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3 years ago

If one stand of DNA reads as AATTCCGGATCG, what would the opposite strand bases be?

Physics

1 answer:

3241004551 [841]3 years ago

4 0

TTAAGGCCTAGC

hope this helps

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What if an object traveled from position -6 m to -1 m, what is the distance traveled?

Oxana [17]

The object traveled +5 m

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3 years ago

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Physical data is often used in the court system. In fact, police officers use radar to determine your speed when you are driving

Mashcka [7]

The driver is telling the truth, the radar gun must have been set incorrectly to record relative velocity.

The given parameters:

Speed of the driver observed by the stationary police officer, Vo = 44.7 m/s
Speed of the driver, V = 26.8 m/s.
Speed limit = 60 mph

The speed limit of the driver in meter per second is calculated as follows;

$= 60 \ \times \frac{miles}{hour} \times \frac{1609.34 \ m}{1 \ mile} \times \frac{1 \ hr}{3600 \ s}\\\\= 26.82 \ m/s$

From the speed limit, it is obvious that the driver's speed is within the limit. Thus, we can conclude that the driver is telling the truth, the radar gun must have been set incorrectly to record relative velocity.

Learn more about relative velocity here: brainly.com/question/17228388

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2 years ago

A backpack weighs 8.2 Newtons and has a mass of 5 kilograms on the moon what is the strength of gravity on the moon

lions [1.4K]

i dont get it so much but

The weight of the bag pack is 8.2 N. g = 1.64 m/s2. Hence, the acceleration due to gravity on moon is 1.64 m/s2. sooo? is it right

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3 years ago

Which describes a function of the thalamus?

Natalka [10]

Answer:

figure it out bot

Explanation:

jhkhllkhlkh

4 0

3 years ago

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Without the wheels, a bicycle frame has a mass of 6.71 kg. Each of the wheels can be roughly modeled as a uniform solid disk wit

dalvyx [7]

Answer:43.311 J

Explanation:

Given

mass of frame $\left ( m_f\right )=6.71 kg$

mass of wheel $\left ( m_w\right )=0.820 kg$

radius of wheel $\left ( r\right )=0.343 m$

v=3.22 m/s

Moment of inertia of each wheel\left ( I\right )=\frac{1}{2}mr^2[/tex]

$I=0.0482 kg-m^2$

kinetic Energy of whole cycle=Kinetic energy of wheels and frame+rotational energy of Wheels

$K.E.=\frac{1}{2}\left ( m_f+m_w\right )v^2+2\times \frac{1}{2}I\omega ^2$

$K.E.=\frac{1}{2}\left ( 6.71+0.820\right )3.22^2+2\times \frac{1}{2}0.0482\times \left ( \frac{3.22}{0.343}\right )^2$

K.E.=39.037+4.274=43.311J

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3 years ago