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Temka [501]
4 years ago
9

Two cars A and B leave an intersection at the same time. Car A travels west at an average speed of x miles per hour and car B tr

avels south at an average speed of y miles per hour. Write a program that prompts the user to enter the average speed of both the cars and the elapsed time (in hours and minutes) and outputs the (shortest) distance between the cars.
Engineering
1 answer:
svet-max [94.6K]4 years ago
7 0

Answer:

Distance between A & B, dAB, is hypotenus.

dAB²=dA²+dB²

A'=70 mph, B'=55mph, (convert t to hours)

dA=A'•t, dB=B'•t

dAB=√((70t)²+(55t)²)

Explanation:

Distance between A & B, dAB, is hypotenus.

dAB²=dA²+dB²

A'=70 mph, B'=55mph, (convert t to hours)

dA=A'•t, dB=B'•t

dAB=√((70t)²+(55t)²)

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Engineers need to be open-ended when dealing with their designs. Why?
melomori [17]
I think the answer would be A if its wrong I’m sorry
7 0
3 years ago
A 15.00 mL sample of a solution of H2SO4 of unknown concentration was titrated with 0.3200M NaOH. the titration required 21.30 m
natima [27]

Answer:

a. 0.4544 N

b. 5.112 \times 10^{-5 M}

Explanation:

For computing the normality and molarity of the acid solution first we need to do the following calculations

The balanced reaction

H_2SO_4 + 2NaOH = Na_2SO_4 + 2H_2O

NaOH\ Mass = Normality \times equivalent\ weight \times\ volume

= 0.3200 \times 40 g \times 21.30 mL \times  1L/1000mL

= 0.27264 g

NaOH\ mass = \frac{mass}{molecular\ weight}

= \frac{0.27264\ g}{40g/mol}

= 0.006816 mol

Now

Moles of H_2SO_4 needed  is

= \frac{0.006816}{2}

= 0.003408 mol

Mass\ of\ H_2SO_4 = moles \times molecular\ weight

= 0.003408\ mol \times 98g/mol

= 0.333984 g

Now based on the above calculation

a. Normality of acid is

= \frac{acid\ mass}{equivalent\ weight \times volume}

= \frac{0.333984 g}{49 \times 0.015}

= 0.4544 N

b. And, the acid solution molarity is

= \frac{moles}{Volume}

= \frac{0.003408 mol}{15\ mL \times  1L/1000\ mL}

= 0.00005112

=5.112 \times 10^{-5 M}

We simply applied the above formulas

4 0
3 years ago
Q1: The first option should always be to get out safely (RUN)
nekit [7.7K]

Answer:

Q1 true

Q2 true

And other I am confuse

6 0
3 years ago
A 3-m-high, 11-m-wide rectangular gate is hinged at the top edge at A and is restrained by a fixed ridge at B. Determine the hyd
serg [7]

Answer:

its a

Explanation:

its a

5 0
3 years ago
A 100 ft long steel wire has a cross-sectional area of 0.0144 in.2. When a force of 270 lb is applied to the wire, its length in
blondinia [14]

Answer:

(a) The stress on the steel wire is 19,000 Psi

(b) The strain on the steel wire is 0.00063

(c) The modulus of elasticity of the steel is 30,000,000 Psi

Explanation:

Given;

length of steel wire, L = 100 ft

cross-sectional area, A = 0.0144 in²

applied force, F = 270 lb

extension of the wire, e = 0.75 in

<u>Part (A)</u> The stress on the steel wire;

δ = F/A

   = 270 / 0.0144

δ  = 18750 lb/in² = 19,000 Psi

<u>Part (B)</u> The strain on the steel wire;

σ = e/ L

L = 100 ft = 1200 in

σ = 0.75 / 1200

σ = 0.00063

<u>Part (C)</u> The modulus of elasticity of the steel

E = δ/σ

   = 19,000 / 0.00063

E = 30,000,000 Psi

4 0
3 years ago
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