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Sphinxa [80]
3 years ago
14

Three wires are connected at a branch point. One wire carries a positive current of 18 A into the branch point, and a second wir

e carries a positive current of 7 A away from the branch point. Find the current carried by the third wire into the branch point.
Physics
1 answer:
madreJ [45]3 years ago
3 0

Answer:

The current in third branch is 11 A.

Explanation:

incoming current in one branch = 18 A

outgoing current in the other branch = 7 A

let the current in the third branch is i.

According to the Kirchoff's fist law in electricity

incoming current = out going current

18 = 7 + i

i = 11 A

The current in third branch is 11 A.

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4 years ago
During a marathon race, a runner’s blood flow increases to 10.0 times her resting rate. Her blood’s viscosity has dropped to 95.
Phoenix [80]

To solve the problem it is necessary to apply the equations related to the Poiseuilles laminar flow law, with which the stationary laminar flow ΦV of an incompressible and uniformly viscous liquid (also called Newtonian fluid) can be determined through a cylindrical tube of constant circular section. Mathematically this can be expressed:

Q = \frac{\Delta P \pi r^4}{8\eta l}

Where:

\eta_i = are the viscosities of the concrete before and after the increase

l = Length of the vessel

r_1, R_2 = Radio of the vessel before and after the increase

\Delta P= Change in the pressure

Q_{1,2} = The rates of flow before and after he increase

Our values are given as:

Q_2 = 10Q_1 \rightarrow 10 times her resting rate

\eta_2 = 0.95\eta_1 95% of its normal value

\Delta P_2 = 1.5\Delta P_1 Increase of 50%

Plugging known information to get

Q_1 = \frac{\Delta P \pi r^4}{8\eta l}

Q_1 8\eta_1 l = \Delta P_1 \pi r_1^4

r_1^4 = \frac{Q_1 8\eta_1 l}{\Delta P_1 \pi}

r_1 = (\frac{Q_1 8\eta_1 l}{\Delta P_1 \pi})^{1/4}

r_2 = (\frac{Q_2 8\eta_2 l}{\Delta P_2 \pi})^{1/4}

r_2 = (\frac{10Q_18 \times 0.95\eta_1 l}{1.5\Delta P_1 \pi})^{1/4}

r_2 = 1.586r_1

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7 0
4 years ago
A horizontal force, F1 = 65 N, and a force, F2 = 12.4 N acting at an angle of θ to the horizontal, are applied to a block of mas
Nezavi [6.7K]

Answer:

(a) FN = 24.18 N

(b) a = 22.87 m/s²

Explanation:

Newton's second law of the  block:

∑F = m*a Formula (1)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Forces acting on the box

We define the x-axis in the direction parallel to the movement of the block on the surface   and the y-axis in the direction perpendicular to it.

F₁ : Horizontal force

F₂ : acting at an angle of θ to the horizontal,

W: Weight of the block  : In vertical direction

FN : Normal force : perpendicular to the direction the surface

fk : Friction force: parallel to the direction to the surface

Known data

m =3.1 kg : mass of the  block

F₁ = 65 N,  horizontal force

F₂ = 12.4 N acting at an angle of θ to the horizontal

θ = 30° angle θ of F₂ with respect to the horizontal

μk = 0.2 : coefficient of kinetic friction between the block and the surface

g = 9.8 m/s² : acceleration due to gravity

Calculated of the weight  of the block

W= m*g  = (3.1 kg)*(9.8 m/s²) = 30.38 N

x-y F₂ components

F₂x = F₂cos θ= (12.4)*cos(30)° = 10.74 N

F₂y = F₂sin θ= (12.4)*sin(30)° = 6.2 N

a)Calculated of the Normal force  (FN)

We apply the formula (1)

∑Fy = m*ay    ay = 0

FN+6.2-30.38 = 0

FN = -6.2+30.38

FN = 24.18 N

Calculated of the Friction force:

fk=μk*N=  0.2* 24.18 N = 4.836 N

b) We apply the formula (1) to calculated acceleration of the block:

∑Fx = m*ax ,  ax= a  : acceleration of the block

F₁ + F₂x -fk = ( m)*a

65 N + 10.74 -4.836 = ( 3.1)*a

70.904 = ( 3.1)*a

a = (70.904 ) / ( 3.1)

a = 22.87 m/s²

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Answer:

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neonofarm [45]

Explanation:

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