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4 years ago

Using the periodic table entry below, how many neutrons does the most common isotope of hydrogen have?

2 answers:

alexandr1967 [171]4 years ago

6 0

Hydrogen is a gas and the first element listed on the periodic table.Elemental hydrogen has three naturally occurring isotopes they are namely protium, deuterium and tritium. Protium is the most common isotope with an occurrence of 99.98 percent. This isotope consists of one proton and one electron.

NikAS [45]4 years ago

3 0

zero...................................zero

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A 6.8 KG object moves with a velocity of 8 M/S what’s it's kinetic energy?

FromTheMoon [43]

Answer:

Explanation:

KE = ½mv² = ½(6.8)8² = 217.6 J

round as appropriate because that result is way too much precision for the inputs provided. Arguably should be 200 J based on the single significant digit of the velocity.

8 0

3 years ago

A uniform 1.4-kg rod that is 0.75 m long is suspended at rest from the ceiling by two springs, one at each end of the rod. Both

svetlana [45]

Answer:

7 deg

Explanation:

$m$ = mass of the rod = $1.4 kg$

$W$ = weight of the rod = $mg = (1.4) (9.8) = 13.72 N$

$k_{L}$ = spring constant for left spring = $59 Nm^{-1}$

$k_{R}$ = spring constant for right spring = $33 Nm^{-1}$

$x_{L}$ = stretch in the left spring

$x_{R}$ = stretch in the right spring

$L$ = length of the rod = 0.75 m

$\theta$ = Angle the rod makes with the horizontal

Using equilibrium of force in vertical direction for left spring

$k_{L} x_{L} = (0.5) W\\(59) x_{L} = (0.5) (13.72)\\x_{L} = 0.116 m$

Using equilibrium of force in vertical direction for right spring

$k_{R} x_{R} = (0.5) W\\(33) x_{R} = (0.5) (13.72)\\x_{R} = 0.208 m$

Angle made with the horizontal is given as

$\theta = tan^{-1}(\frac{(x_{R} - x_{L})}{L} )\\\theta = tan^{-1}(\frac{(0.208 - 0.116)}{0.75} )\\\theta = 7 deg$

3 0

4 years ago

Does anyone have the answers for: Lesson 2: Final Exam Physical Science B Unit &: Cumulative Exam

Sladkaya [172]

Answer: Yep! I made a yt vid on them so look this up:

Video result for rick roll with a different link copy and paste0:10

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Video result for rick roll with a different link copy and paste0:10

Rick roll, but with different link

NEVER GOING TO GIVE YOU UP

Video result for rick roll with a different link copy and paste0:10

Rick roll, but with different link

4 0

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

Calculate the energy released as heat when 51.60 g Cu 2 O ( s ) undergo oxidation at constant pressure.

lesya692 [45]

Answer:

Energy released as heat will be -189.417 Kj

Explanation:

The oxidation is the type of reaction in which reaction occurs in the presence of oxygen. The exothermic reaction is the type of reaction in which heat is released. The enthalpy is the type of physical quantity that is used to measure the energy in the thermodynamic system.

As we know the molar mass of Copper II Oxide is 79.545 g

Also the oxidation of copper(II) oxide, CuO(s), is an exothermic process, 2Cu2O (s) + O2-->4CuO (s) delta H reaction= -292.0 kj/mol.

So for 51.60g of Copper II Oxide,

Heat released will be (-292.0 * 51.60/79.545) = -189.417 Kj

6 0

3 years ago