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2 years ago

What is the matching nitrogen base sequence for the gene below?

Physics

1 answer:

grigory [225]2 years ago

6 0

Answer:

Explanation:

T-G-T is the answer .

the matching Nitrogen base sequence for gene ATC is A

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Explain in terms of impulse how padding reduces forces in a collision. State this in terms of a real example, such as the advant

Lena [83]

Answer:

Impulse = Average force x time of contact

Explanation:

Impulsive force is a force which is very large but applied on a body for a very small duration of time.

Impulse is given by the change in momentum of the body.

Impulse = Average force x small time interval

When padding is there, the time interval of contact is large and thus, the force exerted by the body is small.

So, when a person falls on the tile floor, there is no compression and thus, the time of contact is very small and thus the impulsive force is very large, due to which the body may damage.

So, when a person falls on the carpeted floor, there is a compression and thus, the time of contact is comparatively large and thus the impulsive force is small, due to which the body may safe.

3 0

2 years ago

A thin, rectangular sheet of metal has mass M and sides of length a and b. Find the moment of inertia of this sheet about an axi

Lubov Fominskaja [6]

Answer:

The moment of inertia is $I=\frac{M}{12} a^{2}$

Explanation:

The moment of inertia is equal:

$I=\int\limits^a_b {r^{2} } \, dm$

If r is $-\frac{a}{2}$

and $dm=\frac{M}{a} dr$

$I=\int\limits^a_b {r^{2}\frac{M}{a} } \, dr\\a=\frac{a}{2} \\b=-\frac{a}{2}$

$I=\frac{M}{a} \int\limits^a_b {r^{2} } \, dr\\\\I=\frac{M}{a} (\frac{M}{3} )_{b}^{a}\\ I=\frac{M}{3a} (\frac{a^{3} }{8} +\frac{a^{3} }{8} )\\I=\frac{M}{12} a^{2}$

7 0

2 years ago

What will happen two temperatures if you increase particle motion

nalin [4]

If the temperature is increased the particles gain more kinetic energy or vibrate faster. This means that they move faster and take more space.

8 0

2 years ago

Question 6 Multiple Choice Worth 4 points)

allsm [11]

is iron and aluminium is there

6 0

2 years ago

One end of a thin rod is attached to a pivot, about which it can rotate without friction. Air resistance is absent. The rod has

Mars2501 [29]

Answer:

6.86 m/s

Explanation:

This problem can be solved by doing the total energy balance, i.e:

initial (KE + PE) = final (KE + PE). { KE = Kinetic Energy and PE = Potential Energy}

Since the rod comes to a halt at the topmost position, the KE final is 0. Therefore, all the KE initial is changed to PE, i.e, ΔKE = ΔPE.

Now, at the initial position (the rod hanging vertically down), the bottom-most end is given a velocity of v0. The initial angular velocity(ω) of the rod is given by ω = v/r , where v is the velocity of a particle on the rod and r is the distance of this particle from the axis.

Now, taking v = v0 and r = length of the rod(L), we get ω = v0/ 0.8 rad/s

The rotational KE of the rod is given by KE = 0.5Iω², where I is the moment of inertia of the rod about the axis of rotation and this is given by I = 1/3mL², where L is the length of the rod. Therefore, KE = 1/2ω²1/3mL² = 1/6ω²mL². Also, ω = v0/L, hence KE = 1/6m(v0)²

This KE is equal to the change in PE of the rod. Since the rod is uniform, the center of mass of the rod is at its center and is therefore at a distane of L/2 from the axis of rotation in the downward direction and at the final position, it is at a distance of L/2 in the upward direction. Hence ΔPE = mgL/2 + mgL/2 = mgL. (g = 9.8 m/s²)

Now, 1/6m(v0)² = mgL ⇒ v0 = $\sqrt{6gL}$

Hence, v0 = 6.86 m/s

4 0

2 years ago