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2 years ago

Identify 3 distractions for young drivers and explain how you plan to minimize these distractions

Engineering

2 answers:

Ghella [55]2 years ago

6 0

Answer: devices,overthinking or rush and lack of experience

Explanation: devices: most young drivers tend to be on their devices 50% of time while driving and this could lead to lack of concentration hereby causing an accident.

overthinking or rush: on days where we are late to school ,appointments e.t.c we tend to rush and we end up forgetting some things same with driving when they rush they tend to lose concentration and focus and this eventually needs to overthinking

lack of experience: this is mainly common with people who drive without permission or a driver's license so they eventually have an accident when they do not have experience due to panic and other things.

Veseljchak [2.6K]2 years ago

4 0

Answer:

Phone, Eating, People in the car

Explanation:

your phone should always be off while driving and put away. Eating much like texting takes your attention away from the road so you should wait till you're parked or home to eat. People thrashing around or talking very loudly can be a terrible distraction. Ask them nicely to quiet down so you can focus on the road.

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A bankrupt chemical firm has been taken over by new management. On the property they found a 20,000-m3 brine pond containing 25,

Liula [17]

Answer:

Flow-rate = 0.0025 m^3/s

Explanation:

We need to assume that the flow-rate of pure water entering the pond is the same as the flow-rate of brine leaving the pond, in other words, the volume of liquid in the pond stays constant at 20,000 m^3. Using the previous assumption we can calculate the flow rate entering or leaving the tank (they are the same) building a separable differential equation dQ/dt, where Q is the milligrams (mg) of salt in a given time t, to find a solution to our problem we build a differential equation as follow:

dQ/dt = -(Q/20,000)*r where r is the flow rate in m^3/s

what we pose with this equation is that the variable rate at which the salt leaves the pond (salt leaving over time) is equal to the concentration (amount of salt per unit of volume of liquid at a given time) times the constant rate at which the liquid leaves the tank, the minus sign in the equation is because this is the rate at which salt leaves the pond.

Rearranging the equation we get dQ/Q = -(r/20000) dt then integrating in both sides ∫dQ/Q = -∫(r/20000) dt and solving ln(Q) = -(r/20000)*t + C where C is a constant (initial value) result of solving the integrals. Please note that the integral of dQ/Q is ln(Q) and r/20000 is a constant, therefore, the integral of dt is t.

To find the initial value (C) we evaluate the integrated equation for t = 0, therefore, ln(Q) = C, because at time zero we have a concentration of 25000 mg/L = 250000000 mg/m^3 and Q is equal to the concentration of salt (mg/m^3) by the amount of liquid (always 20000 m^3) -> Q = 250000000 mg/m^3 * 20000 m^3 = 5*10^11 mg -> C = ln(5*10^11) = 26.9378. Now the equation is ln(Q) = -(r/20000)*t + 26.9378, the only thing missing is to find the constant flow rate (r) required to reduce the salt concentration in the pond to 500 mg/L = 500000 mg/m^3 within one year (equivalent to 31536000 seconds), to do so we need to find the Q we want in one year, that is Q = 500000 mg/m^3 * 20000 m^3 = 1*10^10 mg, therefore, ln(1*10^10) = -(r/20000)*31536000 + 26.9378 solving for r -> r = 0.002481 m^3/s that is approximately 0.0025 m^3/s.

Note:

ln() refer to natural logarithm
The amount of liquid in the tank never changes because the flow-rate-in is the same as the flow-rate-out
When solving the differential equation we calculated the flow-rate-out and we were asked for the flow-rate-in but because they are the same we could solve the problem
During the solving process, we always converted units to m^3 and seconds because we were asked to give the answer in m^3/seg

7 0

2 years ago

The Hernandez family budget is shown in the graph.

pishuonlain [190]

Answer:

housing, savings, and food

housing, food, and transportation

Explanation:

Given: Family Budget comprises of 23% housing, 21% food, 10% savings, 16% transportation, 12% medical, 13% clothing and 5% emergency fund.

To find: categories that make up more than half of the Hernandez family budget

Solution:

clothing + medical + emergency fund = $13+12+5=30 \%$

clothing + housing + savings = $13+23+10=46\%$

housing + savings + food = $23+10+21=54\%$

housing + food + transportation = $23+21+16=60\%$

food + transportation + emergency fund = $21+16+5=42\%$

So, 3 categories that make up more than half of the Hernandez family budget are housing, savings, and food, housing, food, and transportation

5 0

2 years ago

Read 2 more answers

A binary star system consists of two stars of masses m1m1m_1 and m2m2m_2. The stars, which gravitationally attract each other, r

d1i1m1o1n [39]

Answer:

$a_c_2=\dfrac{a_c_1\times m_1}{m_2}$

Explanation:

The question is: <em>Find the magnitude of the centripetal acceleration of the star with mass m₂</em>

The <em>centripetal acceleration</em> is the quotient of the centripetal force and the mass.

$a_c=\dfrac{F_c}{m}$

Thus, you can write the equations for each star:

$a_c_1=\dfrac{F_c_1}{m_1}$

$a_c_2=\dfrac{F_c_2}{m_2}$

As per Newton's third law, the centripetal forces are equal in magnitude. Then:

$a_c_1\times m_1=a_c_2\times m_2$

Now you can clear $a_c_2$ :

$a_c_2=\dfrac{a_c_1\times m_1}{m_2}$

6 0

2 years ago

The uniform dresser has a weight of 90 lb and rests on a tile floor for which the coefficient of static friction is 0.25. If the

Juliette [100K]

Answer:

a) $F = 736.065\,lbf$ , b) $\mu_{k} = 0.15$

Explanation:

a) The uniform dresser is modelled by using the following equations of equilibrium:

$\Sigma F_{x} = F - \mu_{k}\cdot N = 0$

$\Sigma F_{y} = N-m\cdot g=0$

After some algebraic manipulation, the following formula is derived:

$F = \mu_{k}\cdot m \cdot g$

$F = (0.25)\cdot (90\,lbm)\cdot (32.714\,\frac{ft}{s^{2}} )$

$F = 22.5\,lbf$

b) The man is described by the following equations of equilibrium:

$\Sigma F_{x} = -F + \mu_{k}\cdot N = 0$

$\Sigma F_{y} = N-m\cdot g=0$

After some algebraic manipulation, the following formula for the static coefficient of friction is:

$\mu_{k} = \frac{F}{m\cdot g}$

$\mu_{k} = \frac{22.5\,lbf}{150\,lbf}$

$\mu_{k} = 0.15$

3 0

2 years ago

You are considering building a residential wind power system to produce 6,000 kWh of electricity each year. The installed cost o

alisha [4.7K]

Answer:

leveled cost of electricity LCOE is 0.1159/kwh

Explanation:

given data:

installation cost of system = $1.50/winstalled

capacity factor = 22%

life = 10 year

rate of interest 8%

operation and maintenance cost = $0.04 / kwh generated

capcaitence recovery factor $CRF = \frac{i(i+1)^n}{(i+1)^n -1}$

i= rate of interest

n = annuity period

$CRF = \frac{0.08(1+0.08)^10}}{(1+0.08)^{10} -1} = 0.14903$

$LCOE = \frac{cost\ of \installation \times CRF}{hours/ available \times capacity\ factor} + variable\ o&M\ cost$

$= \frac{1.50\times 1000/kw \times 0.14903}{8760\times 0.22} + 0.04/kwh$

= 0.1159/kwh + 0.04 /kwh

= 0.1559 /kwh

8 0

2 years ago