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

Consider three force vectors Fi with magni- tude 53 N and direction 116º, F2 with mag- nitude 57 N and direction 217°, and F3 wi

Flauer [41]

Answer:

a. Fnet =37.67N

b. The direction = 133.4 from the x axis counter clockwise.

c. Option 2

Explanation:

Given that F1 is 53N at 116°, then it will be at a direction of 116-90=26° in the second quadrant.

Given that F2 is 57N at 116°, then it will be at a direction of 217-180=37° in the third quadrant..

Given that F1 is 71N at 20°, then it is in the first quadrant.

a. Fnet= F1+F2+F3

Fnet= -F1sin26i+F2cos26j-F2cos37i-F2sin37j+F3cos20i+F3sin20j

Fnet= 53sin26i+53cos26j-57cos37i-57sin37j+71cos20i+71sin20j

Resolving the vectors into x and y components.

Fnet= -2.04i+37.62j

Magnitude of the vector

Fnet= √((-2.04)^2+(37.62)^2)

Fnet= 37.67N

Fnet is approximately 38N.

b. Direction of the Fnet.

Angle=arctan(y/x)

Angle=arctan(-37.61/2.04)

Angle= -43.37°

The angle is in the negative x axis and positive y axis.

Then the direction becomes 180-43.37

Therefore, the direction of the net force is 133.37°.

c. The instantaneous velocity of a body is always in the direction of the net force at that instant. Option 2 is correct.

Fnet=ma

Fnet= mv/t

So the velocity is in the direction of the Fnet.

3 0

3 years ago

a rocket engine can accelerate a rocket launched from rest vertically up with an acceleration of 22.9 m/s2. however, after 50.0

xxTIMURxx [149]

The acceleration of a rocket engine is given here, and after 50 seconds of flight, the engine fails, and we must determine the altitude of the rocket at the time the engine fails. Because the rocket starts from rest, the time taken is 50 seconds, the initial velocity is zero, and the acceleration is 22.9 meters per second square. So we use the kinamatics equation s equal to v. I t plus half 8 square. There is no acceleration at the start. 22.9 and t is 50 seconds, so displacement 2.86 times 10 to the power 4 is met. This is the rocket's displacement in 50 seconds, so this is the rocket's altitude when the engine fails.

<h3>What exactly is accelerate?</h3>

In mechanics, acceleration is defined as the rate of change of an object's velocity with respect to time. Vector quantities are accelerations. The orientation of an object's acceleration is determined by the orientation of its net force.

In his second law of motion, Sir Isaac Newton (1642-1727) defined acceleration as the ratio of a force acting on an object to its mass: a = f/m.

Accelerate is a verb that means to speed up. When you press the gas pedal, the car accelerates. If you know someone who works at the consulate, you can speed up the process.

Acceleration is the rate at which velocity changes over time, both in terms of speed and direction. A point or object moving in a straight line is accelerated if it accelerates or decelerates.

Even if the speed is constant, motion on a circle is accelerated because the direction is constantly changing. Both effects contribute to acceleration in all other types of motion.

Hence, There is no acceleration at the start. 22. This is the rocket's displacement in 50 seconds, so this is the rocket's altitude when the engine fails.

To know more about Accelerate refer to:

brainly.com/question/460763

#SPJ4

3 0

1 year ago

Block A of 5 kg with a speed of 3 m/s collides with block B of 10 kg that is stationary. After the collision, block B travels wi

Lemur [1.5K]

Answer:

-1m/s

Explanation:

We can calculate the speed of block A after collision

According to collision theory:

MaVa+MbVb = MaVa+MbVb (after collision)

Substitute the given values

5(3)+10(0) = 5Va+10(2)

15+0 = 5Va + 20

5Va = 15-20

5Va = -5

Va = -5/5

Va = -1m/s

Hence the velocity of ball A after collision is -1m/s

Note that the velocity of block B is zero before collision since it is stationary

6 0

3 years ago

What type of wave is energy transmitted and a definite direction in with a definite speed

SOVA2 [1]

A surface wave is a wave in which particles of the medium undergo a circular motion.

7 0

3 years ago

Read 2 more answers

A force of 44 N will stretch a rubber band 88 cm (0.080.08 m). Assuming that Hooke's law applies, how far will aa 11-N forc

Setler79 [48]

Answer:

The rubber band will be stretched 0.02 m.

The work done in stretching is 0.11 J.

Explanation:

Force 1 = 44 N

extension of rubber band = 0.080 m

Force 2 = 11 N

extension = ?

According to Hooke's Law, force applied is proportional to the extension provided elastic limit is not extended.

F = ke

where k = constant of elasticity

e = extension of the material

F = force applied.

For the first case,

44 = 0.080K

K = 44/0.080 = 550 N/m

For the second situation involving the same rubber band

Force = 11 N

e = 550 N/m

11 = 550e

extension e = 11/550 = 0.02 m

The work done to stretch the rubber band this far is equal to the potential energy stored within the rubber due to the stretch. This is in line with energy conservation.

potential energy stored = $\frac{1}{2}ke^{2}$

==> $\frac{1}{2}* 550* 0.02^{2}$ = 0.11 J

3 0

3 years ago