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3 years ago

Mainly need help with part A. No links, thanks!

Physics

1 answer:

Vlad1618 [11]3 years ago

4 0

Answer:

R = 4Ω

Explanation:

If we have two resistors with resistances R1 and R2 in series the total resistance is R = R1 + R2

If the resistances are in parallel, the total resistance is given by:

1/R = 1/R1 + 1/R2.

First, we have a resistor with R1 = 1.5Ω

This resistor is connected in series with a parallel part (let's find the resistance of this parallel part), in one branch we have two resistors in series with resistances:

R2 = 8Ω and R3 = 4Ω

Because these are in series, the resistance of that branch is:

R = 8Ω + 4Ω = 12Ω

In the other branch, we have a single resistor of R4 = 4Ω

The resistance of the parallel part is:

1/R = 1/12Ω + 1/4Ω = 1/12Ω + 3/12Ω = 4/12Ω = 1/3Ω

1/R = 1/3Ω

R = 3Ω

Then we have a resistor (the first one, R1 = 1.5Ω) in series with a resistor of 3Ω.

Then the total resistance is:

R = 1Ω + 3Ω = 4Ω

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stira [4]

Answer:

v= 3.18 m/s

Explanation:

Given that

m= 150 g = 0.15 kg

M= 240 g = 0.24 kg

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$\omega =\dfrac{2\pi N}{60}$

$\omega =\dfrac{2\pi \times 150}{60}$

ω = 15.7 rad/s

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3 years ago

Radon-222 transforms into polonium-218 and an alpha particle (he-4). what else is created?

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3 years ago

A 100-kg block being released from rest from a height of 1.0 m. It then takes it 1.40 s to reach the floor. What is the mass m o

dimulka [17.4K]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The mass of the other block is $m_1 = 81.14 \ kg$

Explanation:

From the question we are told that

Mass of the first block is $m_1 = 100 \ kg$

The height is $s = 1.0 \ m$

The time it takes it is $t = 1.40 \ s$

Generally from kinematic equation

$s = ut + \frac{1}{2} at^2$

Here u is the initial velocity which zero given that it was at rest initially

$s = 0 * t + \frac{1}{2} at^2$

=> $s = \frac{1}{2} at^2$

=> $1 = \frac{1}{2}* a * (1.40 )^2$

=> $a = 1.0204 \ m/s^2$

Generally from the diagram the resultant force due to the weight of the first object and the tension on the string is mathematically represented as

$mg - T = ma$

=> $T = m g - ma$

=> $T = m(g - a)$

=> $T = 877.96 \ N$

Generally from the diagram the resultant force due to the weight of the second object and the tension on the string is mathematically represented as

$T - m_1g = m_1 a$

=> $877.96 = m_1 (a + g)$

=> $877.96 = m_1 (1.0204 + 9.8 )$

=> $m_1 = 81.14 \ kg$

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3 years ago

What is the electric field at the point x=6.5 m? positive e-fields point to the right.?

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3 years ago

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Rasek [7]

Answer:

36.1 nC

Explanation:

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So, kq²/r² = mg

q²/r² = mg/k

q² = mgr²/k

taking square-root of both sides,

q = √(mg/k)r

So, substituting the values of the variables into the equation, we have

q = √(mg/k)r

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q = √(13.07 × 10⁻¹² C²/m²)1 × 10⁻² m

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q = 3.61 × 10⁻⁸ C

q = 36.1 × 10⁻⁹ C

q = 36.1 nC

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3 years ago

Mainly need help with part A. No links, thanks!​

Mainly need help with part A. No links, thanks!