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aksik [14]
3 years ago
7

Calculate the total resustance Circut 2ohmls, 2Ohlms In parrelle

Physics
1 answer:
Colt1911 [192]3 years ago
5 0

Explanation:

total resistance=(2×2)/(2+2)= 1ohms

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Check attachment

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What part of the hammer acts as the fulcrum when the hammer is used to remove a nail
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Suppose that a person gets hit by a bus moving at 30 mi/h with a 58,000 lbs of force in the direction of motion. If the mass of
alexandr402 [8]

The impulse of a force is due to the change in the motion of an object

A. The persons speed after impact is approximately 59.38 mi/h

B. The expected speed is <u>29.89 mi/h</u> which is less than the findings

Reason:

Known parameters are;

The speed of the bus, v = 30 mi/h

The force with which the person was hit, F = 58,000 lbs

Mass of the bus, M = 40,000 lbs

Mass of the person, m = 150 lbs

Duration of the impact, Δt = 0.007 seconds

A. The speed of the person at the end of the impact, <em>v</em>, is given as follows;

The impulse of the force = F × Δt = m × Δv

For the person, we get;

58,000 lbf ≈ 1866094.816 lb·ft./s²

58,000 lbf × 0.007 s = 150 lbs × Δv

1,866,094.816 lb·ft./s²

\Delta v = \dfrac{1,866,094.816\ lbs \times 0.007 \, s}{150 \, lbs} \approx  87.084  \ ft./s

Δv = v₂ - v₁

The initial speed of the person at the instant, can be as v₁ = 0

The final speed, v₂ = Δv - v₁

∴ v₂ ≈  87.084 ft./s - 0 = 87.084 ft./s

≈ <u>87.084 ft./s</u>

<u />v_2 \approx \dfrac{87.084 \ ft./s}{y} \times\dfrac{1 \ mi}{5280 \ ft.} \times \dfrac{3,600 \ s}{1 \, hour} \approx 59.38 \ mi/h<u />

The speed of the person at the end of the impact, v₂ ≈ <u>59.38 mi/h</u>

B. Where the momentum is conserved, we have;

m₁·v₁ + m₂v₂ = (m₁ + m₂)·v

v = \dfrac{m_1 \cdot v_1 + m_2 \cdot v_2}{m_2 + m_1}

v = \dfrac{40,000 \times 30  + 150 \times 0}{40,000 + 150} \approx 29.89

The expected speed of the person at the end of the impact is 29.89 mi/h, and therefore, <u>the findings does not agree with the expectation</u>

Learn more here:

brainly.com/question/18326789

3 0
3 years ago
A beryllium-9 ion has a positive charge that is double the charge of a proton, and a mass of 1.50 ✕ 10−26 kg. At a particular in
seropon [69]

Answer:

Magnetic force, F = 3.52\times 10^{-13}\ N

Explanation:

Given that,

A beryllium-9 ion has a positive charge that is double the charge of a proton, q=2\times 1.6\times 10^{-19}\ C=3.2\times 10^{-19}\ C

Speed of the ion in the magnetic field, v=5\times 10^6\ m/s

Its velocity makes an angle of 61° with the direction of the magnetic field at the ion's location.

The magnitude of the field is 0.220 T.

We need to find the magnitude of the magnetic force on the ion. It is given by :

F=qvB\\\\F=3.2\times 10^{-19}\times 5\times 10^6\times 0.22\\\\F=3.52\times 10^{-13}\ N

So, the magnitude of magnetic force on the ion is 3.52\times 10^{-13}\ N.

3 0
3 years ago
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The atom must have gained 1 or more electrons or must have lost 1 or more electrons.
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