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ivolga24 [154]
3 years ago
13

A protein molecule in an electrophoresis gel has a negative charge.The exact charge depends on the pH of the solution, but 30 ex

cesselectrons is typical. What is the magnitude of the electric forceon a protein with this charge in a 1500 N/C electric field?
Physics
1 answer:
Ad libitum [116K]3 years ago
6 0

Answer:

The magnitude of the electric force on a protein with this charge is 7.2\times10^{-15}\ N

Explanation:

Given that,

Electric field = 1500 N/C

Charge = 30 e

We need to calculate the magnitude of the electric force on a protein with this charge

Using formula of electrostatic force

F=Eq

Where, F = force

E = electric field

q = charge

Put the value into the formula

F=1500\times30\times1.6\times10^{-19}

F=7.2\times10^{-15}\ N

Hence, The magnitude of the electric force on a protein with this charge is 7.2\times10^{-15}\ N

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You measure a watch's hour and minute hands to be 7.4 mm and 12.1 mm long, respectively. Part A In one day, by how much does the
pychu [463]

Answer:

109.385m

Explanation:

In 1 day, the hour hand travels 2 circles, or 4π rad in angular. The distance it travels is its angle times the radius

7.4 * 4π = 93 mm

In 1 day, the minute hand travels 24*60 = 1440 circles, or 1440 * 2π = 2880π rad in angular. The distance it travels is

12.1 * 2880π = 109478 mm

So the distance traveled by the tip of the minute hand that exceed the distance traveled by the tip of the hour hand is

109478 - 93 = 109385 mm or 109.385 m

3 0
3 years ago
1. Analyze the number line above to describe the motion of the object. Be sure to include starting position, intermediate positi
Colt1911 [192]

1) Distance: 9 m, displacement: +5 m

2) The average speed is 3 m/s

3) The average velocity is +2 m/s

Explanation:

1)

We start by reminding the definitions of distance and displacement:

- Distance: it is the total length of the path covered by an object, regardless of the direction of motion

- Displacement: is the distance in a straight line between the initial position and final position of motion

Let's now analyze the motion represented in the figure:

Initial position: x_0 = -1 m

Intermediate position: x_1 = +6 m

Final position: x_2 = +4 m

Therefore, the distance covered is

distance = |x_1-x_0|+|x_2-x_1|=|+6-(-1)|+|+4-(+6)|=7+2=9 m

While the displacement is:

displacement = x_2 - x_1 = +4 - (-1) = +5 m

2)

The average speed is a scalar quantity defined as

speed=\frac{d}{t}

where

d is the distance travelled

t is the time taken

For the object in this problem,

d = 6 m

t = 2 s

Substituting, we find the average speed:

speed = \frac{6 m}{2 s}=3 m/s

3)

The average velocity is a vector quantity defined as

velocity = \frac{d}{t}

where

d is the displacement

t is the time interval

For the object in this problem,

d = +12 m

t = 6 s

Substituting, we find the average velocity:

velocity=\frac{+12 m}{6s}=+2 m/s

Where the positive sign + indicates the direction of the velocity vector.

Learn more about average speed and average velocity:

brainly.com/question/8893949

brainly.com/question/5063905

#LearnwithBrainly

7 0
4 years ago
What is the smallest part of an element?<br> An atom
KatRina [158]

Answer:

poop

Explanation:

3 0
3 years ago
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How much heat is needed to melt 16.5kg of silver that is initially at 20*c?
zubka84 [21]
The Specific Heat Capacity of silver is 230 J/kgK, melting point is 961.8 C so the difference is 941.8K. Now we simply do q=230J/kgK*16.5kg*941.8K and that is 3 574 131 J
8 0
4 years ago
At a certain instant, a rotating turbine wheel of radius RR has angular speed ωω (measured in rad/srad/s). What must be the magn
jonny [76]

Answer:

Explanation:

The magnitude of the acceleration makes an angle of 30° with the tangential velocity.

Resolving the acceleration to tangential and radial acceleration

at = aCos30 = √3a/2

ar = aSin30 = ½a

a = 2•ar

Then, the tangential acceleration is the linear acceleration, so the relationship between the tangential acceleration and angular acceleration is given as:

at = Rα

Then, α = at/R

since at = √3a/2

Then, α = √3 at/2R, equation 1

The radial acceleration is given as

ar = ω²R

Note that, at² + ar² = a²

at = √(a²-ar²)

Back to equation 1

α = √3 at/2R

α = √3√(a²-ar²)/2R

α = √3√(a²-(w²R)²)/2R

α = √3(a²-w⁴R²) / 2R

Also, a = 2•ar = 2w²R

Then,

α = √3((2w²R)²-w⁴R²) / 2R

α = √3(4w⁴R²-w⁴R²) / 2R

α = √3(3w⁴R²) / 2R

α = √9w⁴R² / 2R

α = 3w²R / 2R

α = 3w²/2

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