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Sindrei [870]
3 years ago
14

What chemical formula did early instigators assume for water?

Physics
1 answer:
Aloiza [94]3 years ago
4 0

Answer:

Explanation:

 Investigators  thought this meant that oxygen was eight times more massive than hydrogen. They presumed the  chem-ical form-ula for water to be H-O

Hope this helped!!!  .

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During the first stage of a multistage rocket, a satellite is:
lianna [129]
B. sent through the atmosphere 

7 0
3 years ago
Which scientist first made a model of the atom called the plum pudding model? John Dalton J.J. Thomson Neils Bohr Ernest Rutherf
Norma-Jean [14]

Answer: J.J Thomson

Explanation: J. J. Thomson, who discovered the electron in 1897, proposed the plum pudding model of the atom in 1904 before the discovery of the atomic nucleus in order to include the electron in the atomic model.

6 0
2 years ago
A person of mass 70 kg stands at the center of a rotating merry-go-round platform of radius 2.9 m and moment of inertia 900 kg⋅m
Cloud [144]

Explanation:

It is given that,

Mass of person, m = 70 kg

Radius of merry go round, r = 2.9 m

The moment of inertia, I_1=900\ kg.m^2

Initial angular velocity of the platform, \omega=0.95\ rad/s

Part A,

Let \omega_2 is the angular velocity when the person reaches the edge. We need to find it. It can be calculated using the conservation of angular momentum as :

I_1\omega_1=I_2\omega_2

Here, I_2=I_1+mr^2

I_1\omega_1=(I_1+mr^2)\omega_2

900\times 0.95=(900+70\times (2.9)^2)\omega_2

Solving the above equation, we get the value as :

\omega_2=0.574\ rad/s

Part B,

The initial rotational kinetic energy is given by :

k_i=\dfrac{1}{2}I_1\omega_1^2

k_i=\dfrac{1}{2}\times 900\times (0.95)^2

k_i=406.12\ rad/s

The final rotational kinetic energy is given by :

k_f=\dfrac{1}{2}(I_1+mr^2)\omega_1^2

k_f=\dfrac{1}{2}\times (900+70\times (2.9)^2)(0.574)^2

k_f=245.24\ rad/s

Hence, this is the required solution.

5 0
3 years ago
A 110 kg hoop rolls along a horizontal floor so that the hoop's center of mass has a speed of 0.220 m/s. How much work must be d
Softa [21]

Answer:

the work that must be done to stop the hoop is 2.662 J

Explanation:

Given;

mass of the hoop, m = 110 kg

speed of the center mass, v = 0.22 m/s

The work that must be done to stop the hoop is equal to the change in the kinetic energy of the hoop;

W = ΔK.E

W = ¹/₂mv²

W = ¹/₂ x 110 x 0.22²

W = 2.662 J

Therefore, the work that must be done to stop the hoop is 2.662 J

6 0
3 years ago
HEY CAN ANYONE ANSWER DIS QUESTION PLS!!!!!!!!
Valentin [98]
It’s very big and very small numbers
6 0
3 years ago
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