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ASHA 777 [7]
3 years ago
10

A series circuit contains a 9-volt battery, a 3-ohm resistor and a 2-ohm resistor. What is the voltage drop across the 2-ohm res

istor?
3.2 V
3.4 V
3.6 V
3.8 V
Physics
2 answers:
NeX [460]3 years ago
8 0

Answer:

3.6V

Explanation:

BARSIC [14]3 years ago
7 0
Since everything in the circuit is in series .. .

-- The total resistance is  (3 + 2) = 5 ohms.

-- The voltage across the 3-ohm resistor is 3/5 of the total voltage.

-- The voltage across the 2-ohm resistor is 2/5 of the total voltage.

                  (2/5) of (9 volts)  =  18/5  =  3.6 volts .

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Two identical stones, A and B, are thrown from a cliff from the same height and with the same initial speed. Stone A is thrown v
Taya2010 [7]

Answer:

Option A

Explanation:

This can be explained based on the conservation of energy.

The total mechanical energy of the system remain constant in the absence of any external force. Also, the total mechanical energy of the system is the sum of the potential energy and the kinetic energy associated with the system.

In case of two stones thrown from a cliff one vertically downwards the other vertically upwards, the overall gravitational potential energy remain same for the two stones as the displacement of the stones is same.

Therefore the kinetic energy and hence the speed of the two stones should also be same in order for the mechanical energy to remain conserved.

5 0
3 years ago
Read 2 more answers
1) A thin ring made of uniformly charged insulating material has total charge Q and radius R. The ring is positioned along the x
allochka39001 [22]

Answer:

(A) considering the charge "q" evenly distributed, applying the technique of charge integration for finite charges, you obtain the expression for the potential along any point in the Z-axis:

V(z)=\frac{Q}{4\pi (\epsilon_{0}) \sqrt{R^{2} +z^{2}}  }

With (\epsilon_{0}) been the vacuum permittivity

(B) The expression for the magnitude of the E(z) electric field along the Z-axis is:

E(z)=\frac{QZ}{4\pi (\epsilon_{0}) (R^{2} +z^{2})^{\frac{3}{2} }    }

Explanation:

(A) Considering a uniform linear density λ_{0} on the ring, then:

dQ=\lambda dl (1)⇒Q=\lambda_{0} 2\pi R(2)⇒\lambda_{0}=\frac{Q}{2\pi R}(3)

Applying the technique of charge integration for finite charges:

V(z)= 4\pi (ε_{0})\int\limits^a_b {\frac{1}{ r'  }} \, dQ(4)

Been r' the distance between the charge and the observation point and a, b limits of integration of the charge. In this case a=2π and b=0.

Using cylindrical coordinates, the distance between a point of the Z-axis and a point of a ring with R radius is:

r'=\sqrt{R^{2} +Z^{2}}(5)

Using the expressions (1),(4) and (5) you obtain:

V(z)= 4\pi (\epsilon_{0})\int\limits^a_b {\frac{\lambda_{0}R}{ \sqrt{R^{2} +Z^{2}}  }} \, d\phi

Integrating results:

V(z)=\frac{Q}{4\pi (\epsilon_{0}) \sqrt{R^{2} +z^{2}}  }   (S_a)

(B) For the expression of the magnitude of the field E(z), is important to remember:

|E| =-\nabla V (6)

But in this case you only work in the z variable, soo the expression (6) can be rewritten as:

|E| =-\frac{dV(z)}{dz} (7)

Using expression (7) and (S_a), you get the expression of the magnitude of the field E(z):

E(z)=\frac{QZ}{4\pi (\epsilon_{0}) (R^{2} +z^{2})^{\frac{3}{2} }    } (S_b)

4 0
3 years ago
You want to photograph a circular diffraction pattern whose central maximum has a diameter of 1.0 cm. You have a helium-neon las
topjm [15]

Answer:

0.776 m far Pinhole should be placed before the viewing screen

Explanation:

For circular aperture of diameter D will have a bright central maximum of diameter, width is given by

w=\frac{2.44 \lambda L}{D}

where \lambda is wavelength of helium neon laser = 633 nm, D=10.cm, w=0.12 mm

Pinhole should be placed before the viewing screen is

L=\frac{wD}{2.44\lambda}\\L=\frac{0.12\times 10^{-3}\times 0 .01}{2.44\times 633 \times 10^{-9}}\\L=0.776 m

4 0
3 years ago
A proton is released from rest at the positive plate of a parallel-plate capacitor. It crosses the capacitor and reaches the neg
katrin2010 [14]

Answer:

2.1x10^6m/s

Explanation:

One electron has a charge of –1.602e-19 C

mass of electron is 9.1e-31 kg

mass of proton is 1.6726e−27 kg

mass ratio is 1.6726e−27 / 9.1e-31 = 1838

The force is constant, F

distance is constant, d

a = F/m

a increases by a factor 1838, as m decreases by that factor

a = a₀1838

v₀² = 2a₀d

v² = 2a₀d1838

v²/v₀² = 2a₀d1838 / 2a₀d = 1838

v² = 1838v₀² = 1838(45000)²

v = 45000√1838 = 2.1e6 m/s

5 0
3 years ago
The period of a wave is found to be 50 seconds. What is the frequency of the wave? 50 Hz 0.8 Hz 0.02 Hz not enough information g
Tju [1.3M]
0.02 Hertz

Hope this helped!
7 0
3 years ago
Read 2 more answers
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