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2 years ago

If a single strand of DNA were stretched out, how far would it reach?

2 answers:

6 0

If unwound and tied together the strands of DNA in 1 cell would reach almost 6 feet but would only be 50 trillionths of an inch wide.

kipiarov [429]2 years ago

3 0

Answer: 2-3 meters long

Explanation:

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3 years ago

Consider a thin, spherical shell of radius 14.5 cm with a total charge of 29.5 µC distributed uniformly on its surface. (a) Find

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2 years ago

The distance vs. time graph of a car moving at constant speed should be a straight line. Why do the data points in the graph plo

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The measurements are inexact

Explanation:

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2 years ago

In Hooke’s law, what does k represent?

notsponge [240]

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3 years ago

Read 2 more answers

Graphing: Michelle climbs a tree and drops the toy car once she has reached the top. Please create an Energy-Time graph to show

Vlada [557]

Answer:

Refer to the attachment for the graph. The shape of both functions should resemble part of a parabola. Assumption: air resistance on the car is negligible.

Explanation:

The toy car started with a large amount of (gravitational) potential energy (PE) when it is at the top of the tree. Since it wasn't moving (as it was within Michelle's grip,) its kinetic energy (KE) would be equal to zero.
As the car falls to the ground, its PE converts to KE.
When the car was about to reach the ground, its PE is almost zero, while its KE is at its maximum.

The size of gravitational PE depends on both the mass and the height of the object. In this case, assume that the mass of the car stayed the same, PE should be proportional to the height of the car.

Assume that air resistance on the car is negligible. The height $h$ of the car at time $t$ could be found with the equation:

$\displaystyle h = -\frac{1}{2}\, g\, t^2 + h_0 \, (\text{Initial height})$ ,

where

$g \approx \rm 9.81\; m \cdot s^{-2}$ near the surface of the earth, and
$h_0$ is the initial height of the car.

On the other hand, $\displaystyle \text{GPE} = m \, g \, h = -\frac{m \cdot g^2}{2}\, t^2 + \underbrace{m \cdot g \cdot h_0}_{\text{Initial GPE}}$ .

In other words, plotting the gravitational PE of the car against time would give a parabola. Since $\displaystyle -\frac{m \cdot g^2}{2} < 0$ (the quadratic coefficient is smaller than zero,) the parabola should open downwards. Besides, since at $t = 0$ the initial GPE is positive, the $y$ -intercept of this parabola should also be positive.

Assume that the air resistance on the car is negligible. The mechanical energy (ME) of the toy car should conserve (stay the same.) The mechanical energy of an object is the sum of its PE and KE. The PE of the toy car has already been found as a function of time. Therefore, simply subtract the expression of PE from mechanical energy to find an expression for KE.

To find the value of mechanical energy, consider the PE of the toy car before it was dropped. Since initially KE was equal to zero, the mechanical energy of the toy car would be equal to its initial PE. That's $m \cdot g \cdot h_0$ . If there's no air resistance, the value of ME would stay at

Subtract PE from ME to obtain an expression for KE:

$\begin{aligned} \text{KE} &= \text{ME} - \text{PE} \cr &= m \cdot g \cdot h_0 - \left(-\frac{m \cdot g^2}{2}\, t^2 + m \cdot g \cdot h_0\right)\cr &= \frac{m \cdot g^2}{2}\, t^2\end{aligned}$ .

That's also a parabola when plotted against $t$ . Note that since the quadratic coefficient $\displaystyle \frac{m \cdot g^2}{2}$ is positive, the parabola shall open upwards.

7 0

3 years ago