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Alina [70]
2 years ago
15

When you accelerate, the size of the front tire patch becomes____

Engineering
1 answer:
vekshin12 years ago
5 0

Answer:

Larger

Explanation:

When a vehicle is accelerated, the weight of the vehicle and the driver is transferred to the front of the  rear of the car.

This transferred weight is nothing but the  additional weight that presses the car's suspension in the downward direction thereby affecting tire contact patch in the rear due to which the rear tires loses adhesion.

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Why is it reasonable to say that no system is 100% efficient?​
Virty [35]

Generally, frictional losses are more predominant for the machines being not 100% efficient. This friction leads to the loss of energy in the form of heat, into the surroundings. Some of the supplied energy may be utilised to change the entropy (measure of randomness of the particles) of the system.

5 0
3 years ago
How can we calculate the speed of the output gear in a simple gear train? Explain with the help of an example.
Snowcat [4.5K]

Answer:

N_3=\dfrac{T_1}{T_3}N_1

Explanation:

In the diagram there three gears in which gear 1 is input gear ,gear 2 is idle gear and gear 3 is out put gear.

Lets take

Speed\ of\ gear 1=N_1

Number\ of\ teeth\ of\ gear 1=T_1

Speed\ of\ gear 3=N_3

Number\ of\ teeth\ of\ gear 3=T_3

All external matting gears will rotates in opposite direction with respect to each other.

So the speed of gear third can be given as follows

\dfrac{T_1}{T_3}=\dfrac{N_3}{N_1}

N_3=\dfrac{T_1}{T_3}N_1

3 0
3 years ago
Consider casting a concrete driveway 40 feet long, 12 feet wide and 6 in. thick. The proportions of the mix by weight are given
Akimi4 [234]

Answer:

Weight of cement = 10968 lb

Weight of sand = 18105.9 lb

Weight of gravel = 28203.55 lb

Weight of water = 5484 lb

Explanation:

Given:

Entrained air = 7.5%

Length, L = 40 ft

Width,w = 12 ft

thickness,b= 6 inch, convert to ft = 6/12 = 0.5 ft

Specific gravity of sand = 2.60

Specific gravity of gravel = 2.70

The volume will be:

40 * 12 * 0.5 = 240 ft³

We need to find the dry volume of concrete.

Dry volume = wet volume * 1.54 (concrete)

Dry volume will be = 240 * 1.54 = 360ft³

Due to the 7% entarained air content, the required volume will be:

V = 360 * (1 - 0.07)

V = 334.8 ft³

At a ratio of 1:2:3 for cement, sand, and gravel respectively, we have:

Total of ratio = 1+2+3 = 6

Their respective volume will be =

Volume of cement = \frac{1}{6}*334.8 = 55.8 ft^3

Volume of sand = \frac{2}{6}*334.8 = 111.6 ft^3

Volume of gravel = \frac{3}{6}*334.8 = 167.4 ft^3

To find the pounds needed the driveway, we have:

Weight = volume *specific gravity * density of water

Specific gravity of cement = 3.15

Weight of cement =

55.8 * 3.15 * 62.4 = 10968 pounds

Weight of sand =

111.6 * 2.60 * 62.4 = 18105.9 lb

Weight of gravel =

167.4 * 2.7 * 62.4 = 28203.55 lb

Given water to cement ratio of 0.50

Weight of water = 0.5 of weight of cement

= 1/2 * 10968 = 5484 lb

4 0
3 years ago
During his military campaign in what is now Germany, Julius Caesar lead his army of 40,000 soldiers to the western bank of the R
CaHeK987 [17]

Answer:

identifying a problem

Explanation:

its right

5 0
2 years ago
River models are used to study many different types of flow situations. A certain small river has an average width and depth of
slavikrds [6]

Answer: 7ft x21 I’d be right but yes I am

Explanation: because it is Welty

4 0
2 years ago
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