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2 years ago

When you park your car uphill next to a curb, the right front wheel should be:?

1 answer:

6 0

<span> When headed uphill at a </span>curb<span>, turn the front </span>wheels<span> away from the </span>curb<span> and let </span>your vehicle<span> roll backwards slowly until the rear part of the front </span>wheel<span> rests against the </span>curb<span> using it as a block.</span>

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A railroad car is pulled through the distance of 960 m by a train that did 578 kJ of work during this pull.

Wittaler [7]

Answer:

Explanation:

The force supplied by the train can be found by using the formula

$f = \frac{w}{d} \\$

w is the workdone

d is the distance

From the question we have

$f = \frac{578000}{960} \\ = 602.083333...$

We have the final answer as

Hope this helps you

7 0

3 years ago

If Patricia worked 40 hours this week and her paycheck shows that her gross pay is $ 340 , what is her hourly wage?

GrogVix [38]

For 40 hours, she gets paid $340
Therefore, for 1 hour, she gets paid 340/40=$8.50
Her hourly wage is $8.50

8 0

2 years ago

The Balmer series in hydrogen atom produces only infrared emission. O True O False

lorasvet [3.4K]

Answer:

False

Explanation:

As we know that, the Balmer series gives the n values as,

$n_{i}=2$ .

$[tex]n_{f}=3,4,5,.....\infty$ .

Now the value of wavelength can be calculated as,

$\frac{1}{\lambda}=R(\frac{1}{n_{i} }-\frac{1}{n_{f} } )z^{2}$ .

Here, $R=109677 cm^{-1}$ .

And $n_{f}=3$ .

Now,

$\frac{1}{\lambda}=109677 cm^{-1}(\frac{1}{2}-\frac{1}{3})$ .

Therefore,

$\lambda=\frac{6}{109677} cm\\\lambda=547\times 10^{-9} m\\ \lambda=547 nm$

Therefore, the wavelength of Balmer series lies in visible region which is 547 nm.

8 0

3 years ago

Read 2 more answers

What is the definition of motivation?

elixir [45]

Motivation is an encouragement to do or achieve something

5 0

3 years ago

Three conducting plates, each of area A, are connected as shown.

Shkiper50 [21]

You have effectively got two capacitors in parallel. The effective capacitance is just the sum of the two.
Cequiv = ε₀A/d₁ + ε₀A/d₂ Take these over a common denominator (d₁d₂)
Cequiv = ε₀d₂A + ε₀d₁A / (d₁d₂) Cequiv = ε₀A( (d₁ + d₂) / (d₁d₂) )
B) It's tempting to just wave your arms and say that when d₁ or d₂ tends to zero C -> ∞, so the minimum will occur in the middle, where d₁ = d₂
But I suppose we ought to kick that idea around a bit.
(d₁ + d₂) is effectively a constant. It's the distance between the two outer plates. Call it D.
C = ε₀AD / d₁d₂ We can also say: d₂ = D - d₁ C = ε₀AD / d₁(D - d₁) C = ε₀AD / d₁D - d₁²
Differentiate with respect to d₁
dC/dd₁ = -ε₀AD(D - 2d₁) / (d₁D - d₁²)² {d2C/dd₁² is positive so it will give us a minimum} For max or min equate to zero.
-ε₀AD(D - 2d₁) / (d₁D - d₁²)² = 0 -ε₀AD(D - 2d₁) = 0 ε₀, A, and D are all non-zero, so (D - 2d₁) = 0 d₁ = ½D
In other words when the middle plate is halfway between the two outer plates, (quelle surprise) so that
d₁ = d₂ = ½D so
Cmin = ε₀AD / (½D)² Cmin = 4ε₀A / D Cmin = 4ε₀A / (d₁ + d₂)

7 0

3 years ago