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3 years ago

If points a and b are connected by a wire with negligible resistance, find the magnitude of the current in the 12.0 v battery.

1 answer:

7 0

V = I * R
Where V is the voltage, I is the current and R is the resistance. Using Ohm's law, you require resistance to find the current through the wire. Technically, if the wire has a resistance of 0, you will get infinite current. But this isn't possible. Maybe the negligible resistance refers to the battery's internal resistance - not the wire's resistance.

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A Juggler is juggling a uniform rod one end of which is coated in tar and burning. He is holding the rod by the opposite end and

lyudmila [28]

The vertical speed of the rod varies inversely as the rotational speed for a given number of complete rotations

The equation that gives the vertical velocity, v₀ is $\, \underline{v_0= \dfrac{ g \cdot \pi \cdot n}{ \omega_0 }}$

Reason:

From a similar question, the given parameters appear to be correctly given as follows;

The vertical speed of the rod = v₀

Angular velocity of the = ω₀

Required;

The value of, v₀, so that the rod has made exactly, n, number of turns when it returns to his hand

Where;

n = An integer

Solution;

The time the rod spends in the air is given as follows;

$Height, \ h = u \cdot t - \dfrac{1}{2} \cdot g \cdot t^2$

Where;

g = Acceleration due to gravity

t = Time of motion

When the rod returns to his hand, we have, h = 0, therefore;

$0 = u \cdot t - \dfrac{1}{2} \cdot g \cdot t^2$

$u \cdot t = \dfrac{1}{2} \cdot g \cdot t^2$

$t^2 = \dfrac{u \cdot t }{\dfrac{1}{2} \cdot g } = \dfrac{2 \cdot u \cdot t }{g }$

$Time, \ t = \dfrac{2 \cdot u }{g }$

We have;

$Angular \ velocity, \ \omega_0 = \dfrac{ 2 \cdot \pi \cdot n}{t} \ (required)$

$\omega_0 = \dfrac{ 2 \cdot \pi \cdot n}{ \dfrac{2 \cdot v_0 }{g }} = \dfrac{ g \cdot \pi \cdot n}{ v_0 }$

Therefore;

$The \ vertical \ velocity, \, \underline{v_0= \dfrac{ g \cdot \pi \cdot n}{ \omega_0 }}$

Learn more here:

brainly.com/question/14140053

4 0

3 years ago

A U-tube open at both ends is partially filled withwater. Oil

ArbitrLikvidat [17]

Answer:

a) h_w = 0.02139 m , b) v₁ = 9.74 m / s

Explanation:

For this exercise we use the pascal principle that states that the pressure at one point is the same regardless of body shape.

At the initial moment (before emptying the oil), we fix the point on the surface of the liquid, in this case the left and right sides are in balance.

P₁ = P₂ = P₀

Now we add the 5 cm (h₂ = 0.05 m) of oil, in this case the weight of the oil creates an extra pressure that pushes the water, let's look for how much the water moved (h_w). The weight of oil added is equal to the weight of displaced water

W_w = W_oil

m_w g= m_oil g

Density is defined

ρ = m / V

we replace

ρ_w V_w = ρ_oil W_oil

V = A h

ρ_w A h_w = ρ_oil A h_oil

h_w = h_oil ρ_oil / ρ_w

Now let's analyze the pressure at the initial reference height for both sides two had in U

Right side

We have the atmospheric pressure, with its decrease due to the lower height, plus the oil pressure above the reference level

h’= 0.05 cm - h_w

P = (P₀ - ρ_air g (0.05-h_w)) + ρ_oil g (0.05-h_w)

Left side

We have at the same point, the atmospheric pressure with its reduction due to the height change plus the water pressure

P = (P₀ - ρ_air h h_w) + ρ_w g h_w

As we have the same point we can equalize the pressure

(P₀ -ρ_air g (0.05-h_w)) +ρ_oil g (0.05-h_w) = (Po -ρ_air h h_w) +ρ_w g h_w

ρ_air g (h_w - (0.05-h_w)) = ρ_w g h_w - ρ_oil g (0.05-h_w)

- ρ_air g 0.05 = h_w g ( ρ_w + ρ _oil) - rho_oil g 0.05

h_w g ( ρ_w + ρ_oil) = g 0.05 ( ρ_air - ρ_oil)

calculate

h_w = 0.05 (ρ_ oil- ρ_air) / ( ρ_w + ρ_oil)

h_w = 0.05 (750 - 1.29) / (1000-750)

h_w = 0.02139 m

The amount that decreases the height on one side is equal to the amount that increases the other

b) cover the right side and blow the air on the left side, let's use Bernoulli's equation, where index 1 will be for the left side and index 2 for the right side

P₁ + 1/2 ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂

Since they indicate that both sides are at the same height y₁ = y₂, the right side is protected from the wind speed therefore v₂ = 0, let's write the equation

P₁ + ½ ρ_air v₁² = P₂

v₁² = (P₂-P₁) 2 / ρ_air

Let's analyze the pressure on each side of the tube, since the new equilibrium height is the height that was added of oil distributed between the two tubes, bone

h’= 2.5 cm = 0.025 m

P₁ = P₀ - ρ_w g h’

P₂ = P₀ - ρ_oil g h ’

P₂-P₁ = g h’ (rho_w - Rho_oil)

We replace

v₁² = 2g h’ (ρ_w –ρ_oil) / ρ_air

calculate

v₁² = 2 9.8 0.025 (1000 - 750) /1.29

v₁ = √ 94.96

v₁ = 9.74 m / s

3 0

4 years ago

Which of the following is true concerning atoms?Atoms are made of even smaller substances called subatomic particles

olasank [31]

The correct answer of the given question above would be the first option. The statement that is true concerning atoms is that, atoms are made of even smaller substances called subatomic particles. Subatomic particles include neutrons, protons, electrons and atomic nuclei. Hope this answer helps.

3 0

4 years ago

How much energy is required to ionize hydrogen when it is in the n = 5 state?

PtichkaEL [24]

Answer:

0.544 eV

Explanation:

To ionize the hydrogen atom $n_f\rightarrow infinte$