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3 years ago

A keyboard, monitor and mouse are all considered as computer hardware and a program is known as a software.

Engineering

1 answer:

Iteru [2.4K]3 years ago

5 0

Answer:

This should be true because program is a software and mouse and keyboard are specific computer hardware to make your exsperience with a computer more efficent and faster.

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A driver is traveling at 52 mi/h on a wet road. an object is spotted on the road 415 ft ahead and the driver is able to come to

sveticcg [70]

Answer:

G = 0.424

Explanation:

Ds = ( 0.278tr * V ) + (0.278 * V²)/ ( 19.6* ( f ± G))

Where Ds = stopping sight distance = 415miles = 126.5m

G = absolute grade road

V = velocity of vehicle = 52miles/hr

f = friction = 0 because the road is wet

tr = standard perception / reaction time = 2.5s

So therefore:

Substituting to get G

We have

2479.4G = 705.6G + 751.72

1773.8G = 751.72

G = 751.72/1773.8

G = 0.424

8 0

4 years ago

Read 2 more answers

The water of a 14’ × 48’ metal frame pool can drain from the pool through an opening at the side of the pool. The opening is abo

Tresset [83]

Answer:

Explanation:

Height h = 1.03m

Volume v = 3780 gallons = 3780 * 0.0037851m^3 = 14.3073m^3

Time t = 13.5 mins = 13.5 * 60 = 810 seconds

Length of pool L = 14 inch = 14 * 2.54 = 35.56cm

width of pool b = 48 inch = 48 * 2.54 = 121.92 cm

a.) Consider the bernoulli's equation is given as:

$P_1+\rho gh_1 + \frac{1}{2}\rho v_1^2 = P_2 + \rho gh_2 + \frac{1}{2}\rho v_2^2 ...(1)$

consider the equation of bernoulli at the top of the pool

$P_0+\rho gh_1 + \frac{1}{2}\rho v_1^2 =constant ...(2)$

where $P_1=P_0$ atm pressure

At the top of the pool $v_1=0m/s$ , substitute in V_1 in equation (2)

$P_0+\rho gh_1 =constant ...(3)$

Hence equation (3) serves as the bernoullis equation at the top.

b.) Consider the equation of bernoulli's at the opening of the pool

$P_2+\rho gh_2 + \frac{1}{2}\rho v_2^2 =constant ...(4)\\P_0+\rho gh_2 + \frac{1}{2}\rho v_2^2 =constant ...(5)$

where $P_2=P_0$ atm pressure and $h_2=0m$

$P_0+\rho v_1^2 =constant ...(6)$

Hence equation (6) serves as the bernoullis equation of water at the opening of the pool.

c.) Consider the equation (3) and (4)

$P_0+\rho gh_1 =P_0+\rho v_1^2\\\\\frac{1}{2}\rho v_2^2=\rho gh_1\\v_2^2=2gh_1\\v_2=(\sqrt{2gh_1})m/s...(7)$

Hence velocity is $v_2=(\sqrt{2gh_1})m/s$

d.) consider (7)

$v_2=(\sqrt{2(9.81)(1.03)})=4.4954m/s(approx)$

This is the norminal value of velocity

e.) consider the equation of flow rate interval of v and t

flow(t)= $\frac{dv}{dt}(m^3/s)$ hence this is the flow rate

f.) Consider the equation cross sectional area in terms of V,v2 and t

$AV_2=\frac{v}{t}\\\\A=\frac{v}{v_2t}(m^2)...8$

hence this serves as the cross sectional area.

g.) Consider the equation of area from equation (8)

$A=\frac{v}{v_2t}\\=\frac{14.3073}{4.4954\times 810}=0.003929=0.00393m^2=39.3cm^2$

6 0

3 years ago

It is known that the connecting rod AB exerts on the crank BCa 2.5-kN force directed down andto the left along the centerline of

monitta

Answer:

M_c = 61.6 Nm

Explanation:

Given:

F_a = 2.5 KN

Find:

Determine the moment of this force about C for the two casesshown.

Solution:

- Draw horizontal and vertical vectors at point A.

- Take moments about point C as follows:

M_c = F_a*( 42 / 150 ) *88

M_c = 2.5*( 42 / 150 ) *88

M_c = 61.6 Nm

- We see that the vertical component of force at point A passes through C.

Hence, its moment about C is zero.

6 0

3 years ago

A tank with a volume of 8 m3 containing 4 m3 of 20% (by volume) NaOH solution is to be purged by adding pure water at a rate of

lawyer [7]

Answer:

The time necessary to purge 95% of the NaOH is 0.38 h

Explanation:

Given:

vfpure water(i) = 3 m³/h

vNaOH = 4 m³

xNaOH = 0.2

vfpure water(f) = 2 m³/h

pwater = 1000 kg/m³

pNaOH = 1220 kg/m³

The mass flow rate of the water is = 3 * 1000 = 3000 kg/h

The mass of NaOH in the solution is = 0.2 * 4 * 1220 = 976 kg

When the 95% of the NaOH is purged, thus the NaOH in outlet is = 0.95 * 976 = 927.2 kg

The volume of NaOH in outlet after time is = 927.2/1220 = 0.76 m³

The time required to purge the 95% of the NaOH is = 0.76/2 = 0.38 h

4 0

4 years ago

Select the correct answer.<br> Which equation gives you the amount of work performed?

Fantom [35]

Answer:

Hello

this is the answer

5 0

4 years ago