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2 years ago

7

This type of mutation occurs when one or more base pairs are added to the gene sequence.

1 answer:

vekshin12 years ago

3 0

Answer:

Insertion

Explanation:

It causes a frameshift.

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GP A sinusoidal wave traveling in the negative x direction (to the left) has an amplitude of 20.0 cm, a wavelength of 35.0 cm, a

SIZIF [17.4K]

The period of the wave is determined as 0.083 seconds.

<h3>What is period of a wave?</h3>

The period of a wave is the time taken by a particle of the medium to complete one vibration.

<h3>Period of the wave</h3>

The period of the wave is calculated as follows;

T = 1/f

where;

T is the period of the wave
f is frequency of the wave

T = 1/12

T = 0.083 seconds

Thus, the period of the wave is determined as 0.083 seconds.

Learn more about period of a wave here: brainly.com/question/18818486

#SPJ4

8 0

1 year ago

The Mars Curiosity rover was required to land on the surface of Mars with a velocity of 1 m/s. Given the mass of the landing veh

Aliun [14]

Answer:

The value is $A = 39315 \ m^2$

Explanation:

From the question we are told that

The velocity which the rover is suppose to land with is $v = 1 \ m/s$

The mass of the rover and the parachute is $m = 2270 \ kg$

The drag coefficient is $C__{D}} = 0.5$

The atmospheric density of Earth is $\rho = 1.2 \ kg/m^3$

The acceleration due to gravity in Mars is $g_m = 3.689 \ m/s^2$

Generally the Mars atmosphere density is mathematically represented as

$\rho_m = 0.71 * \rho$

=> $\rho_m = 0.71 * 1.2$

=> $\rho_m = 0.852 \ kg/m^3$

Generally the drag force on the rover and the parachute is mathematically represented as

$F__{D}} = m * g_{m}$

=> $F__{D}} = 2270 * 3.689$

=> $F__{D}} = 8374 \ N$

Gnerally this drag force is mathematically represented as

$F__{D}} = C__{D}} * A * \frac{\rho_m * v^2 }{2}$

Here A is the frontal area

So

$A = \frac{2 * F__D }{ C__D} * \rho_m * v^2 }$

=> $A = \frac{2 * 8374 }{ 0.5 * 0.852 * 1 ^2 }$

=> $A = 39315 \ m^2$

8 0

3 years ago

The boom is supported by the winch cable that has a diameter of 0.5 in. and allowable normal stress of σallow=21 ksi. A boom ris

Andrew [12]

Explanation:

Let us assume that forces acting at point B are as follows.

$\sum F_{x} = 0$

$T + F_{AB} Sin 60$ = 0 ...... (1)

$\sum F_{y}$ = 0

$F_{AB} Cos 60 + W$ = 0 .......... (2)

Hence, formula for allowable normal stress of cable is as follows.

$\sigma_{allow} = \frac{T}{A}$

T = $(20 \times 1000) \frac{\pi}{4} \times (0.5)^{2}$

= 3925 kip

From equation (1), $F_{AB}Sin (60^{o})$ = -3925

$F_{AB} \times -0.304 8$ = -3925

$F_{AB}$ = 12877.29 kip

From equation (2), -12877.29 (Cos 60) + W = 0

$-12877.29 kip \times \frac{1}{2} + W$ = 0

W = 6438.64 kip

Thus, we can conclude that greatest weight of the crate is 6438.64 kip.

5 0

2 years ago

The property that compares the mass of an object with its volume is _____.

Licemer1 [7]

The property that compares the mass of an object with its volume is density.

7 0

3 years ago

Read 2 more answers

gayaneshka [121]

This is confusing yes

3 0

2 years ago