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a_sh-v [17]
3 years ago
11

An automotive paint sequence on a suspect's car is as follows: clear coat, silverstone ii metallic, primer. a paint smear is fou

nd on a metal post near a hit-and-run assault. what coat(s) are you most likely to find on the metal post
Physics
2 answers:
ipn [44]3 years ago
6 0
<h2><u>Answer:</u></h2>

A car paint grouping on a speculate's the vehicle is as per the following: clear coat, Silverstone ii metallic, groundwork. a paint smear is found on a metal post close to an attempt at manslaughter attack.

The destined to discover on the metal post is the Color-layer grouping of the automobile.  Paint is generally utilized as a physical proof in an attempt at manslaughter occurrence.

Maru [420]3 years ago
5 0
<span>An automotive paint sequence on a suspect's car is as follows: clear coat, silverstone ii metallic, primer. a paint smear is found on a metal post near a hit-and-run assault. The most likely to find on the metal post is the </span>Color-layer sequence of the automobile.  Paint is usually used as a physical evidence in a hit-and-run incident. 
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A(n) 1100 kg car is parked on a 4◦ incline. The acceleration of gravity is 9.8 m/s 2 . Find the force of friction keeping the ca
igomit [66]

Answer:

Force of friction, f = 751.97 N

Explanation:

it is given that,

Mass of the car, m = 1100 kg

It is parked on a 4° incline. We need to find the force of friction keeping the car from sliding down the incline.

From the attached figure, it is clear that the normal and its weight is acting on the car. f is the force of friction such that it balances the x component of its weight i.e.

f=mg\ sin\theta

f=1100\ kg\times 9.8\ m/s^2\ sin(4)

f = 751.97 N

So, the force of friction on the car is 751.97 N. Hence, this is the required solution.

3 0
3 years ago
2)It is known that the connecting rodS exerts on the crankBCa 2.5-kN force directed down andto the left along the centerline ofA
11111nata11111 [884]

Answer:

M_c = 100.8 Nm

Explanation:

Given:

F_a = 2.5 KN

Find:

Determine the moment of this force about C for the two cases shown.

Solution:

- Draw horizontal and vertical vectors at point A.

- Take moments about point C as follows:

                        M_c = F_a*( 42 / 150 ) *144

                        M_c = 2.5*( 42 / 150 ) *144

                        M_c = 100.8 Nm

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

Hence, its moment about C is zero.

5 0
4 years ago
PLZ HELP MEEEEEEEEE ASAP
mafiozo [28]
The answer is “Impulse acting on it” according to the impulse-momentum theorem.
6 0
3 years ago
Ed biked home from school in 400 seconds. His home is located 2000 m south of the school. What was his velocity?
Naily [24]

<em>answer =  5 \:  \: metre \: per \: second\\ distance = 2000 \: m \\ time = 400 \: seconds \\ velocity =  \:  \frac{distance \:t ravelled}{time \: taken}  \\  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  =  \frac{2000}{400}  \\  \:  \:  \:  \:  \:  \:  \:  \:  \:  = 5 \: metre \: per \: second \\ hope \: it \: helps</em>

7 0
3 years ago
A 1.10 kg box slides down a rough incline plane from a height h of 1.75 m. The box had a speed of 2.99 m/s at the top and a spee
madam [21]

Answer:

20.18 J

Explanation:

We are given that

Mass of bx=m=1.1 kg

Height=h=1.75 m

Initial speed of box=u=2.99 m/s

Final speed,v=2.56 m/s

We have to find the mechanical energy lost due to friction.

Energy at top=K.E+P.E=\frac{1}{2}mu^2+mgh=\frac{1}{2}(1.1)(2.99)^2+1.1\times 9.8\times 1.75=23.78 J

Where g=9.8 m/s^2

At bottom,h'=0

Energy at bottom=\frac{1}{2}mv^2+mgh'=\frac{1}{2}(1.1)(2.56)^2+0=3.6 J

Energy lost=Energy(Top)-Energy(bottom)=23.78-3.6=20.18 J

5 0
3 years ago
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