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3 years ago

Help with these questions please , 15 PTS & brainliest

Physics

1 answer:

myrzilka [38]3 years ago

6 0

For these two questions, first you need to know that the voltage across each branch of a parallel circuit is the same.

So, for Q5, we can first find out the voltage across R₂ by V=IR.

Voltage across R₂ = 2.5 × 8 = 20V

Since R₂ and R₃ are in parallel circuit, their voltage should be the same. Thus, voltage across R₃ is 20V.

So, by V=IR,

current of R₃ = $\frac{20}{4}$ = 5A

Q6. voltage across R₁ = 2 × 4 = 8V

∴voltage across R₂ = 8V

current of R₂ = $\frac{8}{8}$ = 1A

<h3><u>Alternative method</u></h3>

From these two examples, you can find out that the current of each branch of the parallel circuit is inversely proportional to the resistance of the branch.

ie. for Q5,

$\frac{R2}{R3}$ = $\frac{I3}{I2}$

$\frac{8}{4}$ = $\frac{I3}{2.5}$

I₃ = 5A

Q6. $\frac{R1}{R2}$ = $\frac{I2}{I1}$

$\frac{4}{8}$ = $\frac{I2}{2}$

I₂ = 1A

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Answer:

The answer to your question is: F = 0.4375 N. The force will be 16 times lower than with the first conditions.

Explanation:

Data

F = 7 N

F = ? if the masses is quartered

Formula

$F = \frac{Km1m2}{r2}$

Process

Normal conditions F = Km₁m₂/r² = 7

When masses quartered F = K(m₁/4)(m₂/4)/r² = ?

F = K(m₁m₂/16)/r²

F = K(m₁m₂/16r² = 7/16 = 0.4375 N

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3 years ago

The electric field in a particular thundercloud is 3.8 x 105 N/C. What is the acceleration (in m/s2) of an electron in this fiel

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Answer:

6.68 x 10^16 m/s^2

Explanation:

Electric field, E = 3.8 x 10^5 N/C

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Let a be the acceleration of the electron.

The force due to electric field on electron is

F = q E

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F = 1.6 x 10^-19 x 3.8 x 10^5

F = 6.08 x 10^-14 N

According to Newton's second law

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Explanation:

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3 years ago

A horizontal pipe of diameter 1.03m has a smooth constriction to a section of diameter 0.618 m. The density of oil flowing in th

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Velocity of the oil in the pipe: 0.76 m/s, in the constricted section: 5.87 m/s

Explanation:

We can solve this problem by using Bernoulli's equation:

$p_1 + \frac{1}{2}\rho v_1^2 = p_2 + \frac{1}{2}\rho v_2^2$ (1)

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$v_2$ is the velocity of the oil in the constricted section

Also, according to the continuity equation,

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So we can write

$v_2=\frac{r_1^2}{r_2^2}v_1$

Substituting into eq.(1),

$p_1 + \frac{1}{2}\rho v_1^2 = p_2 + \frac{1}{2}\rho (\frac{r_1^2}{r_2^2}v_1)^2$

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And the velocity in the constricted section is

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3 years ago

A capacitor with a capacitance of 50µf when connected to a battery of 400 V. The charge and energy stored on it is? a. 0.05 C an

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Answer:

c. 0.02 C and 4 J

Explanation:

Applying,

Q = CV................ Equation 1

Where Q = Charge, C = Capacitance of the capacitor, V = Voltage.

From the question,

Given: C = 50 μF = 50×10⁻⁶ F, V = 400 V

Substitute these values into equation 1

Q = (50×10⁻⁶)(400)

Q = 0.02 C.

Also Applying

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Where E = Energy stored.

Therefore,

E = (50×10⁻⁶ )(400²)/2

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3 years ago