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3 years ago

What is the maximum possible resultant of two vectors with magnitudes of 2 and 4 units? what is the minimum possible resultant?

1 answer:

5 0

When two vectors are in same direction then we add up to get their resultant and in this case resultant would be maximum

R = A + B

R = 4 + 2

R = 6

If vectors are in opposite direction then they provide minimum resultant and in this case we subtract them

R = A - B

R = 4 - 2

R = 2

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What did j.J. Thomson discover about the composition of atoms?

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Answer:

J.J. Thomson discovered <u><em>the electron</em></u> by experimenting with a Crookes, or cathode ray, tube. He demonstrated that cathode rays were negatively charged. In addition, he also studied positively charged particles in neon gas.

Explanation:

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3 years ago

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2 years ago

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A package is pushed across the floor a distance of 90 feet by exerting a force of 32 lbs downward at an angle of 24 ∘ with the h

Crazy boy [7]

Answer:

2632 foot-pound

Explanation:

Work done: Work is said to be done when ever a force moves a body through a given distance. The S.I unit of force is Newton (N).

From the question,

The expression for work done is given as,

W = Fdcos∅......................... Equation 1

Where W = work done, F = force, d = distance, ∅ = angle between the force and the horizontal.

Given: F = 32 lbs, d = 90 feet, ∅ = 24°

substitute into equation 1

W = 32×90×cos24

W = 2880(0.914)

W = 2632.32

W = 2632 foot-pound

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4 years ago

Who is responsible for developing the three laws of planetary motion?

Ivenika [448]

Johannes Kepler- he did it by observing the ‘Tycho Brahe’. His 3rd law was published 10 years later to his first two laws.

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3 years ago

A 8.0-cm-diameter horizontal pipe gradually narrows to 5.0 cm . When water flows through this pipe at a certain rate, the gauge

adelina 88 [10]

Answer:

A 8.0 cm diameter horizontal pipe gradually narrows to 5.0 cm. The the water flows through this pipe at certain rate, the gauge pressure in these two sections is 31.0 kPa and 24.0 kPa, respectively. What is the volume of rate of flow?

The flow rate is 3.1175×10⁻³ m³/s

Explanation:

To solve the question we rely on Bernoulli's principle as follows $P_{1} +\frac{1}{2}\rho v^{2} _{1} + \rho gz_{1} = P_{2} +\frac{1}{2}\rho v^{2} _{2} + \rho gz_{2}$

thus where the pipe is horizontal we have

z₁ = z₂ hence the above equattion becomes

$P_{1} +\frac{1}{2}\rho v^{2} _{1} = P_{2} +\frac{1}{2}\rho v^{2} _{2}$

since the flow rate is constant then

Q = v₁A₁ = v₂A₂

Where is the area of the two sections given by A₁ = π·D₁²÷4 and

A₂ = π·D₂²÷4

Thereffore A₁ = π·0.08²÷4 = 5.02×10⁻³ m²

and A₂ = π·0.05²÷4 = 1.96×10⁻³ m²

v₁ = v₂A₂/A₁ =0.391×v₂

The given pressures are P₁ = 31.0 kPa and P₂ = 24.0 pKa and

ρ = 1000 kg/m³

Plugging the values into the above equation we get

31.0 kPa +0.5× 1000 kg/m³× (0.391×v₂)² = 24.0 pKa +0.5×1000 kg/m³×v₂²

= 31000+76.3·v₂² =24000+500·v₂²

or 423.706·v₂² = 7000

v₂² = 7000/423.706 = 16.52 or v₂ = 4.065 m/s and v₁ 0.391×4.065 = 1.59 m/s

The flow rate = v₂A₂ = 1.59×1.96×10⁻³ = 3.1175×10⁻³ m³/s

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4 years ago