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3 years ago

If you are setting up a race car. What is the cross weight? Does it matter?

Engineering

1 answer:

lara31 [8.8K]3 years ago

7 0

Answer:

cross-weight is used to tighten it up.

Explanation:

and yes this is important because Cross-weight percentage compares the diagonal weight totals to the car's total weight.

hope this help

(mark this answer as an brainliest answer)

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John has just graduated from State University. He owes $35,000 in college loans, but he does not have a job yet. The college loa

IRISSAK [1]

Answer:

The correct response is "821.88". A further explanation is given below.

Explanation:

According to the question,

The largest amount unresolved after five years would have been:

= $35000\times (\frac{F}{P}, 4 \ percent,5 )$

= $35000\times 1.216 7$

= $42584.50$

Now,

time (t) will be:

= $5\times 12$

= $60 \ monthly \ payments$

So, monthly payment will be:

= $42582.85\times (\frac{A}{P}, 0.5 \ percent,60 )$

= $42584.50\times 0.0193$

= $821.88$

6 0

3 years ago

X+3=2 x=?? No spamming

PtichkaEL [24]

Answer:

x+3=2

x=2-3꧁

꧁꧁꧂꧂꧂꧂

x=-1

4 0

3 years ago

How to study thermodynamics?

expeople1 [14]

It is study of the relationships between heat, temprature, work and energy

7 0

2 years ago

Read 2 more answers

The end of a large tubular workpart is to be faced on a NC vertical boring mill. The part has an outside diameter of 38.0 in and

nata0808 [166]

Answer:

(a) the cutting time to complete the facing operation = 11.667mins

b) the cutting speeds and metal removal rates at the beginning= 12.89in³/min and end of the cut. = 8.143in³/min

Explanation:

check attached files below for answer.

5 0

3 years ago

A 1 m3 rigid tank initially contains air whose density is 1.18kg/m3. The tank is connected to a high pressure supply line throug

Elanso [62]

To solve this problem it is necessary to apply the concepts related to density in relation to mass and volume for each of the states presented.

Density can be defined as

$\rho = \frac{m}{V}$

Where

m = Mass

V = Volume

For state one we know that

$\rho_1 = \frac{m_1}{V}$

$m_1 = \rho_1 V$

$m_1 = 1.18*1$

$m_1 = 1.18Kg$

For state two we have to

$\rho_2 = \frac{m_2}{V}$

$m_2 = \rho_2 V$

$m_1 = 7.2*1$

$m_1 = 7.2Kg$

Therefore the total change of mass would be

$\Delta m = m_2-m_1$

$\Delta m = 7.2-1.18$

$\Delta m = 6.02Kg$

Therefore the mass of air that has entered to the tank is 6.02Kg

5 0

3 years ago