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3 years ago

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The parents of a new baby believe they brought the wrong child home from the hospital. Gel electrophoresis was performed using D

NA samples from the parents and the child. A section of the gel electrophoresis results is shown below.Which conclusion is valid based on the gel electrophoresis results?

1 answer:

polet [3.4K]3 years ago

7 0

Would you be able to post the gel electrophoresis results?

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Determine the minimum amount of energy (in MeV) needed to remove a neutron from a 88Sr nucleus. (For all masses, keep six places

Sindrei [870]

Answer:

11.106 Mev.

Explanation:

Sr⁸⁸ → Sr ⁸⁷ + ₀n¹

87.905612 86.908877 1.008664 ( atomic masses in amu )

mass defect = ( 86.908877 + 1.008664) - 87.905612 = .011929 amu.

Equivalent energy in MeV = .011929 x 931 = 11.106 MeV

6 0

3 years ago

Two large thin metal plates are parallel and close to each other. On their inner faces, the plates have excess surface charge of

wariber [46]

Answer:

For left = 0 N/C

For right = 0 N/C

At middle = $-7.6836 * 10^{-11} \vec{i}$ N/C

Explanation:

Given data :-

б = $6.8 * 10^{-22}$ C/ m²

Considering the two thin metal plates to be non conducting sheets of charges.

Electric field is given by

$E = \frac{\sigma }{2\varepsilon }$

1) To the left of the plate

$\vec{E}= (\frac{\sigma }{2\varepsilon })(-\vec{i})+ (\frac{\sigma }{2\varepsilon })(\vec{i})$ = 0 N/C.

2) To the right of them.

$\vec{E}= (\frac{\sigma }{2\varepsilon })(-\vec{i})+ (\frac{\sigma }{2\varepsilon })(\vec{i})$ = 0 N/C.

3) Between them.

$\vec{E}= (\frac{\sigma }{2\varepsilon })(-\vec{i})+ (\frac{\sigma }{2\varepsilon })(-\vec{i})$ = $(\frac{\sigma }{\varepsilon })(-\vec{i})$ = $-\frac{6.8 * 10^{-22} }{8.85 * 10 ^{-12} } \vec{i}$ = $-7.6836 * 10^{-11} \vec{i}$ N/C

5 0

3 years ago

Sapting Leng You have a string with a mass of 13.7 q. You stretch the string with a force of 8.39 N, giving it a length of 1.87

marshall27 [118]

Answer:

a) $\lambda=0.935\ \textup{m}$

b) $f=36.19\approx 36\ \textup{Hz}$

Explanation:

Given:

String vibrates transversely fourth dynamic, thus n = 4

mass of the string, m = 13.7 g = 13.7 × 10⁻¹³ kg

Tension in the string, T = 8.39 N

Length of the string, L = 1.87 m

a) we know

$L= n\frac{\lambda}{2}$

where,

$\lambda$ = wavelength

on substituting the values, we get

$1.87= 4\times \frac{\lambda}{2}$

or

$\lambda=0.935\ \textup{m}$

b) Speed of the wave (v) in the string is given as:

$v =f\lambda$

also,

$v=\sqrt\frac{T}{(\frac{m}{L})}$

equating both the formula for 'v' we get,

$f\lambda=\sqrt\frac{T}{(\frac{m}{L})}$

on substituting the values, we get

$f\times 0.935=\sqrt\frac{8.39}{(\frac{13.7\times 10^{3}}{1.87})}$

or

$f=\frac{33.84}{0.935}$

or

$f=36.19\approx 36\ \textup{Hz}$

5 0

3 years ago

Who will give me answer I will give you brainliest.....

kondaur [170]

1 . W=mass times acceleration due to gravity
60kg times 9.8m/s2
= 588N

2. W=mg
1176N=m times 9.8
m=120kg

3. 1 hour=3600s
24 hours=?
24 times 3600
= 86400 seconds

4. 1000g=1kg
25000g=?
25000 times 1 divide by 1000
=25kg

5. 1000000mg=1kg
123000000=?
123000000 times 1 divide by 1000000
=123 kg

7 0

3 years ago

Marisol is designing a system to detect when the door to her room is opened. She has observed that the magnetic field in the vic

KIM [24]

Answer:

The minimum time interval required is 18.1 sec

Explanation:

Given:

Magnetic field $B = 5.50 \times 10^{-5}$ T

Minimum induced emf $\epsilon = 1 \times 10^{-6}$ V

Angle between magnetic field and area vector $\theta =$ 34°

Area of door $A = 0.895 \times 2.15 = 1.924$ $m^{2}$

According to the faraday's law,

$\epsilon = -\frac{d\phi}{dt}$

Where $\phi =$ magnetic flux

From the formula of magnetic flux,

$\phi = BA\cos \theta$

Initially,

$\phi _{i} = BA\cos 34$

$= 5.50 \times 10^{-5} \times 1.924 \times 0.8290$

$= 8.773 \times 10^{-5}$ Wb

When door is closed,

$\phi _{f} = BA\cos 0$

$= 5.50 \times 10^{-5} \times 1.924$

$\phi _{f} = 10.58 \times 10^{-5}$ Wb

Hence $d\phi = \phi _{f} - \phi _{i}$

$d\phi = (10.58 - 8.773 ) \times 10^{-5}$

$d \phi = 1.81 \times 10^{-5}$ Wb

Put the values in above equation,

$dt = \frac{d \phi}{\epsilon}$

$dt = \frac{1.81 \times 10^{-5} }{1 \times 10^{-6} }$

$dt = 18.1$ sec

Therefore, the minimum time interval required is 18.1 sec

6 0

3 years ago