3 years ago

A steel beam is rotated in a horizontal plane to provide the centripetal acceleration for

Physics

1 answer:

oksian1 [2.3K]3 years ago

7 0

Answer:

12 m/s

Explanation:

a = v² / r

78 m/s² = v² / 2.0 m

v = 12.49 m/s

Rounding to two significant figures, the speed is 12 m/s.

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F an object has a mass of 200 kg and a weight of 1000 N, what is g?

jeka94

Answer:

g = 5 m/s square

Explanation:

Weight(W), Mass(m), Gravity(g)

W = mg

1,000N = 200g

g = 1000/200

g = 5 m/s square

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2 years ago

if the pressure on a gas increases, the volume of the gas also increases is that statement true or false?

Arada [10]

False, according to Boyle's law it's pressure increases, volume decreases

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3 years ago

PLEASE HELP!!! GIVING BRAINLIEST!! ill also answer questions that you have posted if you answer these correctly!!!! (40pts)

denpristay [2]

Answer:

False, true

Explanation:

3 0

3 years ago

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Identify each part of this chemical equation that describes the burning of methane and oxygen. B (blue box): D (number): E (purp

Allushta [10]

Answer:

The correct answer is -

A (the entire green box): Chemical Equation

B (the blue box): Reactants

C (the arrow): Reacts to Form

D (the number): Coefficient

E (the purple box): Products

Explanation:

The chemical reaction of burning methane and oxygen is as follows;

Here, the green part A is the chemical equation that includes various parts that are reactants B, methane, and oxygen, C is an arrow that indicates the formation of products.

2 is here coefficient that indicates the moles of the oxygen which forms carbon dioxide and water in box E is products

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2 years ago

A metallic wire has a diameter of 4.12mm. When the current in the wire is 8.00A, the drift velocity is 5.40×10−5m/s.What is the

podryga [215]

Answer:

$6.9\times 10^{28}m^{-3}$

Explanation:

We are given that

Diameter of wire=d=4.12 mm

Radius of wire=r $r=\frac{d}{2}=\frac{4.12}{2}=2.06mm=2.06\times 10^{-3} m$

$1mm=10^{-3} m$

Current=I=8 A

Drift velocity= $v_d=5.4\times 10^{-5} m/s$

We have to find the density of free electrons in the metal

We know that

Density of electron= $n=\frac{I}{v_deA}$

Using the formula

Density of free electrons= $\frac{8}{5.4\times 10^{-5}\times 1.6\times 10^{-19}\times 3.14\times (2.06\times 10^{-3})^2}$

By using Area of wire= $\pi r^2$

$\pi=3.14\\e=1.6\times 10^{-19} C$

Density of free electrons= $6.9\times 10^{28}m^{-3}$

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2 years ago

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