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harina [27]
3 years ago
12

Define a combination circuit.

Physics
1 answer:
Neporo4naja [7]3 years ago
7 0

Answer:

a circuit that is a blend of series paths and parallel paths. See Figure for a visual explanation. Most circuits are of this variety. Don't be afraid to tackle these circuits as far as the math goes.

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I'm giving out a lot of points for whoever helps me and answers correctly! 3 questions to answer! Please answer them ALL!!! :D
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5 0
4 years ago
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How are mixtures and pure substances related?
ddd [48]

A pure substance is only composed of one element for example pure gold which as the only element present Gold (Ag). So the properties of these kinds of substances is also dependent on only one element e.g. ductility, conductance etc.

On the other hand mixture is made of many different pure elements such as Bronze. This gives an object different properties. So mixtures contain many different pure elements

4 0
4 years ago
Unlike other states of matter, what expand to fill their containers
padilas [110]

Answer:

gas

Explanation:

8 0
3 years ago
Read 2 more answers
Two concentric conducting spherical shells produce a radially outward electric field of magnitude49,000 N/C at a point 4.10 m fr
Korolek [52]

To solve this problem it is necessary to apply the concepts related to the electric field according to the definition of Coulomb's law.

 The electric field is defined mathematically as a vector field that associates to each point in space the (electrostatic or Coulomb) force per unit of charge exerted on an infinitesimal positive test charge at rest at that point.

Mathematically this can be described as:

E = \frac{1}{4\pi \epsilon_0}\frac{q}{r^2}

Where,

\epsilon_0 = permittivity of free space

r = Distance

q = Charge

E = Electric Field

Our values are given as,

E= 49.000N/C

r= 4.1m

\epsilon_0=8.854*10{-12}C^2N^{-1}m^{-2}

Replacing we have,

E = \frac{1}{4\pi \epsilon_0}\frac{q}{r^2}

49000 = \frac{1}{4\pi (8.854*10{-12})}\frac{q}{(4.1)^2}

q= 9.16*10^{-5} C

q= 91.6\mu C

Therefore the amoun of charge on the outer surface of the larger shell is 91.6 \mu C

6 0
4 years ago
On a rectangle with length 100 m, width 50m and 2 vehicles stationed together. Find the time that they meet given that car A tra
zhannawk [14.2K]

Answer:

T=35.625sec

Explanation:

From the question we are told that:

Length L=100 m

Width W=50m

Velocity of Car A V_A=5m/s

Velocity of Car B V_B=3m/s

Distance traveled by car A before car B moves

d_l=5*3

d_l=15

Therefore total distance traveled at same time interval

D=total\ distance-d_l

Where

Total distance=Perimeter of rectangle

P=2(L+B)

P=2(100+50)

P=300

Therefore

D=total\ distance-d_l

D=300-15\\D=285m

Generally the equation for time taken to meet is mathematically given by

T=\frac{Distance D}{Relative\ speed V_r}

Where

Relative speed = Speed of car A +Speed of car B

V_r=V_A+V_B

V_r=5+3

V_r=8m/s

Therefore the time taken to meet

T=\frac{ D}{ V_r}

T=\frac{ 285}{ 8}

T=35.625sec

8 0
3 years ago
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