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netineya [11]
3 years ago
6

Within a DNA molecule, which two nitrogen bases are correctly paired with one another?

Physics
2 answers:
fomenos3 years ago
3 0
The answer is B. adenine (A) and thymine (T)
slavikrds [6]3 years ago
3 0

a is the correct answer

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Which of the following units would be used for measuring volume?
zalisa [80]

i am right their with you i do believe it is A or D but i am more certain of D

6 0
3 years ago
An emf is induced by rotating a 1060 turn, 20.0 cm diameter coil in the Earth's 5.25 ✕ 10−5 T magnetic field. What average emf (
lara31 [8.8K]

Answer:

The average emf induced in the coil is 175 mV

Explanation:

Given;

number of turns of the coil, N = 1060 turns

diameter of the coil, d = 20.0 cm = 0.2 m

magnitude of the magnetic field,  B = 5.25 x 10⁻⁵ T

duration of change in field, t = 10 ms = 10 x 10⁻³ s

The average emf induced in the coil is given by;

E = N\frac{\delta \phi}{dt} \\\\E = N\frac{\delta B}{\delta t}A

where;

A is the area of the coil

A = πr²

r is the radius of the coil = 0.2 /2 = 0.1 m

A = π(0.1)² = 0.03142 m²

E = \frac{NBA}{t} \\\\E = \frac{1060*5.25*10^{-5}*0.03142}{10*10^{-3}} \\\\E = 0.175 \ V\\\\E = 175 \ mV

Therefore, the average emf induced in the coil is 175 mV

3 0
3 years ago
The amount of charge flowing through a particular point in a conductor is represented by the equation Q = at^3 + bt + c, where a
stira [4]

Explanation:

Below is an attachment containing the solution.

3 0
4 years ago
What is the energy Q released when 131 53Idecays and 131 54Xe is formed? The atomic mass of 131 53I is 130.906118 u and the atom
DanielleElmas [232]

Answer:0.967meV

Explanation:

-Find the difference in u, so 130.906118-130.90508= 0.001038u

- convert to meV

1 u = 931.494meV

multiply 0.001038 by 931.494

=0.001038 X 931.494

0.967 meV

5 0
4 years ago
A frictionless piston–cylinder device contains 5 kg of nitrogen at 100 kPa and 250 K. Nitrogen is now compressed slowly accordin
Arte-miy333 [17]

Answer:

The work input during this process is -742 kJ

Explanation:

Given;

Initial temperature of nitrogen T₁ = 250 K

final temperature of nitrogen T₂ = 450 K

mass of nitrogen, m = 5 kg

PV^{1.4} = constant

The work input during the process is calculated as;

W = \frac{m*R(T_2-T_1)}{1-n}

where;

R is gas constant = 0.2968 kJ/kgK

substitute given values in above equation.

W = \frac{m*R(T_2-T_1)}{1-n} = \frac{5*0.2968(450-250)}{1-1.4} = -742 \ kJ

Therefore, the work input during this process is -742 kJ

8 0
4 years ago
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