All categories

1 year ago

This mutation blank alter the protein function

Physics

1 answer:

mariarad [96]1 year ago

4 0

Answer:

magsagot kau ng inyo wag kau mag hanap tseee hahahahhahahahahabababab

You might be interested in

17-<br> Find the magnitude of vector product \BxĀ| for A=– 23 +3Â and B = 2î – 3+ Å<br> vectors.

aksik [14]

Answer:

alam ko sagot pero mataas

8 0

2 years ago

What charge accumulates on the plates of a 2.0-μF air-filled capacitor when it is charged until the potential difference across

enot [183]

Answer:

0.0002 C.

Explanation:

Charge: This can be defined as the ratio of current to time flowing in a circuit. The S.I unit of charge is Coulombs (C)

Mathematically, charge can be expressed as

Q = CV ................................. Equation 1

Where Q = amount of charge, C = capacitance of the capacitor, V = potential difference across the plates.

Given: C = 2.0-μF = 2×10⁻⁶ F, V = 100 V.

Substitute into equation 1

Q = 2×10⁻⁶× 100

Q = 2×10⁻⁴ C

Q = 0.0002 C.

The amount of charge accumulated = 0.0002 C

7 0

2 years ago

Read 2 more answers

Arcsin 0.9331 in degrees

lisabon 2012 [21]

Answer:

68.9233231661

Explanation:

Just put it into your calculator, shift sin should do it but it will come up like this: $sin^{-1}$ which is the same as arcsin

6 0

2 years ago

Which one does not affect the gravitational force between the 1 point

Zinaida [17]

Answer:

D. The net electrical charge of the Earth.

Explanation:

All the other three choices affect the gravitational force. I remember from middle school.

6 0

2 years ago

A truck is traveling at 27 m/s down the interstate highway where you are changing a flat tire. frequency of 185 Hz.

Reil [10]

Answer:

(a) the observed frequency is 200 Hz

(b) the observed frequency is 188 Hz.

Explanation:

speed of the truck, Vs = 27 m/s

frequency of the truck as it approaches, Fs = 185 Hz

(a) Apply Doppler effect to determine the frequency you will hear.

As the truck approaches you, the observed frequency will be higher than the source frequency because of decrease in distance.

$F_s = F_o [\frac{V}{V_S + V} ]$

Where;

Fo is the observed frequency which is the frequency you will hear.

V is speed of sound in air

$F_s = F_o [\frac{V}{V_S + V} ]\\\\185 = F_o [\frac{340}{27 + 340} ]\\\\185 = F_o (0.926)\\\\F_o = \frac{185}{0.926}\\\\F_o = 199.78 \ Hz$

$F_o = 200 \ Hz$

(b) Apply the following formula for a moving observer and a moving source;

$F_o = F_s[\frac{V-V_o}{V} ](\frac{V}{V-V_S} )$

The observed frequency is negative since you are driving away from the truck and the source frequency is also negative since it is driving towards you.

$F_o = F_s[\frac{V-V_o}{V} ](\frac{V}{V-V_S} )\\\\F_o = 185[\frac{340-22}{340} ](\frac{340}{340-27} )\\\\F_o = 185(0.9353)(1.0863)\\\\F_o = 188 \ Hz$

5 0

2 years ago