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3 years ago

A process occurs in which a system's potential energy increases while the environment does work on the system.

Physics

1 answer:

iogann1982 [59]3 years ago

4 0

Ok, now what do you want to know about it?

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a body starts from rest and accelerates uniformly at 5ms‐2. Calculate the time taken by the body to cover a distance of 1km

mixas84 [53]

The body will take 20 seconds to cover a distance of 1000 m i.e. 1 km

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3 years ago

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What type of motion occurs when an object spins around an axis without altering its linear position?

myrzilka [38]

<h2><em>C. translational motion</em></h2><h2><em>HOPE IT HELPS !!!!!</em></h2>

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3 years ago

Question 1 of 10

Marianna [84]

Answer:

Option D. ²²²₉₀Th

Explanation:

Let the unknown be ⁿₘZ. Thus, the equation becomes:

²²⁶₉₂U —> ⁴₂He + ⁿₘZ

Next, we shall determine n, m and Z. This can be obtained as follow:

For n:

226 = 4 + n

Collect like terms

226 – 4 = n

222 = n

n = 222

For m:

92 = 2 + m

Collect like terms

92 – 2 = m

90 = m

m = 90

For Z:

ⁿₘZ => ²²²₉₀Z => ²²²₉₀Th

Therefore, the complete equation becomes:

²²⁶₉₂U —> ⁴₂He + ⁿₘZ

²²⁶₉₂U —> ⁴₂He + ²²²₉₀Th

Thus, the unknown is ²²²₉₀Th

6 0

3 years ago

What is the rabbit's displacement from t = 0s to 3 s?<br> Answer with two significant digits.

Inessa05 [86]

Answer: i think the answer is 20.0s

Explanation:

3 0

3 years ago

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The temperature and time t given in hours from 0 to 24 after midnight in downtown mathville is given by t=10-5 sin(pi/12 t) degr

Dvinal [7]

Answer:

$T_A_v_g=9.918192559$^{\circ}C$

Explanation:

The problem tell us that the temperature as function of time in downtown mathville is given by:

$T(t)=10-5*sin(\frac{\pi}{12t})$

The average temperature over a given interval can be calculated as:

$T_a_v_g=\frac{T_o+T_f}{2}$

Where:

$T_o=Initial\hspace{3}temperature\\T_f=Final\hspace{3}temperature$

So, the initial temperature in this case, would be the temperature at noon, and the final temperature would be the temperature at midnight:

Therefore:

$T_o=T(12)=10-5*sin(\frac{\pi}{12*12}) =9.890925575$^{\circ}C$

$T_f=T(24)=10-5*sin(\frac{\pi}{12*24}) =9.945459543$^{\circ}C$

Hence, the average temperature between noon and midnight is:

$T_A_v_g=\frac{9.890925575+9.945459543}{2}=9.918192559$^{\circ}C$

3 0

3 years ago