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3 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]3 years ago

3 0

Answer: 2-3 meters long

Explanation:

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Calculate the pressure on a man’s foot when a woman who weighs 520 N steps on his foot with her heel which has an area of 0.001

Vinvika [58]

Answer:

520000 or 520000 pa

Force = 520N

Area of contact = 0.001

Pressure: 520000 or 520000

6 0

3 years ago

Your grandmother enjoys creating pottery as a hobby. She uses a potter's wheel, which is a stone disk of radius R-0.520 m and ma

Lesechka [4]

Answer:

0.54454

104.00902 N

Explanation:

m = Mass of wheel = 100 kg

r = Radius = 0.52 m

t = Time taken = 6 seconds

$\omega_f$ = Final angular velocity

$\omega_i$ = Initial angular velocity

$\alpha$ = Angular acceleration

Mass of inertia is given by

$I=\dfrac{mr^2}{2}\\\Rightarrow I=\dfrac{100\times 0.52^2}{2}\\\Rightarrow I=13.52\ kgm^2$

Angular acceleration is given by

$\alpha=\dfrac{\tau}{I}\\\Rightarrow \alpha=\dfrac{\mu fr}{I}\\\Rightarrow \alpha=\dfrac{\mu 50\times 0.52}{13.52}$

Equation of rotational motion

$\omega_f=\omega_i+\alpha t\\\Rightarrow \omega_f=\omega_i+\dfrac{\mu (-50)\times 0.52}{13.52}t\\\Rightarrow 0=60\times \dfrac{2\pi}{60}+\dfrac{\mu (-50)\times 0.52}{13.52}\times 6\\\Rightarrow 0=6.28318-11.53846\mu\\\Rightarrow \mu=\dfrac{6.28318}{11.53846}\\\Rightarrow \mu=0.54454$

The coefficient of friction is 0.54454

At r = 0.25 m

$\omega_f=\omega_i+\dfrac{0.54454 (-50)\times 0.52}{13.52}6\\\Rightarrow 0=60\times \dfrac{2\pi}{60}+\dfrac{0.54454 f\times 0.25}{13.52}6\\\Rightarrow 2\pi=0.06041f\\\Rightarrow f=\dfrac{2\pi}{0.06041}\\\Rightarrow f=104.00902\ N$

The force needed to stop the wheel is 104.00902 N

5 0

3 years ago

2. What would be the acceleration of the clown at 5 s? (A) 1.6 m/s2 (B) 8.0 m/s2 (C) 2.0 m/s2 (D) 3.4 m/s2 3. After 12 seconds,

Neko [114]

Is there an image that goes with this question?

7 0

2 years ago

An electric field exerts a force of 3.00 E -4 N on a positive test charge of 7.20 E-4 C. The magnitude of the field at this loca

Step2247 [10]

3.00 E-4 / 7.20 E-4 = 0.417
answer is b

3 0

3 years ago

Read 2 more answers

A person wants to determine the spring constant of an exercise stretch cord. He pulls the cord with a force probe that exerts a

amid [387]

Answer: 180N/m(to 2 significant figures)

Explanation:

According to hooked law which states that the extension of a material is directly proportional to the applied force provided that the elastic limit is not exceeded. Mathematically, F = ke where;

F is the applied force in newtons

k is the elastic/spring constant in N/m

e is the extension in meters

Given applied force = 35N

extension = 20cm = 0.2m

Since F = ke,

k = F/e = 35/0.2

k = 175N/m

The spring constant is 175N/m

= 180N/m (to 2significant figures)

8 0

3 years ago