All categories

2 years ago

6

Engineering

2 answers:

grin007 [14]2 years ago

8 0

Answer:

(1)speed at which ball B was thrown upward-13.95m/s (2) velocity pf a=14m/s and b=7.25 m/s

Explanation:

Eduardwww [97]2 years ago

7 0

Answer:

ball a is released from rest at a height of 20 m

Explanation:

You might be interested in

There are 10 vehicles in a queue when an attendant opens a toll booth. Vehicles arrive at the booth at a rate of 4 per minute. T

dmitriy555 [2]

Answer:

as slated in your solution, if delay time is 2.30 mins, hence 9 vehicle will be on queue as the improved service commenced.

Explanation:

4 vehicle per min, in 2 mins of the delay time 8 vehicles while in 0.3 min average of 1 vehicle join the queue. making 9 vehicle maximum

3 0

2 years ago

The interior wall of a building is made from 2×4 wood studs, plastered on one side. If the wall is 13 ft high, determine the loa

Elanso [62]

Answer:

load = 156 lb/ft

Explanation:

given data

interior wall of a building = 2×4 wood studs

plastered = 1 side

wall height = 13 ft

solution

we get here load so first we get wood stud load and that is

we know here from ASCE-7 norm

dead load of 2 x 4 wood studs with 1 side plaster = 12 psf

and we have given height 13 ft

so load will be = 12 psf × 13 ft

load = 156 lb/ft

7 0

2 years ago

Determine the mass of a car that weight 3,500 lbs both in slugs and kilograms. The answer should be in kilograms.

Andreas93 [3]

Answer:

The mass of car in slug =108.5 slug.

The mass of car in kilogram =1575 Kg.

Explanation:

Given that

Mass of car = 3500 lbs

We know that

1 lbs =0.031 slug

3500 lbs = 0.031 x 3500 slug

So the mass of car in slug =108.5 slug.

We know that

1 lbs =0.45 kilograms

3500 lbs = 3500 x 0.45 Kg

So the mass of car in kilogram =1575 Kg.

4 0

2 years ago

Water flows through a pipe of 100 mm at the rate of 0.9 m3 per minute at section A. It tapers to 50mm diameter at B, A being 1.5

pochemuha

Answer:

The velocities in points A and B are 1.9 and 7.63 m/s respectively. The Pressure at point B is 28 Kpa.

Explanation:

Assuming the fluid to be incompressible we can apply for the continuity equation for fluids:

$Aa.Va=Ab.Vb=Q$

Where A, V and Q are the areas, velocities and volume rate respectively. For section A and B the areas are:

$Aa=\frac{pi.Da^2}{4}= \frac{\pi.(0.1m)^2}{4}=7.85*10^{-3}\ m^3$

$Ab=\frac{pi.Db^2}{4}= \frac{\pi.(0.05m)^2}{4}=1.95*10^{-3}\ m^3$

Using the volume rate:

$Va=\frac{Q}{Aa}=\frac{0.9m^3}{7.85*10^{-3}\ m^3} = 1.9\ m/s$

$Vb = \frac{Q}{Ab}= \frac{0.9m^3}{1.96*10^{-3}\ m^3} = 7.63\ m/s$

Assuming no losses, the energy equation for fluids can be written as:

$Pa+\frac{1}{2}pa.Va^2+pa.g.za=Pb+\frac{1}{2}pb.Vb^2+pb.g.zb$

Here P, V, p, z and g represent the pressure, velocities, height and gravity acceleration. Considering the zero height level at point A and solving for Pb:

$Pb=Pa+\frac{1}{2}pa(Va^2-Vb^2)-pa.g.za$

Knowing the manometric pressure in point A of 70kPa, the height at point B of 1.5 meters, the density of water of 1000 kg/m^3 and the velocities calculated, the pressure at B results:

$Pb = 70000Pa+ \frac{1}{2}*1000\ \frac{kg}{m^3}*((1.9m/s)^2 - (7.63m/s)^2) - 1000\frac{kg}{m^3}*9,81\frac{m}{s^2}*1.5m$