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andrew11 [14]
3 years ago
9

Find the standard deviation of 7, 11, 8, 10, 9 if necessary round your answer to the nearest tenth

Physics
1 answer:
hoa [83]3 years ago
3 0
1.6 is the answer you are looking for.
You might be interested in
How do you move the decimal from megameters to meters?
Deffense [45]
The prefix "mega" means million.
Therefore
1 megameter = 10⁶ meters

That is,
1.0 megameter = 1,000,000.0 meters.

Answer:
The decimal is moved right by 6 places to convert a megameter to meters.
8 0
4 years ago
If 5000Pa a pressure is exerted on an object with the contact area 0.04m^2. Calculate the mass of an object
Whitepunk [10]
  • Pressure = 5000 Pa
  • Contact Area = 0.04 m^2
  • Acceleration due to gravity = 9.8 m/s^2
  • Let the force be F.
  • We know, Force = Pressure × Contact Area
  • Therefore, Force = 5000 Pa × 0.04 m/s^2
  • or, Force = 200 N
  • We know, force = mass × acceleration
  • Therefore, mass = force ÷ acceleration
  • or, mass = 200 N ÷ 9.8 m/s^2 = 20.4 Kg

<u>Answer</u><u>:</u>

<u>2</u><u>0</u><u>.</u><u>4</u><u> </u><u>Kg</u>

Hope you could understand.

If you have any query, feel free to ask.

6 0
2 years ago
A bucket of water experiencing a gravitational force of 525 N is pulled from a water well. Net force in the Y direction is 45 N
vivado [14]

Answer:

T = 570 N

Explanation:

Given that,

The gravitational force acting on a bucket of water = 525 N

Net force in the Y direction is 45 N

We need to find the magnitude of the force of tension. It can be calculated as :

45 = T - 525

T = 525 + 45

T = 570 N

Hence, the force of tension is 570 N.

7 0
3 years ago
When something is moving, it is in motion. Pressing the gas pedal in a car will put the car in motion. Two important properties
Alex
A They will run an equal distance
5 0
3 years ago
Read 2 more answers
A speaker at the front of a room and an identical speaker at the rear of the room are being driven at 456 Hz by the same sound s
vampirchik [111]

Answer:

2.97795 Hz

Explanation:

v = Speed of sound in air = 343 m/s

v_r = Relative speed between the speakers and the student = 1.12 m/s

f' = Actual frequency of sound = 456 Hz

Frequency of sound heard as the student moves away from one speaker

f_1=f'\dfrac{v-v_r}{v}\\\Rightarrow f_1=456\dfrac{343-1.12}{343}\\\Rightarrow f_1=454.51102\ Hz

Frequency of sound heard as the student moves closer to the other speaker

f_2=f'\dfrac{v+v_r}{v}\\\Rightarrow f_2=456\dfrac{343+1.12}{343}\\\Rightarrow f_2=457.48897\ Hz

The difference in the frequencies is

f_2-f_1=457.48897-454.51102=2.97795\ Hz

The student hears 2.97795 Hz

6 0
3 years ago
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