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natta225 [31]
2 years ago
10

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

Physics
1 answer:
Cloud [144]2 years ago
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:

<h2>602.08 N</h2>

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

<h3>602.08 N</h3>

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
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