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sasho [114]
3 years ago
7

A variable that is not changed.

Physics
1 answer:
AleksAgata [21]3 years ago
5 0

Answer:

A controlled variable does not change during a experiment

Explanation:

it's c

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If there is no oxygen in outer space, and, at the same time, oxygen must be present for fire, where does the sun come from?​
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The sun is a huge ball of gas held together by gravity.

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If the mass of a material is 111 grams and the volume of the material is 23 cm3, what would the density of the material be? g/cm
Butoxors [25]
Density can be kg/m^3 or g/cm3
In g/cm3 density =mass /volume =111g/23cm3
=4.826g/cm3.

In kg/m3,density=mass/volume. converting mass in grams to kg, 1000g=1kg,111g=0.111kg.
cm3 to m3, 1cm3=10^-6m3, 23cm3=0.000023m3
density=0.111kg/0.000023m3 or 2.3*10^-5=4,826.1kg/m3.

the other is a long process.
4 0
3 years ago
Creates an image that appears upside down behind the focal point
zheka24 [161]
An image that appears upside down behind the focal point is an image that is reflected on a concave mirror. Mirrors reflect different kinds of images based on the placement of an object that is reflected towards it. There are two kinds of mirrors, concave and a convex mirrors, the latter makes objects seem smaller and farther than where it is exactly.
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3 years ago
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
2 years ago
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