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ludmilkaskok [199]
3 years ago
8

Activity log - Standing Whole Body Record your time and activity below and in your unit fitness log. If a particular category do

es not apply, leave it blank. Date: Warm-up: Cool-down: Type of Activity: (Cardiovascular, Muscular, or Flexibility) Description of Activity and Time: Intensity Level: (Moderate or Vigorous) Total Time:
Physics
1 answer:
sasho [114]3 years ago
6 0

Answer:

Date: N/A

Warm-up: 5 minutes

Cool-down: 6 minutes

Type of Activity: Cardiovascular

Description of Activity and Time: Running in place for 10 minutes

Intensity Level: Moderate

Total Time: 21 minutes

Explanation: Mark me as brainiest hope this helps I just did it

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By what factors does the speed of the elctron exceed that of the proton?<br>​
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Answer:

» An electron is lighter than a proton.

<u>explanation</u><u>:</u>

{  =  \: \sf{an \: electron \: has \: formula \:  \: }}{ \bf{ {}^{0}_{  - 1}e }}

hence it's mass number is zero

{  =  \: \sf{an \: electron \: is \: helium \: particle \:  \: }}{ \bf{ {}^{4} _{2}He  }}

hence it's mass number is 4

<u>Therefore</u><u>,</u><u> </u><u>proton</u><u> </u><u>is</u><u> </u><u>heavier</u><u> </u><u>than</u><u> </u><u>electron</u>

» An electron has a small charge magnitude than a proton.

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An electron has charge of -1 while proton has charge of +2, therefore electron is less deflected by any energetic fields than a proton

8 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
One hundred jumping beans are placed along the center line of a gymnasium floor at six-inch intervals. Twelve hours later, the d
stira [4]

Answer:

a) Diffusion  coefficient, D = 1.5 in/hr

b) Mean jump frequency, f = 0.0833 Hz

Explanation:

a) The relationship between the diffusion coefficient, time and mean displacement and can be given by the expression:

^{2} = 2Dt..........(1)

Where <r> = mean displacement

D = Diffusion coefficient

t = time = 12 hrs

sum of the squares of the distance divided by 100 is 36 in2.

<r>²= 36 in²

Substituting these values into equation (1) above

36 = 2 * D *12\\36 = 24 D\\D = 36/24\\D = 1.5 in/hr

b) Mean jumping distance, <r> = 0.1 inches

Applying equation (1) again

Where D = 1.5 in/hr

^{2} = 2Dt

0.1^{2}  = 2 * 1.5t\\0.01 = 3t\\t = 0.01/3\\t = 0.0033 hrs\\t = 0.0033 * 3600\\t = 12 seconds

The mean jump frequency, f = 1/t

f = 1/12

f = 0.0833 Hz

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3 years ago
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The ear drum vibrates

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