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ankoles [38]
3 years ago
13

Item 1 Circuit A has a resistance of 7.5 ohms, and circuit B has a resistance of 5 ohms. Which statement about the circuits must

be correct? (1 point) Item 2 Item 3 If the current in circuit A is less than the current in circuit B, then the voltage in circuit A is less than the voltage in circuit B. Item 4 If the voltage of circuit A is equal to the voltage of circuit B, then the current in circuit A is less than the current in circuit B. If the current in circuit A is equal to the current in circuit B, then the voltage in circuit A is less than the voltage in circuit B. If the voltage of circuit A is equal to the voltage of circuit B, then the current in circuit A is equal to the current in circuit B. 0 Type here to search е c​
Physics
1 answer:
kirza4 [7]3 years ago
3 0

Answer:

e4atetetsgts4tgesetse

Explanation:

ger54ttagcuit B, then the voltage in circuit A is less than the voltage in circuit B. Item 4 If the voltage of circuit A is equal to the voltage of circuit B, then the current in circuit A is less than the current in circuit B. If the current in circuit A is equal to the current in circuit B, then the voltage in circuit A is less than the voltage in circuit B. If the voltage of circuit A is equal to the voltage of circuit B, then the current in circuit A is equal to the current in circuit B. 0 Type here to search е c​

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klio [65]

Answer:18/2

Explanation:

its half

6 0
3 years ago
Refraction:
Novay_Z [31]
Refraction is caused by any wave crossing the boundary out of one substance and into another one. The two different substances may be no more than two parcels of air at different temperatures.
4 0
3 years ago
A coaxial cable has a charged inner conductor (with charge +8.5 µC and radius 1.304 mm) and a surrounding oppositely charged con
Tcecarenko [31]

Complete question:

A 50 m length of coaxial cable has a charged inner conductor (with charge +8.5 µC and radius 1.304 mm) and a surrounding oppositely charged conductor (with charge −8.5 µC and radius 9.249 mm).

Required:

What is the magnitude of the electric field halfway between the two cylindrical conductors? The Coulomb constant is 8.98755 × 10^9 N.m^2 . Assume the region between the conductors is air, and neglect end effects. Answer in units of V/m.

Answer:

The magnitude of the electric field halfway between the two cylindrical conductors is 5.793 x 10⁵ V/m

Explanation:

Given;

charge of the coaxial capable, Q = 8.5 µC = 8.5  x 10⁻⁶ C

length of the conductor, L = 50 m

inner radius, r₁ = 1.304 mm

outer radius, r₂ = 9.249 mm

The magnitude of the electric field halfway between the two cylindrical conductors is given by;

E = \frac{\lambda}{2\pi \epsilon_o r} = \frac{Q}{2\pi \epsilon_o r L}

Where;

λ is linear charge density or charge per unit length

r is the distance halfway between the two cylindrical conductors

r = r_1 + \frac{1}{2}(r_2-r_1) \\\\r = 1.304 \ mm \ + \  \frac{1}{2}(9.249 \ mm-1.304 \ mm)\\\\r = 1.304 \ mm \ + \ 3.9725 \ mm\\\\r = 5.2765 \ mm

The magnitude of the electric field is now given as;

E = \frac{8.5*10^{-6}}{2\pi(8.85*10^{-12})(5.2765*10^{-3})(50)} \\\\E = 5.793*10^5 \ V/m

Therefore, the magnitude of the electric field halfway between the two cylindrical conductors is 5.793 x 10⁵ V/m

5 0
3 years ago
Four identical particles of mass 0.980 kg each are placed at the vertices of a 4.14 m x 4.14 m square and held there by four mas
Zielflug [23.3K]

Answer:

a) The rotational inertia when it passes through the midpoints of opposite sides and lies in the plane of the square is 16.8 kg m²

b) I = 50.39 kg m²

c) I = 16.8 kg m²

Explanation:

a) Given data:

m = 0.98 kg

a = 4.14 * 4.14

The moment of inertia is:

I=mr^{2} \\r=\frac{a}{2} \\I=m(a/2)^{2} \\I=\frac{ma^{2} }{4}

For 4 particles:

I=4(\frac{ma^{2} }{4} )=ma^{2} =0.98*(4.14)^{2} =16.8kgm^{2}

b) Distance from top left mass = x = a/2

Distance from bottom left mass = x = a/2

Distance from top right mass = x = √5 (a/2)

The total moment of inertia is:

I=m(\frac{a}{2} )^{2} +m(\frac{a}{2} )^{2}+m(\frac{\sqrt{5a} }{2} )^{2}+m(\frac{\sqrt{5a} }{2} )^{2}=\frac{12ma^{2} }{4} =3ma^{2} =3*0.98*(4.14)^{2} =50.39kgm^{2}

c)

I=2m(\frac{m}{\sqrt{2} } )^{2} =ma^{2} =0.98*(4.14)^{2} =16.8kgm^{2}

8 0
3 years ago
a 90 kg architect is standing 2 meters from the center of a scaffold help up by a rope on both sides. the scaffold is 6m long an
Mademuasel [1]
We can solve the problem by requiring the equilibrium of the forces and the equilibrium of torques.

1) Equilibrium of forces:
T_1 - W_p - W_s + T_2 =0
where
W_p = (90kg)(9.81 m/s^2)=883 N is the weight of the person
W_s = (200kg)(9.81 m/s^2)=1962 N is the weight of the scaffold
Re-arranging, we can write the equation as
T_1 = 2845 N-T_2 (1)

2) Equilibrium of torques:
T_1 \cdot 3 m - W_p \cdot 2 m - T_2 \cdot 3m =0
where 3 m and 2 m are the distances of the forces from the center of mass of the scaffold.
Using W_p = 883 N and replacing T1 with (1), we find
2845 N \cdot 3 m - T_2 \cdot 3 m - 833 N \cdot 2 m - T_2 \cdot 3 m=0
from which we find
T_2 = 1128 N

And then, substituting T2 into (1), we find
T_1 = 1717 N
8 0
3 years ago
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