubricant therefore has a smooth motion over the ice.

8 0
2 years ago
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