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k0ka [10]
3 years ago
11

When you float an ice cube in water, you notice that 90% of it is submerged beneath the surface. Now suppose you put the same ic

e cube in a glass of some liquid whose density is less than that of water.
How much of the ice cube will be submerged below the surface of this liquid?
Physics
1 answer:
Kruka [31]3 years ago
8 0

Answer:

more than 90%

Explanation:

In the unknown liquid the buoyant force and weight relation is

V\rho_lg=mg\\\Rightarrow V\rho_l=m

It can be seen that if the density decreases the buoyant force decreases.

If the object is already 90% submerged in water then, if the other liquid has density less than that of water the object will be submerged more than 90%.

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A space station orbiting Earth.

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alexdok [17]

Answer:

30 N \cdot m

Explanation:

The torque applied by a force can be calculated as

\tau = F d sin \theta

where

F is the magnitude of the force

d is the length of the arm

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In this problem, we have

F = 15 N

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Substituting into the equation, we find

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3 years ago
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VLD [36.1K]

Answer:

Wilhelm Conrad Roentgen (1845-1923)

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Explanation:

Wilhelm Conrad Roentgen (1845-1923)

Contribution:  Discovery of x-rays in 1901.

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Contribution: He discovered that radioactivity is the separation of x-rays and document and the difference between two.

Pierre (1859-1906) and Marie (1867-1934) Curie

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5 0
3 years ago
A train starts at rest in a station and accelerates at a constant 0.987 m/s2 for 182 seconds. Then the train decelerates at a co
algol [13]

Answer:

\displaystyle X_T=66.6\ km

Explanation:

<u>Accelerated Motion </u>

When a body changes its speed at a constant rate, i.e. same changes take same times, then it has a constant acceleration. The acceleration can be positive or negative. In the first case, the speed increases, and in the second time, the speed lowers until it eventually stops. The equation for the speed vf at any time t is given by

\displaystyle V_f=V_o+a\ t

where a is the acceleration, and vo is the initial speed .

The train has two different types of motion. It first starts from rest and has a constant acceleration of 0.987 m/s^2 for 182 seconds. Then it brakes with a constant acceleration of -0.321 m/s^2 until it comes to a stop. We need to find the total distance traveled.

The equation for the distance is

\displaystyle X=V_o\ t+\frac{a\ t^2}{2}

Our data is

\displaystyle V_o=0,a=0.987m/s^2,\ t=182\ sec

Let's compute the first distance X1

\displaystyle X_1=0+\frac{0.987\times 182^2}{2}

\displaystyle X_1=16,346.7\ m

Now, we find the speed at the end of the first period of time

\displaystyle V_{f1}=0+0.987\times 182

\displaystyle V_{f1}=179.6\ m/s

That is the speed the train is at the moment it starts to brake. We need to compute the time needed to stop the train, that is, to make vf=0

\displaystyle V_o=179.6,a=-0.321\ m/s^2\ ,V_f=0

\displaystyle t=\frac{v_f-v_o}{a}=\frac{0-179.6}{-0.321}

\displaystyle t=559.5\ sec

Computing the second distance

\displaystyle X_2=179.6\times559.5\ \frac{-0.321\times 559.5^2}{2}

\displaystyle X_2=50,243.2\ m

The total distance is

\displaystyle X_t=x_1+x_2=16,346.7+50,243.2

\displaystyle X_t=66,589.9\ m

\displaystyle \boxed{X_T=66.6\ km}

8 0
3 years ago
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