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krek1111 [17]
3 years ago
9

certain conductor with a resistance of 10 Ohms is crossed by an electric current of intensity 100 mA. What is the potential diff

erence at the terminals of that conductor?
Chemistry
1 answer:
Morgarella [4.7K]3 years ago
8 0

Answer:

1.0 V

Explanation:

Step 1: Given data

  • Resistance of certain conductor (R): 10 Ω
  • Current's intensity (I): 100 mA
  • Potential difference at the terminals of that conductor (V): ?

Step 2: Convert "I" to Ampere

We will use the conversion factor 1 A = 1000 mA.

100 mA × 1 A/1000 mA = 0.100 A

Step 3: Calculate the potential difference

We will use Ohm's law.

V = I × R

V = 0.100 A × 10 Ω = 1.0 V

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Find the density of a piece of chocolate with these measurements: 2.4 g and 5.12 mL
Dahasolnce [82]

Answer:

0.46875g/ml

Explanation:

Density(p) = m / v unit - g/ml or Kg/m^3

Given

mass = 2.4g

volume = 5.12ml

p = m / v

= 2.4g / 5.12ml

= 0.46875g/ml

5 0
2 years ago
80g of sodium hydroxide reacts with excess sulfuric acid to produce 142g of sodium sulfate. What mass of sodium hydroxide would
Ostrovityanka [42]

Answer:

Mass of sodium hydroxide needed  = 28.2 g

Explanation:

Given data:

Mass of sodium hydroxide = 80 g

Mass of sodium sulfate produced = 142 g

Mass of sodium hydroxide needed = ?

Mass of sodium sulfate produced = 50 g

Solution:

Mass of sodium sulfate produced / Mass of sodium hydroxide = Mass of sodium sulfate produced / x

142 g/ 80 g = 50 g / x

x = 50 g × 80 g / 142 g

x =4000 g/ 142

x = 28.2 g

6 0
3 years ago
Find the de Broglie wavelength lambda for an electron moving at a speed of 1.00 \times 10^6 \; {\rm m/s}. (Note that this speed
masya89 [10]

(A) 7.28\cdot 10^{-10} m

The De Broglie wavelength of an electron is given by

\lambda=\frac{h}{p} (1)

where

h is the Planck constant

p is the momentum of the electron

The electron in this problem has a speed of

v=1.00\cdot 10^6 m/s

and its mass is

m=9.11\cdot 10^{-31} kg

So, its momentum is

p=mv=(9.11\cdot 10^{-31} kg)(1.00\cdot 10^6 m/s)=9.11\cdot 10^{-25}kg m/s

And substituting into (1), we find its De Broglie wavelength

\lambda=\frac{6.63\cdot 10^{-34}Js}{9.11\cdot 10^{-25} kg m/s}=7.28\cdot 10^{-10} m

(B) 1.16\cdot 10^{-34}m

In this case we have:

m = 0.143 kg is the mass of the ball

v = 40.0 m/s is the speed of the ball

So, the momentum of the ball is

p=mv=(0.143 kg)(40.0 m/s)=5.72 kg m/s

And so, the De Broglie wavelength of the ball is given by

\lambda=\frac{h}{p}=\frac{6.63\cdot 10^{-34} Js}{5.72 kg m/s}=1.16\cdot 10^{-34}m

(C) 9.02\cdot 10^{-9}m

The location of the first intensity minima is given by

y=\frac{L\lambda}{a}

where in this case we have

y=0.492 cm = 4.92\cdot 10^{-3} m

L = 1.091 is the distance between the detector and the slit

a=2.00\mu m=2.00\cdot 10^{-6}m is the width of the slit

Solving the formula for \lambda, we find the wavelength of the electrons in the beam:

\lambda=\frac{ya}{L}=\frac{(4.92\cdot 10^{-3}m)(2.00\cdot 10^{-6} m)}{1.091 m}=9.02\cdot 10^{-9}m

(D) 7.35\cdot 10^{-26}kg m/s

The momentum of one of these electrons can be found by re-arranging the formula of the De Broglie wavelength:

p=\frac{h}{\lambda}

where here we have

\lambda=9.02\cdot 10^{-9}m is the wavelength

Substituting into the formula, we find

p=\frac{6.63\cdot 10^{-34}Js}{9.02\cdot 10^{-9}m}=7.35\cdot 10^{-26}kg m/s

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ycow [4]

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Explanation:

Thermal energy is released in this process

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