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oee [108]
3 years ago
14

A 1100 kg automobile is at rest at a traffic signal. At the instant the light turns green, the automobile starts to move with a

constant acceleration of 5.0 m/s2. At the same instant a 2000 kg truck, traveling at a constant speed of 7.0 m/s, overtakes and passes the automobile. A)How far is the com of the automobile-truck system from the traffic light at t=3.0 s? B)What is the speed of the com then?
Physics
1 answer:
Viktor [21]3 years ago
6 0

Answer:

a ) 21 m b) 7 m/s

Explanation:

For this case, we can consider for our reference system, that point zero (this is, at the rest of the traffic signal) is when X₀ = 0, and direction of movement is from left to right (positive sign)

For the car:  

a= acceleration is constant = 5 m/s² m1 , mass = 1100 Kg , V = not constant

This is an uniformly accelerated rectilinear movement, and applicable formulas are:

V = V₀ + at , X = X₀ + V₀t + 1/2at²

For the truck:

V = speed is constant = 7 m/s , mass = 2000 Kg, a = 0

This is an uniform rectilinear movement, and the applicable formula is:

X = X₀ + Vt

a) At t = 3.0 sec, we can use the formula for the track, considering that X₀ = 0 (when it pass the rest of the traffic signal)

X = 0 + (7 m/s)x(3 s) = 21 m  

So, at 3 sec, truck will be at 21 m from the rest of traffic light and, as truck has a constant speed; at that exact second; truck and car will be together as a com, so both will be 21 m away from point zero

b) At the exact time (3 sec, or in distance words, 21 m) car and truck will be together as a com, so they will both have the exact speed, hence, speed of car at that point will be 7 m/s

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laila [671]

Answer:

Three, the liquid in the man's cup, the stove is solid, and the air around them is a gas

Explanation:

3 0
4 years ago
What is the final velocity of a body if it is moving with 13 m/s in 300 seconds and its acceleration is 30 m/s 2.
Nutka1998 [239]

Answer:

9013m/s

Explanation:

acceleration= v- u/ t

=> at = v-u

=> v = at + u

=> v =30*300+13

= 9013m/s

6 0
3 years ago
A reservoir located in the mountain 250 m above sea level flows through a pipe to a hydroelectric plant in a town at sea level.
Pavlova-9 [17]

Answer:

     v₂ = 70 m / s

Explanation:

For this exercise let's use Bernoulli's equation

where subscript 1 is for the top of the mountain and subscript 2 is for Tuesday's level

 

          P₁ + ½ ρ v₁² + ρ g y₁ = P₂ +1/2 ρ v₂² + ρ g y₂

indicate that the pressure in the two points is the same, y₁ = 250 m, y₂ = 0 m, the water in the upper part, because it is a reservoir, is very large for which the velocity is very small, we will approximate it to 0 (v₁ = 0), we substitute

         ρ g y₁ = ½ ρ v₂²

         v₂ = \sqrt {2g \ y_1}

let's calculate

         v₂ = √( 2 9.8 250)

         v₂ = 70 m / s

6 0
3 years ago
What kind of symmetry do you have
Keith_Richards [23]
During the daytime, I have mostly line symmetry.

During the night, I often have almost spherical symmetry.
5 0
3 years ago
Read 2 more answers
determine the pressure exerted on the surface of a submarine cruising 175 ft below the free surface of the sea. assume that the
Alenkasestr [34]

Answer:

92.81 psia.

Explanation:

The density of water by multiplying its specific gravity by the density of sea water.

SG = density of sea water/density of water

ρ = SG x ρw

1 kg/m3 = 62.4 lbm/ft^3

= 1.03 * 62.4

= 64.27lbm/ft^3.

The absolute pressure at 175 ft below sea level as this is the location of the submarine.

P = Patm +ρgh

= 14.7 + 64.27 * 32.2 * 175

Converting to pound force square inch,

= 14.7 + 64.27 * (32.2ft/s^2) * (175ft) * (1lbf/32.2lbm⋅ft/s^2) * (1ft^2/144in^2 )

= 14.7 + 78.11 psia

= 92.81 psia.

8 0
3 years ago
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