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3 years ago

Help me please I will thank you

Physics

1 answer:

Degger [83]3 years ago

3 0

Answer:

DNA= 2 strands, sugar is deoxyribose, and the Nitrogen Bases are adenine, uracil, cytosine, and thymine.

RNA= 1 strands, sugar is ribose, and the Nitrogen Nases are adenine, uracil, cytosine, and guanine.

Hope this helps :)

Explanation:

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How can a driver steer a car traveling at constant speed so that the magnitude of the acceleration remains constant. (Assume tha

ratelena [41]

<h2>Circular path i.e 1</h2>

Explanation:

If the driver steer into circular path , the acceleration of car will be

acceleration = velocity²/radius of circle

Because the velocity and radius of circle are both constant at all points . Thus the acceleration will also remain constant .

In case 3 and 4 , the radius of path changes . Because velocity is constant . Therefore the acceleration will not remain constant

3 0

3 years ago

If 27 J of work are needed to stretch a spring from 15 cm to 21 cm and 45 J are needed to stretch it from 21 cm to 27 cm, what i

kupik [55]

Answer:

9 cm.

Explanation:

The energy used for stretch the spring from $15 cm$ to $21 cm$ will be , $E_{1}=27J$

The energy used for stretch the spring from $21 cm$ to $27 cm$ will be , $E_{2}=45J$

using the energy of spring formula ,we find that

$27 = \frac{1}{2}K((21-L^{2})-(15-L^{2}))$

$45 = \frac{1}{2}K((27-L^{2})-(21-L^{2}))$

Dividing both the equation will get,

$\frac{3}{5}=\frac{(21-L)^{2}-(15-L)^{2}}{(27-L)^{2}-(21-L)^{2}}\\5((21-L)^{2}-(15-L)^{2})=3((27-L)^{2}-(21-L)^{2})\\3(729 - 54L + L^{2}- 441 + 42L - L^{2} ) = 5(441 - 42L + L^{2} - 225 + 30L - L^{2} )\\3(288 - 12L) = 5(216 - 12L)\\24L = 216\\L = 9 cm$

Therefore, the natural length of the spring is, 9 cm.

4 0

3 years ago

A football field lies so one endzone is to the east and one is to the west. A running back gets the ball and starts running 20o

Jobisdone [24]

Answer:

( 80.87 i - 12.96 j ) m

Explanation:

Running back movements :

20⁰ north of east for 10 meters,

Runs straight east for 60 meters

runs 35⁰ east of south for 20 meters

A) show vector pointing from the starting point to the end ( where he scored )

The final vector displacement : ( 80.87 i - 12.96 j ) m

which is : 81.90 m, 9.10⁰ south of east

attached below is the required diagram

8 0

3 years ago

I need help plz!!! Please help...

RSB [31]

Answer: 60

Explanation:

4 0

3 years ago

A mine car (mass=390 kg) rolls at a speed of 0.50 m/s on a horizontal track, as the drawing shows. A 250-kg chunk of coal has a

Nady [450]

Answer:

v=0.60 m/s

Explanation:

Given that

m ₁= 390 kg ,u ₁= 0.5 m/s

m₂ = 250 kg ,u₂ = 0.76 m/s

As we know that if there is no any external force on the system the total linear momentum of the system will be conserve.

Pi = Pf

m ₁u ₁+m₂u₂ = (m₂ + m ₁ ) v

Now putting the values in the above equation

390 x 0.5 + 250 x 0.76 = (390 + 250 ) v

$v=\dfrac{390\times 0.5+250\times 0.76}{390+250}\ m/s$

v=0.60 m/s

Therefore the velocity of the system will be 0.6 m/s.

8 0

4 years ago

Help me please I will thank you​

Help me please I will thank you