3 years ago

Hdhshabxbxhxhjdhdhsjshs

Physics

1 answer:

8_murik_8 [283]3 years ago

6 0

Yes sassaaaaaaaaaaaaaaaaap

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PLEASE HELP!!!!!!!!!

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I think it’s Malleability

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3 years ago

A person standing 180m from the foot of a high building claps hi

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Answer:

C) 200 m/s

Explanation:

The sound travels a total distance of 360 m in 0.03 minutes.

v = (360 m) / (0.03 min × 60 s/min)

v = 200 m/s

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3 years ago

Carbon dioxide enters an adiabatic compressor at 100 kPa and 300K at a rate of 0.5 kg/s and leaves at 600 kPa and 450K. Neglecti

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Explanation:

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3 years ago

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A long solid conducting cylinder with radius a = 12 cm carries current I1 = 5 A going into the page. This current is distributed

jenyasd209 [6]

Answer:

Explanation:

We shall use Ampere's circuital law to find magnetic field at required point.

The point is outside the circumference of two given wires so whole current will be accounted for .

Ampere's circuital law

B = ∫ Bdl = μ₀ I

line integral will be over circular path of radius r = 41 cm .

Total current I = 5A -3A = 2A .

∫ Bdl = μ₀ I

2π r B = μ₀ I

2π x .41 B = 4π x 10⁻⁷ x 2

B = 2 x 10⁻⁷ x 2 / .41

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3 years ago

A 2. 5 kg block is launched along the ground by a spring with a spring constant of 56 N/m. The spring is initially compressed 0.

OLga [1]

The speed of the spring when it is released is 3.5 m/s.

The given parameters:

Mass of the block, m = 2.5 kg
Spring constant, k = 56 N/m
Extension of the spring, x = 0.75 m

The speed of the spring when it is released is calculated by applying the principle of conservation of energy as follows;

$K.E = U_x\\\\\frac{1}{2} mv^2 = \frac{1}{2} kx^2\\\\mv^2 = kx^2\\\\v^2 = \frac{kx^2}{m} \\\\v = \sqrt{\frac{kx^2}{m} } \\\\v = \sqrt{\frac{56 \times 0.75^2}{2.5} } \\\\v = 3.5 \ m/s$

Thus, the speed of the spring when it is released is 3.5 m/s.

Learn more about conservation of energy here: brainly.com/question/166559

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3 years ago

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