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3 years ago

You know that ice floats in water. This is because

Physics

2 answers:

pochemuha3 years ago

7 0

A BECAUSE WHEN ITS MORE DENSE IT CAN FLOAT AND THE OTHER CHOICES DONT MAKE SENSE

AlladinOne [14]3 years ago

4 0

Anything that floats in some fluid must be less dense than the fluid.

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What type of radioactive decay is shown in this equation?

svp [43]

There is no <span>radioactive decay</span>

6 0

3 years ago

Being able to see a square in the middle of the image despite not having lines to form a square represents the Gestalt principle

vredina [299]

Answer: The correct answer for the blank is -

C. closure.

Gestalt principle of closure describes how we perceive complete figures even when the information that form the figure is missing.

This is due the fact that our brain responds to the familiar patterns inspite of getting incomplete information.

For instance, in the given question, we are able to perceive an image of square in the center despite not having actual lines that form a square.

Thus, it represents principle of closure.

7 0

3 years ago

Read 2 more answers

1. Two forces F~ 1 and F~ 2 are acting on a block of mass m=1.5 kg. The magnitude of force F~ 1 is 12N and it makes an angle of

VLD [36.1K]

Answer:

Normal force=7.48 N

Explanation:

N+F~1 sinθ-mg=0

=>N=1.5*9.8-12 sin37◦

=>N=14.7-7.22=7.48 N

6 0

3 years ago

Suppose that a planet were discovered between the sun and Mercury, with a circular orbit of radius equal to 2/3 of the average o

Musya8 [376]

Explanation:

It is given that,

A planet were discovered between the sun and Mercury, with a circular orbit of radius equal to 2/3 of the average orbit radius of Mercury.

Mass of the Sun, $M=1.99\times 10^{30}\ kg$

Radius of Mercury's orbit, $r=5.79\times 10^{10}\ m$

Radius of discovered planet, $R=\dfrac{2}{3}r$

$R=\dfrac{2}{3}\times 5.79\times 10^{10}\ m=3.86\times 10^{10}\ m$

Let T is the orbital period of such a planet. Using Kepler's third law of planetary motion as :

$T^2\propto R^3$

$T^2=\dfrac{4\pi^2R^3}{GM}$

$T^2=\dfrac{4\pi^2\times (3.86\times 10^{10})^3}{6.67\times 10^{-11}\times 1.99\times 10^{30}}$

$T=\sqrt{1.71\times 10^{13}}$

T = 4135214.625 s

T = 47.86 days

So, the orbital period of such a planet is 47.86 days. Hence, this is the required solution.

6 0

3 years ago

One knife is longer from the cutting edge to the back of the blade. It's easier to push the longer knife through a pear and more

AlladinOne [14]

By the formula of torque balance we know that

$\tau = r\times F$

here we know that

r = distance of force

F = applied force

so here we know that as we will take large edge knife then the distance of force will increase

due to this the torque will also increase in it so it is easy to pear it

So here we can say

$F_{in}r_1 = F_{out}r_2$

$F_{out} = \frac{r_1}{r_2} F_{in}$

so output force will increase in this case

So here the correct answer would be

<em>b. the longer the knife, the stronger the output force</em>

6 0

3 years ago