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3 years ago

What is the result of a mutation during replication

Physics

2 answers:

Sonbull [250]3 years ago

8 0

The correct answer is A. The strands are different. Hope this helps! :)

ss7ja [257]3 years ago

6 0

Answer:

Its A

Explanation: I took the quiz ewe

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El chofer de una ruta de transporte turístico tiene que informar a sus tripulantes cuanto tiempo tardara en llegar a su destino.

lana66690 [7]

Explanation:

total answer is 119 km

this is the answer for the question

8 0

3 years ago

A boy with a mass of 140 kg and a girl with a mass of 120 kg are on a merry go round. Th merry go round has a radius of 5 meters

White raven [17]

Answer:

we all are the human being we all dont no the all of 5he answer dont take tension beacause other one will give your answer

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3 years ago

In science I have to make a project that can solve any real world problems [Grade 6]

Kryger [21]

You can start with a recycling project , it can solve any problems including saving animals, giving clean water, and cleaner environment ofcourse

6 0

2 years ago

What is the moment of inertia of an object that rolls without slipping down a 3.5-m- high incline starting from rest, and has a

Daniel [21]

Answer:

I = 0.287 MR²

Explanation:

given,

height of the object = 3.5 m

initial velocity = 0 m/s

final velocity = 7.3 m/s

moment of inertia = ?

Using total conservation of mechanical energy

change in potential energy will be equal to change in KE (rotational) and KE(transnational)

PE = KE(transnational) + KE (rotational)

$mgh = \dfrac{1}{2}mv^2 + \dfrac{1}{2}I\omega^2$

v = r ω

$mgh = \dfrac{1}{2}mv^2 + \dfrac{1}{2}\dfrac{Iv^2}{r^2}$

$I = \dfrac{m(2gh - v^2)r^2}{v^2}$

$I = \dfrac{mr^2(2\times 9.8 \times 3.5 - 7.3^2)}{7.3^2}$

$I =mr^2(0.287)$

I = 0.287 MR²

3 0

3 years ago

?a wire is stretched 30% what is the percentage change in resistance

Marat540 [252]

Answer:

The percentage change in resistance of the wire is 69%.

Explanation:

Resistance of a wire can be determined by,

R = (ρl) ÷ A

Where R is its resistance, l is the length of the wire, A its cross sectional area and ρ its resistivity.

When the wire is stretched, its length and area changes but its volume and resistivity remains constant.

$l_{o}$ = 1.3l, and $A_{o}$ = $\frac{A}{1.3}$

So that;

$R_{o}$ = (ρ $l_{o}$ ) ÷ $A_{o}$ = (ρ × 1.3l) ÷ ( $\frac{A}{1.3}$ )

= (1.3lρ) ÷ ( $\frac{A}{1.3}$ )

= $(1.3)^{2}$ × [(ρl) ÷ A]

= 1.69R (∵ R = (ρl) ÷ A)

$R_{o}$ = 1.69R

Where $R_{o}$ is the new resistance, $l_{o}$ is the new length, and $A_{o}$ is the new area after stretching the wire.

The change in resistance of the wire = $R_{o}$ - R

= 1.69R - 1R

= 0.69R

The percentage change in resistance = $\frac{0.69R}{R}$ × 100

= 0.69 × 100

= 69%

The percentage change in resistance of the wire is 69%.

3 0

3 years ago