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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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What is the mass of an object if a force of 34 N produces an acceleration of 4.0 m/s squared​
mrs_skeptik [129]

Answer:

8.5 kg

Explanation:

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6 0
2 years ago
The design speed of a multilane highway is 60 mi/hr. What is the minimum stopping sight distance that should be provided on the
kicyunya [14]

Answer:

Part a: When the road is level, the minimum stopping sight distance is 563.36 ft.

Part b: When the road has a maximum grade of 4%, the minimum stopping sight distance is 528.19 ft.

Explanation:

Part a

When Road is Level

The stopping sight distance is given as

SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}

Here

  • SSD is the stopping sight distance which is to be calculated.
  • u is the speed which is given as 60 mi/hr
  • t is the perception-reaction time given as 2.5 sec.
  • a/g is the ratio of deceleration of the body w.r.t gravitational acceleration, it is estimated as 0.35.
  • G is the grade of the road, which is this case is 0 as the road is level

Substituting values

                              SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35-0)}\\SSD=220.5 +342.86 ft\\SSD=563.36 ft

So the minimum stopping sight distance is 563.36 ft.

Part b

When Road has a maximum grade of 4%

The stopping sight distance is given as

SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}

Here

  • SSD is the stopping sight distance which is to be calculated.
  • u is the speed which is given as 60 mi/hr
  • t is the perception-reaction time given as 2.5 sec.
  • a/g is the ratio of deceleration of the body w.r.t gravitational acceleration, it is estimated as 0.35.
  • G is the grade of the road, which is given as 4% now this can be either downgrade or upgrade

For upgrade of 4%, Substituting values

                              SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35+0.04)}\\SSD=220.5 +307.69 ft\\SSD=528.19 ft

<em>So the minimum stopping sight distance for a road with 4% upgrade is 528.19 ft.</em>

For downgrade of 4%, Substituting values

                              SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35-0.04)}\\SSD=220.5 +387.09 ft\\SSD=607.59ft

<em>So the minimum stopping sight distance for a road with 4% downgrade is 607.59 ft.</em>

As the minimum distance is required for the 4% grade road, so the solution is 528.19 ft.

3 0
3 years ago
Plz! How to solve this question? The answer is B.
LekaFEV [45]

dhvjmdsgyddsfjdbfydgbgdfvdfshfgsdbfhdf nfdfh htiu4ewr huirhfhwref

4 0
3 years ago
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