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lora16 [44]
3 years ago
14

A track star runs 100 meters in 10 seconds. What is the star's average speed?

Physics
1 answer:
kakasveta [241]3 years ago
4 0

Answer:

10m/s

Explanation:

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An undersea research chamber is spherical with an external diameter of 4.70 m . The mass of the chamber, when occupied, is 54300
Ira Lisetskai [31]

Answer: a) 557.3kN b) 543kN

Explanation:

Buoyant force = weight of water displaced.

Formula for buoyant force = mg×(density of fluid/density of object)

Density of fluid = density of sea water = 1025kg/m³

Mass of the chamber(m) = 54300kg

g = acceleration due to gravity = 10m/s²

Density of chamber = mass/volume

Where volume of sphere = 4/3Πr³

radius = 4.70/2 = 2.35m

Volume = 4/3×Π(2.35)³

Volume = 54.37m³

Density of chamber = 54300/54.37

= 998.7kg/m³

Substituting the values into the formula for buoyant force

Buoyant force= 54300×10×{1025/998.7}

= 557299N

= 557.3kN

b) Tension in the cable = mass of cable × acceleration due to gravity

= 54300×10 = 543000N or 543kN.

7 0
3 years ago
The centre of mass of a metre rule is at the 50cm mark. state what is meant by Centre of mass​
Tanya [424]

Answer:

Centre of mass of any body is a point where all mass of a body is supposed to be concentrated

it lies in geometrical centre....

3 0
3 years ago
The kinetic energy of an object can sometimes be greater than a potential energy a originally possessed, true or false?
Mamont248 [21]

Answer:

true

Explanation:

Energy stored in the nuclei of atoms can be used to generate electricity. ... Most of the energy of food is converted to heat.

4 0
3 years ago
Read 2 more answers
The dragster has a mass of 1.3 Mg and a center of mass at G. A parachute is attached at C provides a horizontal braking force of
adell [148]

Answer:

The deceleration of the dragster upon releasing the parachute such that the wheels at B are on the verge of leaving the ground is  16.33 m/s²

Explanation:

The additional information to the question is embedded in the diagram attached below:

The height between the dragster and ground is considered to be 0.35 m since is not given ; thus in addition win 0.75 m between the dragster and the parachute; we have: (0.75 + 0.35) m = 1.1 m

Balancing the equilibrium about point A;

F(1.1) - mg (1.25) = ma_a (0.35)

1.8v^2(1.1) - 1200(9.8)(1.25) = 1200a(0.35)

1.8v^2(1.1) - 14700 = 420 a   ------- equation (1)

F_x = ma_x \\ \\ = 1.8v^2 = 1200 \ a             --------- equation (2)

Replacing equation 2 into equation 1 ; we have :

{1.1 * 1200 \ a} - 14700 = 420 a

1320 a - 14700 = 420 a

1320 a -  420 a =14700

900 a = 14700

a = 14700/900

a = 16.33 m/s²

The deceleration of the dragster upon releasing the parachute such that the wheels at B are on the verge of leaving the ground is  16.33 m/s²

5 0
3 years ago
LetterA., please see attachment
Anika [276]
  • The air of atmosphere contains oxygen which helps us in breathing and digestion and respiration of food
  • The nitrogen in air is helpful for plants and fertilizers
  • The humidity or water in atmosphere helps in rain
7 0
3 years ago
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