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Thepotemich [5.8K]

3 years ago

12

What is the water potential of pure water?

1 answer:

DanielleElmas [232]3 years ago

3 0

Hey :)

The water potential of pure water<span> in an open container is zero because there is no solute and the pressure in the container is zero</span>

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List 1 chemical property of a metal a nail?

timurjin [86]

Iron...................... hope this helpes

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3 years ago

Does a solid sit by itself

Gekata [30.6K]

Answer: yes i think

Explanation:

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3 years ago

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A car with a mass of 2.0×10^3 kg is traveling at 15m/s .what is the momentum of the car ?

Maurinko [17]

Hello,

<span>A car with a mass of 2.0×10^3 kg is traveling at 15m/s. We need to find the momentum of the car. To do so, follow this formula:

p=mv

Where,

p = momentum
m = mass
v = </span>velocity

The cars mass is 2.0E3 and its velocity is 15m/s. Therefore:

p=2.0 x 10^3 *15 or 2000(15)

p=30000

Thus, the cars momentum is 30000 kg m/s

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7 0

3 years ago

The rectangular plates in a parallel-plate capacitor are 0.063 m x 5.4 m. A distance of 3.5 x 10^-5m separate the plates. The pl

Aleks04 [339]

The capacitance of the capacitor is $1.8\cdot 10^{-9}F$

Explanation:

The capacitance of a parallel-plate capacitor is given by the equation

$C=\frac{k\epsilon_0 A}{d}$

where

k is the dielectric constant of the medium

$\epsilon_0 = 8.85\cdot 10^{-12} F/m$ is the vacuum permittivity

A is the area of the plates

d is the separation between the plates

For the capacitor in this problem, we have:

k = 2.1 is the dielectric constant

$d=3.5\cdot 10^{-5}m$ is the separation between the plates

$A=0.063m \cdot 0.054m = 0.0034 m^2$ (I assumed that 5.4 m is a typo, since it is not a realistic size for the side of the plate)

Therefore, the capacitance of the capacitor is

$C=\frac{(2.1)(8.85\cdot 10^{-12})(0.0034)}{3.5\cdot 10^{-5}}=1.8\cdot 10^{-9}F$

Learn more about capacitors:

brainly.com/question/10427437

brainly.com/question/8892837

brainly.com/question/9617400

#LearnwithBrainly

5 0

3 years ago

castortr0y [4]

Answer: $Q=3000 cal$

Explanation:

We are given the following formula:

$Q=m. c. \Delta T$ (1)

Where:

$Q=3000 cal$ is the amount of heat

$m=300g$ is the mass of water

$c=1 cal/g \°C$ is the specific heat of water

$\Delta T$ is the variation in temperature, which in this case is $\Delta T=30\°C-20\°C=10\°C$

Rewriting equation (1) with the known values at the right side, we will prove the result is $3000 cal$ :

$Q=(300g)(1 cal/g \°C)(10\°C)$ (2)

$Q=3000 cal$ This is the result

8 0

3 years ago