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3 years ago

PLEASE ANSWER ASAP!

Engineering

2 answers:

Butoxors [25]3 years ago

7 0

Answer:

Please check the explanation.

Explanation:

While driving, white lines would require us to stay within the lane, also marking the shoulder of the roadway.

While yellow lines generally mark the center of a two-way road being used for two-way traffic.

We generally can pass on a two-way road if the yellow centerline is broken.

tigry1 [53]3 years ago

5 0

Answer:

Solid White Line: requires you to stay within the lane and also marks the shoulder of the roadway. YELLOW LINES mark the center of a two-way road used for two-way traffic. You may pass on a two-way road if the yellow centerline is broken.

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What are the well-known effects of electricity

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Answer:

Hence, the three effects of electric current are heating effect, magnetic effect and chemical effect.

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3 years ago

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Gas chromatography separates compounds depending on their__________ . Benzene, m-xylene, and toluene have similar_________ , the

amm1812

Answer and Explanation:

Gas chromatography separates compounds depending on their **polarity and volatility**. Benzene, m-xylene, and toluene have similar **polarities**, therefore, the main basis for separation is **volatility**. The more volatile a component the ** higher its vapor pressure**, hence the more time it spends in the **gaseous mobile phase**, giving it a **shorter** retention time. Therefore, components of a liquid mixture will elute in order of **increasing boiling points/decreasing volatilities/increasing polarities with the stationary phase**.

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3 years ago

A closed, rigid tank is filled with a gas modeled as an ideal gas, initially at 27°C and a gage pressure of 300 kPa. If the gas

ch4aika [34]

Answer:

gauge pressure is 133 kPa

Explanation:

given data

initial temperature T1 = 27°C = 300 K

gauge pressure = 300 kPa = 300 × 10³ Pa

atmospheric pressure = 1 atm

final temperature T2 = 77°C = 350 K

to find out

final pressure

solution

we know that gauge pressure is = absolute pressure - atmospheric pressure so

P (gauge ) = 300 × 10³ Pa - 1 × $10^{5}$ Pa

P (gauge ) = 2 × $10^{5}$ Pa

so from idea gas equation

$\frac{P1*V1}{T1} = \frac{P2*V2}{T2}$ ................1

so ${P2} = \frac{P1*T2}{T1}$

${P2} = \frac{2*10^5*350}{300}$

P2 = 2.33 × $10^{5}$ Pa

so gauge pressure = absolute pressure - atmospheric pressure

gauge pressure = 2.33 × $10^{5}$ - 1.0 × $10^{5}$

gauge pressure = 1.33 × $10^{5}$ Pa

so gauge pressure is 133 kPa

4 0

3 years ago

. Two rods, with masses MA and MB having a coefficient of restitution, e, move

GarryVolchara [31]

Answer:

a) $V_A = \frac{(M_A - eM_B)U_A + M_BU_B(1+e)}{M_A + M_B}$

$V_B = \frac{M_AU_A(1+e) + (M_B - eM_A)U_B}{M_A + M_B}$

b) $U_A = 3.66 m/s$

$V_B = 4.32 m/s$

c) Impulse = 0 kg m/s²

d) percent decrease in kinetic energy = 47.85%

Explanation:

Let $U_A$ be the initial velocity of rod A

Let $U_B$ be the initial velocity of rod B

Let $V_A$ be the final velocity of rod A

Let $V_B$ be the final velocity of rod B

Using the principle of conservation of momentum:

$M_AU_A + M_BU_B = M_AV_A + M_BV_B$ ............(1)

Coefficient of restitution, $e = \frac{V_B - V_A}{U_A - U_B}$

$V_A = V_B - e(U_A - U_B)$ ........................(2)

Substitute equation (2) into equation (1)

$M_AU_A + M_BU_B = M_A(V_B - e(U_A - U_B)) + M_BV_B$ ..............(3)

Solving for $V_B$ in equation (3) above:

$V_B = \frac{M_AU_A(1+e) + (M_B - eM_A)U_B}{M_A + M_B}$ ....................(4)

From equation (2):

$V_B = V_A + e(U_A -U_B)$ ......(5)

Substitute equation (5) into (1)

$M_AU_A + M_BU_B = M_AV_A + M_B(V_A + e(U_A -U_B))$ ..........(6)

Solving for $V_A$ in equation (6) above:

$V_A = \frac{(M_A - eM_B)U_A + M_BU_B(1+e)}{M_A + M_B}$ .........(7)

$M_A = 2 kg\\M_B = 1 kg\\U_B = -3 m/s( negative x-axis)\\e = 0.65\\U_A = ?$

Rod A is said to be at rest after the impact, $V_A = 0 m/s$

Substitute these parameters into equation (7)

$0 = \frac{(2 - 0.65*1)U_A - (1*3)(1+0.65)}{2+1}\\U_A = 3.66 m/s$

To calculate the final velocity, $V_B$ , substitute the given parameters into (4):

$V_B = \frac{(2*3.66)(1+0.65) - (1 - (0.65*2))*3}{2+1}\\V_B = 4.32 m/s$

c) Impulse, $I = M_AV_A + M_BV_B - (M_AU_A + M_BU_B)$

$I = (2*0) + (1*4.32) - ((2*3.66) + (1*-3))$

I = 0 $kg m/s^2$

d) % $\triangle KE = \frac{(0.5 M_A V_A^2 + 0.5 M_B V_B^2) - ( 0.5 M_A U_A^2 + 0.5 M_B U_B^2)}{0.5 M_A U_A^2 + 0.5 M_B U_B^2} * 100\%$

% $\triangle KE = \frac{((0.5*2*0) + (0.5 *1*4.32^2)) - ( (0.5 *2*3.66^2) + 0.5*1*(-3)^2))}{ (0.5 *2*3.66^2) + 0.5*1*(-3)^2)} * 100\%$

% $\triangle KE = -47.85 \%$

7 0

3 years ago

A driver traveling in her 16-foot SUV at the speed limit of 30 mph was arrested for running a red light at 15th and Main, an int

Lynna [10]

Answer:

(a) Yes

(b) 102.8 ft

Explanation:

(a)First let convert mile per hour to feet per second

30 mph = 30 * 5280 / 3600 = 44 ft/s

The time it takes for this driver to decelerate comfortably to 0 speed is

t = v / a = 44 / 10 = 4.4 (s)

given that it also takes 1.5 seconds for the driver reaction, the total time she would need is 5.9 seconds. Therefore, if the yellow light was on for 4 seconds, that's not enough time and the dilemma zone would exist.

(b) At this rate the distance covered by the driver is

$s = v_0t + \frac{at^2}{2}$

$s =44*1.5 + 44(4.4) - \frac{10*4.4^2}{2} = 162.8 (ft)$

Since the intersection is only 60 feet wide, the dilemma zone must be

162.8 - 60 = 102.8 ft

4 0

3 years ago