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3 years ago

Help please !

Physics

1 answer:

Yuri [45]3 years ago

5 0

Answer:

1) Adenine (A) 2. Uracil (U) 4) Cytosine (C)

Explanation:

"RNA consists of four nitrogenous bases: adenine, cytosine, uracil, and guanine."

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What are the two most abundant gases in the atmosphere

HACTEHA [7]

Nitrogen (like70ish % and oxygen like 20%)

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3 years ago

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Which of the following is the correct unit to measure the current in a radio?

CaHeK987 [17]

Answer:

watts

Explanation:

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3 years ago

An object has 90,000 J of kinetic energy and is moving at 12 m/s. What is the objects mass?

trapecia [35]

Answer:

plug it into your calculator

Explanation:

Kinetic energy = KE = Joules = J

KE= 1/2mv^2

90,000 = 1/2 m (12^2)

180,000 = m (144)

m = 180,000 / 144

4 0

4 years ago

A 90.0-kg man runs at a constant speed of 11.5 m/s. what is his kinetic energy?

harina [27]

The kinetic energy possessed by the man is 517.5 Joules.

Given the following data:

Mass = 90.0 kg

Velocity = 11.5 m/s.

To find the kinetic energy of the man:

Kinetic energy is an energy that is possessed by a physical object or body due to its motion.

Mathematically, kinetic energy is calculated by using the formula;

$K.E = \frac{1}{2} MV^2$

Where:

K.E is the kinetic energy.

M is the mass of an object.

V is the velocity of an object.

Substituting the given values, we have;

$K.E = \frac{1}{2}$ × $90$ × $11.5$

$K.E = 45$ × $11.5$

Kinetic energy = 517.5 Joules.

Therefore, the kinetic energy possessed by the man is 517.5 Joules.

8 0

3 years ago

Two conductors are made of the same material and have the same length. Conductor A is a solid wire of diameter 1.0 mm.

pantera1 [17]

Answer:

R_a/R_b=3

Explanation:

The resistance in terms of the area and the length of the wire is given by:

R=pL/A

if we have two wires, the first one is a solid wire with a diameter of d_A = 1 x 10^-3 m, and the second one is a hollow wire with inner diameter of d_B,i = 1 x 10^-3 m and outer diameter of d_B,σ= 2 x 10^-3 m, so the cross sectional area of the first wire is:

A_a=πr^2_a

A_a=πd^2_A/4

hence the resistance is:

R_a=(4*p*L_a)/π*d^2_A (1)

the area of the second wire is:

A_b=π*r^2_B,σ-π*r^2_B,i

A_b=π/4(d^2_B,σ-d^2_B,i)

hence the resistance is:

R_b=(4*p*L_b)/π(d^2_B,σ-d^2_B,i) (2)

To find the ratio between the resistances R_a/R_b, we divide (1) over (2) to get:

R_a/R_b=(d^2_B,σ-d^2_B,i)*L_a/(d^2_a*L_b)

but the wires have the same length, therefore:

R_a/R_b=(d^2_B,σ-d^2_B,i)/(d^2_a)

substitute with the given values to get:

R_a/R_b=3

6 0

3 years ago

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