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3 years ago

When the program reads the 0 and 1's it is called _______________________.

Physics

1 answer:

inysia [295]3 years ago

5 0

ANSWER:

When the program reads the 0 and 1's it is called _____binary______.

EXPLANATION:

That language of 1's and 0's is called binary. Computers speak in binary because of how they are built.

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What happens to a bowl of hot soup when it is placed into a freezer where the air Temperature is below freezing. How does this a

Artyom0805 [142]

It does not it when put in the freezer it freezes the steam

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3 years ago

D) A tank with dimensions: 3m x 2m x 2m contains water upto its half

lorasvet [3.4K]

Answer:

P = 9800 [Pa]

Explanation:

In order to calculate the pressure at the bottom, we must use the following formula.

P = Ro*g*h

where:

P = pressure [Pa] (units of pascals)

Ro = density of the water = 1000 [kg/m³]

g = gravity acceleration = 9.8 [m/s²]

h = height = 1 [m] (because its half of the portion, the full height is 2 m)

P = 1000*9.8*1

P = 9800 [Pa]

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3 years ago

The unit of energy is a derived unit why

olga2289 [7]

Answer:

The unit of energy is joule which depends upon the fundamental unit kg, m and sec. So, the unit of energy is a derived unit.

Explanation:

7 0

3 years ago

Read 2 more answers

I'll Vote Brainliest

aliina [53]

B) only in animal cells

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3 years ago

A tube of mercury with resistivity 9.84 × 10 -7 Ω ∙ m has an electric field inside the column of mercury of magnitude 23 N/C tha

slava [35]

Answer:

The current through the tube is 73.39A.

Explanation:

The relationship between the resistivity $\rho$ , the electric field $E$ , and the current density $J$ is given by

$\rho = \dfrac{E}{J}$

This equation can be solved for $J$ to get:

$J = \dfrac{E}{\rho}$

Since the current is $I = J\cdot A$

$I= J\cdot A = \dfrac{E}{\rho} \cdot A$

Now, for the tube of mercury $\rho = 9.84*10^{-7}\: \Omega \cdot m$ , $E = 23N/C$ , and the area is $A = \pi r^2 = \pi (1.0*10^{-3}m)^2 = 3.14*10^{-6}m^2$ ; therefore,

$I= \dfrac{23N/C}{9.84*10^{-7}\Omega\cdot m } *3.14*10^{-6}m^2$

$\boxed{I = 73.39A.}$

Hence, the current through the mercury tube is 73.39A.

5 0

4 years ago