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sesenic [268]
4 years ago
14

9. The baseball catcher throws a ball vertically upward and catches it in the same spot as it returns to themitt. At what point

in the ball’s path does it experience zero velocity ?a. midway on the way upD. at the top of its trajectoryc. the instant it leaves the catcher’s handd. the instant before it an'ives in the catcher’s mitt
Engineering
2 answers:
SSSSS [86.1K]4 years ago
5 0

Answer:

Explanation:

  • At what point in the ball’s path does it experience zero velocity ; This will occur at maximum height.
  • Because at maximum height, acceleration due to gravity comes into play with negligible air resistance.
ololo11 [35]4 years ago
3 0

Answer:

At the top of its trajectory

Explanation:

When the base ball catcher throws the ball vertically upwards, the ball is acting with a negative acceleration due to gravity (-g) . from the equations of motion where v =u +at,

As the body continues going up until it gets to its maximum height , the velocity tends towards zero.

The total displacement ,upwards and downwards is also equal to zero.

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Briefly explain the term soil sampling and outline its importance in building engineering works.
Nataliya [291]

Answer:

For civil engineering, the importance is that you want to ensure that your construction is in a good sturdy location.

Explanation:

Im at UNCC as a civil engineering student

4 0
2 years ago
A traffic flow has density 61 veh/km when the speed is 59 veh/hr. If a flow has a jam density of 122 veh/km, what is the maximum
antoniya [11.8K]

Since this traffic flow has a jam density of 122 veh/km, the maximum flow is equal to 3,599 veh/hr.

<u>Given the following data:</u>

  • Density = 61 veh/km.
  • Speed = 59 km/hr.
  • Jam density = 122 veh/km.

<h3>How to calculate the maximum flow.</h3>

According to Greenshield Model, maximum flow is given by this formula:

q_{max}=\frac{V_f \times K_i}{4}

<u>Where:</u>

  • V_f is the free flow speed.
  • K_i is the Jam density.

In order to calculate the free flow speed, we would use this formula:

V_f =2 V\\\\V_f =2\times 59\\\\V_f=118\;km/hr

Substituting the parameters into the model, we have:

q_{max}=\frac{118 \times 122}{4}\\\\q_{max}=\frac{14396}{4}

Max flow = 3,599 veh/hr.

Read more on traffic flow here: brainly.com/question/15236911

6 0
2 years ago
The Stefan-Boltzmann law can be employed to estimate the rate of radiation of energy H from a surface, as in
Mazyrski [523]

Explanation:

A.

H = Aeσ^4

Using the stefan Boltzmann law

When we differentiate

dH/dT = 4AeσT³

dH/dT = 4(0.15)(0.9)(5.67)(10^-8)(650)³

= 8.4085

Exact error = 8.4085x20

= 168.17

H(650) = 0.15(0.9)(5.67)(10^-8)(650)⁴

= 1366.376watts

B.

Verifying values

H(T+ΔT) = 0.15(0.9)(5.67)(10)^-8(670)⁴

= 1542.468

H(T+ΔT) = 0.15(0.9)(5.67)(10^-8)(630)⁴

= 1205.8104

Error = 1542.468-1205.8104/2

= 168.329

ΔT = 40

H(T+ΔT) = 0.15(0.9)(5.67)(10)^-8(690)⁴

= 1735.05

H(T-ΔT) = 0.15(0.9)(5.67)(10^-8)(610)⁴

= 1735.05-1059.83/2

= 675.22/2

= 337.61

5 0
3 years ago
Which of the following tape measure techniques can be used to achieve accurate measurements? Choose all that apply.
zepelin [54]

Answer: Pull.

Because it's all about height width and Breadth!

5 0
3 years ago
3. A 4-m × 5-m × 7-m room is heated by the radiator of a steam-heating system. The steam radiator transfers heat at a rate of 10
Natali [406]

Answer:

14.52 minutes

<u>OR</u>

14 minutes and 31 seconds

Explanation:

Let's first start by mentioning the specific heat of air at constant volume. We consider constant volume and NOT constant pressure because the volume of the room remains constant while pressure may vary.

Specific heat at constant volume at 27°C = 0.718 kJ/kg*K

Initial temperature of room (in kelvin) = 283.15 K

Final temperature (required) of room = 293.15 K

Mass of air in room= volume * density= (4 * 5 * 7) * (1.204 kg/m3) = 168.56kg

Heat required at constant volume: 0.718 * (change in temp) * (mass of air)

Heat required = 0.718 * (293.15 - 283.15) * (168.56) = 1,210.26 kJ

Time taken for temperature rise: heat required / (rate of heat change)

Where rate of heat change = 10000 - 5000 = 5000 kJ/hr

Time taken = 1210.26 / 5000 = 0.24205 hours

Converted to minutes = 0.24205 * 60 = 14.52 minutes

4 0
4 years ago
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