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3 years ago

Which of the formulas represents anonpolar molecular?

1 answer:

8 0

<span>CH4 I assume you're doing a test or something but I'm not good at explaining things so hopefully it's multiple choice! </span>

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A 6,000N is applied to a formula one car that weighs 500kg. What is the car's acceleration?

Vesna [10]

Answer:

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

$a = \frac{f}{m} \\$

f is the force

m is the mass

From the question

f = 6000 N

m = 500 kg

We have

$a = \frac{6000}{500} = \frac{60}{5} = 12 \\$

We have the final answer as

Hope this helps you

8 0

2 years ago

A 0.325 g wire is stretched between two points 57.7 cm apart. The tension in the wire is 650 N. Find the frequency of first harm

Galina-37 [17]

Answer:

f = 931.1 Hz

Explanation:

Given,

Mass of the wire, m = 0.325 g

Length of the stretch, L = 57.7 cm = 0.577 m

Tension in the wire, T = 650 N

Frequency for the first harmonic = ?

we know,

$v =\sqrt{\dfrac{T}{\mu}}$

μ is the mass per unit length

μ = 0.325 x 10⁻³/ 0.577

μ = 0.563 x 10⁻³ Kg/m

now,

$v =\sqrt{\dfrac{650}{0.563\times 10^{-3}}}$

v = 1074.49 m/s

The wire is fixed at both ends. Nodes occur at fixed ends.

For First harmonic when there is a node at each end and the longest possible wavelength will have condition

λ=2 L

λ=2 x 0.577 = 1.154 m

we now,

v = f λ

$f = \dfrac{v}{\lambda}$

$f = \dfrac{1074.49}{1.154}$

f = 931.1 Hz

The frequency for first harmonic is equal to f = 931.1 Hz

7 0

3 years ago

How much does the earth/sky weigh?

Talja [164]

Calculated weight (by experimentally) of Earth is 5.972 × 10²⁴ N

Hope this helps!

6 0

2 years ago

A 95 kg clock initially at rest on a horizontal floor requires a 650 n horizontal force to set it in motion. after the clock is

Tamiku [17]

The μs between the clock and floor is 650(M*g) and the μk between the clock and the floor is 560(M*g)

3 0

3 years ago

Read 2 more answers

Water, which we can treat as ideal and incompressible, flows at 12 m/s in a horizontal pipe with a pressure of 3.0 × 10^4 Pa.

Triss [41]

Answer:

9.8 × 10⁴Pa

Explanation:

Given:

Velocity V₁ = 12m/s

Pressure P₁ = 3 × 10⁴ Pa

From continuity equation we have

ρA₁V₁ = ρA₂V₂

A₁V₁ = A₂V₂

making V₂ the subject of the equation;

$V_{2} = \frac{A_{1}V_{1}}{A_{2}}$

the pipe is widened to twice its original radius,

r₂ = 2r₁

then the cross-sectional area A₂ = 4A₁

⇒ $V_{2}= \frac{A_{1}V_{1}}{4A_{1}}$

$V_{2}= \frac{V_{1}}{4}$

This implies that the water speed will drop by a factor of $\frac{1}{4}$ because of the increase the pipe cross-sectional area.

The Bernoulli Equation;

Energy per unit volume before = Energy per unit volume after

p₁ + $\frac{1}{2}$ ρV₁² + ρgh₁ = p₂ + $\frac{1}{2}$ ρV₂² + ρgh₂

Total pressure is constant and $P_{T}$ = P = $\frac{1}{2}$ ρV₂²ρV²

p₁ + $\frac{1}{2}$ ρV₁² = p₂ + $\frac{1}{2}$ ρV₂²

Making p₂ the subject of the equation above;

p₂ = p₁ + $\frac{1}{2}$ ρV₁² - $\frac{1}{2}$ ρV₂²

But $V_{2} = \frac{V_{1}}{4}$ so,

p₂ = p₁ + $\frac{1}{2}$ ρV₁² - $\frac{1}{2}$ ρ $\frac{V_{1}^{2}}{4^{2}}$

p₂ = 3.0 x 10⁴ + ( $\frac{1}{2}$ × 1000 × 12²) - ( $\frac{1}{2}$ × 1000 × 12²/4² )

P₂ = 3.0 x 10⁴ + 7.2 × 10⁴ - 4.05 x 10³

P₂ = 9.79 × 10⁴Pa

P₂ = 9.8 × 10⁴Pa

4 0

2 years ago