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Alekssandra [29.7K]
4 years ago
11

3. What is the relationship between the kinetic energy of molecules in an object and the object’s temperature? a. As the tempera

ture increases the kinetic energy of the molecules increases b. The kinetic energy always increase whether the temperature increases or decreases c. As the kinetic energy of the molecules decreases, the temperature increases d. The total kinetic energy of the molecules is not affected by a change in temperature
Physics
2 answers:
Novosadov [1.4K]4 years ago
5 0

Answer:

since temperature is the average kinetic energy of the molecules, so increasing temperature

Explanation:

evablogger [386]4 years ago
4 0

I think the answer A since temperature is the average kinetic energy of the molecules, so increasing temperature must increase kinetic energy

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Write your answer to the question below.
ElenaW [278]

Answer:

Hi... Potential energy is converted to kinetic energy and kinetic energy is converted to potential energy

8 0
3 years ago
a concrete slab 20 m long and weighing 400,000 N is supported by one pillar. A 19,600 N car is parked 8 meters from one end, whe
Elden [556K]

let the distance of pillar is "r" from one end of the slab

So here net torque must be balance with respect to pillar to be in balanced state

So here we will have

Mg(r - L/2) = mg(L/2 - 8)

here we know that

mg = 19600 N

Mg = 400,000 N

L = 20 m

from above equation we have

400,000(r - 10) = 19,600 (10 - 8)

r - 10 = 0.098

r = 10.098 m

so pillar is at distance 10.098 m from one end of the slab

3 0
3 years ago
In a women's 100-m race, accelerating uniformly, Laura takes 1.82 s and Healan 3.07 s to attain
Sav [38]

Answer:

Laura is ahead and for a distance of 3.22 m

Explanation:

To solve this problem of one-dimensional kinematics, we have to find the acceleration and the final speed of each runner. Let's start with Laura

Lura data is acceleration time 1.82s, total run time 10.4 s and total distance 100m.  In all the races the  rest starts, so the initial speed is zero (Vo = 0)

   Vf1= Vo + a1 t1    

   Vf1 = x/t                

   XT  = X1 + X2

   X1 = Vo t1 + ½ a1 t1²  

   X1 = ½ a1 t1²  

   X2 = Vf1 (t-t1)

This is the remaining time of the race after the acceleration is over.

    XT = ½ a1 t1² + Vf1 (t-t1)

We remplace the expression of Vf1

     XT = ½ a1 t1² + a1 t1 (t-t1)

Laura's aceleration (a1) is

   a1= XT / [ ½  t1² + t1 (t-t1)]

   a1= 100/ [ ½ 1.82²+ 1.82 (10.4 -1.82)]

   a1=  5.79m/s2  

We repeat the same calculation for the other Healan runner, whose data are: total distance 100m, acceleration time 3.07 s and total time 10.4 s

Vf2= Vo + a2 t2    

Vf2 = x/t                

XT  = X3 + X4

X3 = Vo t2 + ½ a2 t2²  

X3 = ½ a2 t2²  

X4 = Vf2 (t-t2)

XT = ½ a2 t2² + Vf2 (t- t2)

XT = ½ a2 t2² + a2 t2 (t-t2)

The aceleration of Healan (a2)

a2 = XT / [½ t2² + t2 (t-t2)]

a2 = 100 / [½ 3,07²+ 3.07 (10.4 -3.07)]

a2 = 3.67 m / s2

We also need the final speeds of each runner

Laura Vf1 = Vo + a1 t1

          Vf1 = 0 + 5.79 1.82

          Vf1 = 10.54 m / s

Healan Vf2 = Vo + a2 t2

            Vf2 = 0 + 3.67 3.07

            Vf2 = 11.27 m / s

Having the acceleration and speed of each runner, you can start answering the questions

a) For t3 = 6.15s

Laura

The time to stop with constant speed is what remains after accelerating

XL= ½ a1 t1² + Vf1 (t3-t1)

XL= ½ 5.79 1.82² + 10.54 (6.15 – 1.82)    

XL= 55.23 m

Healan  

XH= ½ a2 t2² + Vf2 (t3-t2)

             XH= ½  3.67 3.07² + 11.27 (6.15-3.07)

             XH= 52.01 m

             (XL -XH)= 55.23- 52.01

             (XH -XL)=  3.22 m

It is appreciated from these results that Laura is ahead and for a distance of 3.22 m

b) If we analyze the acceleration values ​​of each runner, knowing that they leave the rest and that Healan at the end has a speed greater than Laura, the point of maximum distance difference is when Laura stops accelerating t = 1.82 s

      XL= ½  a1 t12  

      XL= ½ 5.79 1.822

      XL= 9.59 m

      XH = ½ a2 t12

      XH= ½ 3.67 1.822

      XH= 6.08 m

The maximum distance difference is 3.51 m

c) Already analyzed in the previous part 1.82 s, since the Laura stop accelerating and Heala continue with acceleration will travel greater distances in equal time units

6 0
4 years ago
An automobile engine takes in 4000 j of heat and performs 1100 j of mechanical work in each cycle. (a) calculate the engine's ef
Semmy [17]
(a) The efficiency of an engine is defined as the ratio between the work done by the engine and the heat it takes in:
\eta= \frac{W}{Q_{in}}
The engine in this problem does a work of W=1100 J and it takes in Q_{in}=4000 J of heat, therefore its efficiency is
\eta= \frac{1100 J}{4000 J}=0.275 = 27.5 \%

(b) The heat taken by the machine is 4000 J; of this amount of heat, only 1100 J are converted into useful work. This means that the rest of the heat is wasted. Therefore, the wasted heat is the difference between the heat in input and the work done by the engine:
Q_{wasted}=Q_{in}-W=4000 J-1100 J=2900 J
7 0
3 years ago
Potential energy is the energy an object has because of its
Naya [18.7K]
Radiation compared to other potential
3 0
2 years ago
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