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stealth61 [152]
3 years ago
10

An 800-kg drag racer accelerates from rest to 390 km/hr in 5.8 s. What is the net impulse applied to the racer in the first 5.8

seconds? If the net tangential force applied to the racer is constant, what is its value?
Engineering
1 answer:
marissa [1.9K]3 years ago
6 0

Answer:

Impulse =14937.9 N

tangential force =14937.9 N

Explanation:

Given that

Mass of car m= 800 kg

initial velocity u=0

Final velocity v=390 km/hr

Final velocity v=108.3 m/s

So change in linear momentum P= m x v

           P= 800 x 108.3

 P=86640 kg.m/s

We know that impulse force F= P/t

So F= 86640/5.8 N

F=14937.9 N

Impulse force F= 14937.9 N

We know that

v=u + at

108.3 = 0 + a x 5.8

a=18.66\ m/s^2

So tangential force F= m x a

F=18.66 x 800

F=14937.9 N

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Explanation:

check the attached files for answer.

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3 years ago
1. The construction process begins with which of the following stages?
Firdavs [7]

Answer:

c) site preparation

Explanation:

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Generally, in the construction of any physical asset such as offices, hospitals, schools, stadiums etc, the first step of the construction process is site preparation. Site preparation refers to processes such as clearing, blasting, levelling, landfilling, surveying, cutting, excavating and demolition of all unwanted objects on a piece of land, so as to make it ready for use.

This ultimately implies that, site preparation should be the first task to be accomplished in the construction process.

Hence, the construction process typically begins with site preparation before other activities such as the laying of foundation can be done.

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7 0
3 years ago
What is the value of the work interaction in this process?
Cloud [144]

Answer:

The answer is "-121\  \frac{KJ}{Kg}".

Explanation:

Please find the correct question in the attachment file.

using formula:

\to W=-P_1V_1+P_2V_2 \\\\When \\\\\to W= \frac{P_1V_1-P_2V_2}{n-1}\ \   or \ \  \frac{RT_1 -RT_2}{n-1}\\\\

W =\frac{R(T_1 -T_2)}{n-1}\\\\

    =\frac{0.287(25 -237)}{1.5-1}\\\\=\frac{0.287(-212)}{0.5}\\\\=\frac{-60.844}{0.5}\\\\=-121.688 \frac{KJ}{Kg}\\\\=-121 \frac{KJ}{Kg}\\\\

7 0
3 years ago
Joe Bruin has a big lawn in front of his house that is 30 meters wide and 20 meters long. Josephine makes him go out and mow the
zysi [14]

<u>Explanation:</u>

5 Horsepower for 30 mins,

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General Efficiency of IC engine = 20%

Power required = \frac{3 \cdot 7285}{0 \cdot 2}=18 \cdot 6425 kw

Energy required per week,

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let sunlight hours be 8 hours

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energy input in lawn = (600) (20232) (7)

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Trash cans repaired  

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.......................

7 0
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