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1 year ago

Loss of traction between the rear wheels and road surfaces like ice, sand, or gravel results in what is called _______________.

Physics

1 answer:

Tatiana [17]1 year ago

8 0

Answer:

Fish tailing

Explanation:

Fish tailing is the vehicle handling problem which occurs due to the decrease in friction between the the rear wheels and the surface. So we mostly observe that rear wheels slide out of control when there is rain or ice covered roads, sand etc. They reduce the friction of the surface resulting in over-steering. This problem is named as fish tailing.

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The center of the Hubble space telescope is 6940 km from Earth’s center. If the gravitational force between Earth and the telesc

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The answer is 11,121 kg

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1 year ago

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The diameters of bolts produced by a certain machine are normally distributed with a mean of 0.30 inches and a standard deviatio

enot [183]

Answer:

2.25 %

Explanation:

65-95-99.7 is a rule to remember the precentages that lies around the mean.

at the range of mean ( $\mu$ ) plus or minus one standard deviation ( $\sigma$ ), $P([\mu-\sigma \leq X \leq \mu+\sigma])\approx 68.3%$

at the range of mean plus or minus two standard deviation, $P([\mu -2\sigma \leq X \leq \mu+2\sigma])\approx 95.5%$

at the range of mean plus or minus three standard deviation, $P([\mu - 3\sigma\leq X \leq \mu+3\sigma])\approx 99.7%$

So, note that they are asking about the probability that it is greater than 0.32, that is the mean (0.3) plus two times the standard deviation (0.1) ( $P(X \leq \mu+2\sigma)$ )

So we know that the 95.5% is between $\mu - 2\sigma = 0.3 -2*0.1 = 0.28$ and $\mu + 2\sigma = 0.3 +2*0.1 = 0.32$ , hence approximately the 4.5% (100%-95.5%) is greater than 0.32 or less than 0.28. But half (4.5%/2=2.25%) is greater than 0.32 and the other half is less than 0.28.

So $P(X \leq \mu+2\sigma) \approx 2.25%$

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1 year ago

A teacher sets up a stand carrying a convex lens of focal length 15 cm at 20.5 cm mark on the optical bench. She asks the studen

Brums [2.3K]

We get the clearest image if there is no magnification. When we have no magnification the image and real object have the same size.
If we look at the diagram that I attached we can see that:
$\frac{h_i}{h_0}=\frac{d_i}{d_0}$
Two triangles that I marked are similar and from this we get:
$\frac{h_i}{h_0}=\frac{d_i-f}{f}$
The image and the object must have the same height so we get:
$\frac{h_i}{h_0}=\frac{d_i-f}{f};h_i=h_0\\
1=\frac{d_i-f}{f}\\
d_i=2f$
This tells how far the screen should be from the lens.
The position of the screen on the optical bench is:
$S=20.5cm+2f=20.5+2\cdot 15cm=50.5cm$

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1 year ago

A small block of wood of inertia mb is released from rest a distance h above the ground, directly above your head. you decide to

Zinaida [17]

Since I'm assuming that its perfectly elastic, considering there's not enough information given, so I think that no energy is dissipated in the collision

hmax = h - d + { [ mpvp - mb√(2gd) ] / (mp+mb) }² / (2g)

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1 year ago

Air at 3 104 kg/s and 27 C enters a rectangular duct that is 1m long and 4mm 16 mm on a side. A uniform heat flux of 600 W/m2 is

ad-work [718]

Answer:

$T_{out}=27.0000077$ ºC

Explanation:

First, let's write the energy balance over the duct:

$H_{out}=H_{in}+Q$

It says that the energy that goes out from the duct (which is in enthalpy of the mass flow) must be equals to the energy that enters in the same way plus the heat that is added to the air. Decompose the enthalpies to the mass flow and specific enthalpies:

$m*h_{out}=m*h_{in}+Q\\m*(h_{out}-h_{in})=Q$

The enthalpy change can be calculated as Cp multiplied by the difference of temperature because it is supposed that the pressure drop is not significant.

$m*Cp(T_{out}-T_{in})=Q$

So, let's isolate $T_{out}$ :

$T_{out}-T_{in}=\frac{Q}{m*Cp}\\T_{out}=T_{in}+\frac{Q}{m*Cp}$

The Cp of the air at 27ºC is 1007 $\frac{J}{kgK}$ (Taken from Keenan, Chao, Keyes, “Gas Tables”, Wiley, 1985.); and the only two unknown are $T_{out}$ and Q.

Q can be found knowing that the heat flux is 600W/m2, which is a rate of heat to transfer area; so if we know the transfer area, we could know the heat added.

The heat transfer area is the inner surface area of the duct, which can be found as the perimeter of the cross section multiplied by the length of the duct:

Perimeter:

$P=2*H+2*A=2*0.004m+2*0.016m=0.04m$