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nlexa [21]
3 years ago
15

You throw a 20-n rock vertically into the air from ground level. you observe that when it is a height 16.0 m above the ground, i

t is traveling at a speed of 24.7 m/s upward.
Physics
1 answer:
timama [110]3 years ago
6 0

The problem is missing some parts. But here are the questions and answers.

a.       Use the work-energy theorem to find its speed just as it left the ground 

Initial Kinetic Energy = GPE + KE

1/2 * m * V^2 = m * g * h + 1/2 * m * v^2

V^2 = 2 * 9.8 * 15.6 + 25.6^2

=sqrt 968.96

V (initial speed) ≈ 31.130 m/s

 

b.      Use the work-energy theorem to look for its maximum height. 

H = 968.96 / 2 * 9.8 ≈ 49.44 <span>m</span>

You might be interested in
8892 ml to grams then to mg ​
andrew11 [14]

Answer:

8892 ml = 8892 gm = 8892000 mg

1 ml = 1 gram

8892 ml = 8892 gram

1 gram or ml = 1000 milligram

8892 ml = 8892 × 1000 = 8892000 milligram

hope this helps

have a good day :)

Explanation:

remember: 1 kilogram = 1000 gram = 1000000 milligram.

Milliliter is expressed same as gram and liter is expressed same as kilogram.

1 meter = 100 cm, 1 kilometer = 1000 meter,

1 cm = 10 millimeter.

7 0
3 years ago
Read 2 more answers
What is the distance traversed by the particle between 0 seconds and 6 seconds?
ad-work [718]

Answer:

I dont know to be honest

Explanation:

i dont know to be honest

5 0
3 years ago
If two 100 ohms resistors are placed in series, their total resistance is what?
Sindrei [870]

Answer:

add the Resistance

2(100)= 200

2) 200 ohms

3 0
3 years ago
A container is filled with an ideal diatomic gas to a pressure and volume of P1 and V1, respectively. The gas is then warmed in
lilavasa [31]

Answer:

Explanation:

The  change is as follows

P₁ V₁ to 3P₁, V₁ ( constt volume )  --- first process

3P₁,V₁ to 3P₁ , 5V₁ ( constt pressure ) ---- second process

In the first process Temperature must have been increased 3 times . So if initial temperature is T₁ then final temperature will be 3 T₁

P₁V₁ = n R T₁ , n is no of moles of gas enclosed.

nRT₁ = P₁V₁

Heat added at constant volume  = n Cv ( 3T₁ - T₁)

= n x 5/3 R X 2T₁ ( for diatomic gas Cv = 5/3 R)

= 10/3 x nRT₁

= 10/3x P₁V₁

In the second process,  Temperature must have been increased 5 times . So if initial temperature is 3T₁ then final temperature will be 15 T₁

Heat added at constant pressure in second case  

= n Cp ( 15T₁ - 3T₁)

= n x 7/3 R X 12T₁ ( For diatomic gas Cp = 7/3 R)

= 28 x nRT₁

= 28 P₁V₁

6 0
4 years ago
4. Assume a multiple level queue system with a variable time quantum per queue, and that the incoming job needs 50ms to run to c
Troyanec [42]

Answer:

Explanation:

For the completion of incoming job it will take 50ms

First queue takes 5ms quantum time and the subsequent queue takes double of the previous question

So,

First queue T_1 = 5ms

Second queue T_2 = 2 × T_1 = 2 × 5 = 10ms

Third queue T_3 = 2 × T_2 = 2 × 10 = 20ms

Fourth queue T_4 = 2 × T_3 = 2 × 20 = 40ms

Fifth queue T_5= 2 × T_4 = 2 × 40 = 80ms.

Now, the job will be done after the fifth queue.

So, after the first queue, the job is not completed, so, we have first interruption

After the second queue, the job is not completed, so, we have second interruption

After the third queue, the job is not completed, so we have third interruption.

After the fourth queue, the job is not yet completed, so we have the fourth interruption

And in the fifth queue the job is completed, so we don't have any interruption here.

So, the job will be interrupted 4 times and it will finished on the fifth queue.

6 0
3 years ago
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