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3 years ago

WHAT IS THE MESURE OF THE AVERAGE KINETIC ENERGY OF THE PARTICLES IN A NATERIAL KNOWN AS

1 answer:

4 0

Temperature is a measure of the average kinetic energy of the particles in a substance. It is the kinetic energy of a typical particle

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To meet the minimum required instrument flight experience to act as a pilot in command of an aircraft under IFR, you must have l

lora16 [44]

I think its a tbh bc it seems to be the best answer out of a b c and d

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4 years ago

How is the acceleration of an object in uniform circular motion constant?

nevsk [136]

In both magnitude and direction since acceleration is a vector quantity

5 0

3 years ago

Read 2 more answers

Given the information below, estimate the total distance travelled during these 6 seconds using a left endpoint approximation. t

Nutka1998 [239]

Answer:

184 feets

Explanation:

Given the data:

time (sec) __ velocity (ft/sec)

0 __________30

1 __________ 54

2 __________56

3 __________34

4 __________ 8

5 __________ 2

6 __________22

Using left end approximation:

(0,1) ___ f(0) = 30

(1,2) ___ f(1) = 54

(2,3) ___f(2) = 56

(3,4) ___f(3) = 34

(4,5) ___f(4) = 8

(5,6) __ f(5) = 2

Hence, the Total distance traveled during the 6 second interval is:

Change ; dT = 1

1 * (30 + 54 + 56 + 34 + 8 + 2) = 184

4 0

3 years ago

A 300-n force acts on a 25-kg object. the acceleration of the object is?

Over [174]

To solve the answer use the equation: a = fnet / m

a = 300 N / 25 kg

300 N / 25 kg = 12m/s

The acceleration of the object is 12m/s

8 0

3 years ago

Read 2 more answers

The roller coaster car has a mass of 700 kg, including its passenger. If it is released from rest at the top of the hill A, dete

sweet [91]

Answer:

$h = 18.75 m$

Now when it will reach at point B then its normal force is just equal to ZERO

$N_B = 0$

$F_n = 1.72 \times 10^4$

Explanation:

Since we need to cross both the loops so least speed at the bottom must be

$v = \sqrt{5 R g}$

also by energy conservation this is gained by initial potential energy

$mgh = \frac{1}{2}mv^2$

$v = \sqrt{2gh}$

so we will have

$\sqrt{2gh} = \sqrt{5Rg}$

now we have

$h = \frac{5R}{2}$

here we have

R = 7.5 m

so we have

$h = \frac{5(7.5)}{2}$

$h = 18.75 m$

Now when it will reach at point B then its normal force is just equal to ZERO

$N_B = 0$

now when it reach point C then the speed will be

$mgh - mg(2R_c) = \frac{1}{2]mv_c^2$

$v_c^2 = 2g(h - 2R_c)$

$v_c = 13.1 m/s$

now normal force at point C is given as

$F_n = \frac{mv_c^2}{R_c} - mg$

$F_n = \frac{700\times 13.1^2}{5} - (700 \times 9.8)$

$F_n = 1.72 \times 10^4$

7 0

4 years ago