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PtichkaEL [24]
3 years ago
10

State two condition necessary for a solid to float in a liquid

Physics
1 answer:
hoa [83]3 years ago
4 0

Answer:

it's density must be less than water

law of floatation

wt of the immerged body = wt of the water displaced by it

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GarryVolchara [31]
What are ur answers questions
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2 years ago
Coulomb's law is expressed mathematically as
nataly862011 [7]
Force between two charges  = 

     ( 1/4πε₀ ) · (Charge #1) · (Charge #2) / (Distance between them)²

  in the direction away from each other.

In other words, if the force is positive, the charges are repelling.
If the force is negative, the charges are attracting.
4 0
3 years ago
A father racing his son has 1/3 the kinetic energy of the son, who has 1/4 the mass of the father. The father speeds up by 1.5 m
Feliz [49]

Explanation:

Let the speeds of father and son are v_f\ and\ v_s. The kinetic energies of father and son are K_f\ and\ K_s. The mass of father and son are  m_f\ and\ m_s

(a) According to given conditions, K_f=\dfrac{1}{3}K_s

And m_s=\dfrac{1}{4}m_f

Kinetic energy of father is given by :

K_f=\dfrac{1}{2}m_fv_f^2.............(1)

Kinetic energy of son is given by :

K_s=\dfrac{1}{2}m_sv_s^2...........(2)

From equation (1), (2) we get :

\dfrac{v_f^2}{v_s^2}=\dfrac{1}{12}..............(3)

If the speed of father is speed up by 1.5 m/s, so the ratio of kinetic energies is given by :

\dfrac{K_f}{K_s}=\dfrac{1/2m_f(v_f+1.5)^2}{1/2m_sv_s^2}

v_s^2=4(v_f+1.5)^2

Using equation (3) in above equation, we get :

v_f=\dfrac{1.5}{\sqrt3-1}=2.04\ m/s

(b) Put the value of v_f in equation (3) as :

v_s=7.09\ m/s

Hence, this is the required solution.

8 0
3 years ago
What is used to block the UV light during screen development process to create a stencil in the emulsion
coldgirl [10]

Answer:

Invisible UV energy reacts with emulsion sensitizer and hardens the stencil so it won't dissolve with water and rinse down the drain

Explanation:

4 0
1 year ago
Each of the following diagrams shows a spaceship somewhere along the way between Earth and the Moon (not to scale); the midpoint
djyliett [7]

Answer:

F_5 >F_4>F_1 >F_2>F_3

Where F_i represent the force for each of the 5 cases i -1,2,3,4,5 presented on the figure attached.

Explanation:

For this case the figure attached shows the illustration for the problem

We have an inverse square law with distance for the force, so then the force of gravity between Earth and the spaceship is lower when the spaceship is far away from Earth.

Th formula is given by:

F = G \frac{m_{Earth} m_{Spaceship}}{r^2}

Where G is a constant G = 6.674 x10^{-11} m^2/ (ks s^2)

m_{Earth} represent the mass for the earth

m_{spaceship} represent the mass for the spaceship

r represent the radius between the earth and the spaceship

For this reason when the distance between the Earth and the Spaceship increases the Force of gravity needs to decrease since are inversely proportional the force and the radius, and for the other case when the Earth and the spaceship are near then the radius decrease and the Force increase.

Based on this case we can create the following rank:

F_5 >F_4>F_1 >F_2>F_3

Where F_i represent the force for each of the 5 cases i -1,2,3,4,5 presented on the figure attached.

6 0
2 years ago
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