A protein molecule in an electrophoresis gel has a negative charge. The exact charge depends on the pHpH of the solution, but 30 excess electrons is typical. What is the magnitude of the electric forceon a protein with this charge in a 1500 N/C electric field?
Answer:
The magnitude of the force = 7.2 × 10⁻¹⁵ C
Explanation:
The total quantization of charge q on an electron = n × e
where;
n = 30
e = 1.6 × 10⁻¹⁸ C
q = 30 × 1.6 × 10⁻¹⁸ C
q = 4.8 × 10⁻¹⁸ C
Now, the magnitude of the force is determined by using the formula:
F = qE
F = ( 4.8 × 10⁻¹⁸ C) ( 1500 N/C)
F = 7.2 × 10⁻¹⁵ C
A safety plug is designed to melt when the pressure inside a metal tank becomes too high. A gas
at 51.0 atm and a temperature of 23.0°C is contained in the tank, but the plug melts when the
pressure reaches 75.0 atm. What temperature did the gas reach?
Do it in order.
from smallest to largest
Answer:
The earth, The sun, the solar system and the milky way.
PLEASEEEEEE HELPPPPPPP
Define resistance and discuss how it affects current.
Answer:
Resistance is the opposing of the flow of current through a conductor.
An Egyptian pyramid contains approximately 1.95 million stone blocks. The average weight of each block is 2.55 tons. What is the weight of the pyramid in pounds?
Answer:
More than 2,300,000 limestone and granite blocks were pushed, pulled, and dragged into place on the Great Pyramid. The average weight of a block is about 2.3 metric tons (2.5 tons).
Assuming no friction, how does the initial gravitational potential energy of
the marble on a downward slope compare to the final kinetic energy?
a) they are the same
b) the initial gravitational potential energy is greater than the final kinetic energy
c) the initial gravitational potential energy is less then the final kinetic energy
Answer:
a) They are the same.
Explanation:
Assuming no friction, there should be no energy transfer and thus the Law of Conservation of Energy says:
[tex]PE=KE,\\mgh=\frac{1}{2}mv^2[/tex]
These types of problems also disregard any air resistance the surface of the object may cause. Therefore, no energy is transferred and from the Law of Conservation of Energy, [tex]100\%[/tex] of energy is preserved.
Artificial satellites in space can help you find locations on
Earth. True or false?
calculate the average speed of talias car during the trip
Answer:
We're no strangers to love
You know the rules and so do I
A full commitment's what I'm thinking of
You wouldn't get this from any other guy
I just wanna tell you how I'm feeling
Gotta make you understand
Never gonna give you up
Never gonna let you down
Never gonna run around and desert you
Never gonna make you cry
Never gonna say goodbye
Never gonna tell a lie and hurt you
We've known each other for so long
Your heart's been aching but you're too shy to say it
Inside we both know what's been going on
We know the game and we're gonna play it
And if you ask me how I'm feeling
Don't tell me you're too blind to see
Never gonna give you up
Never gonna let you down
Never gonna run around and desert you
Never gonna make you cry
Never gonna say goodbye
Never gonna tell a lie and hurt you
No, I'm never gonna give you up
No, I'm never gonna let you down
No, I'll never run around and hurt you
Never, ever desert you
We've known each other for so long
Your heart's been aching but
Never gonna give you up
Never gonna let you down
Never gonna run around and desert you
Never gonna make you cry
Never gonna say goodbye
Never gonna tell a lie and hurt you
No, I'm never gonna give you up
No, I'm never gonna let you down
No, I'll never run around and hurt you
I'll never, ever desert you
Explanation:
RICK ROLLED
At a certain instant, a ball is thrown downward with a velocity of 8.0 m/s from a height of 40 m. At the same instant, another ball is thrown upward from ground level directly in line with the first ball with a velocity of 12 m/s. Find (a) the time when the balls collide and (b) the height at which they collide. Take g = 10 m/s2
Answer:
(a) The two balls collide [tex]2\; \rm s[/tex] after launch.
(b) The height of the collision is [tex]4\; \rm m[/tex].
(Assuming that air resistance is negligible.)
Explanation:
Let vector quantities (displacements, velocities, acceleration, etc.) that point upward be positive. Conversely, let vector quantities that point downward be negative.
The gravitational acceleration of the earth points dowards (towards the ground.) Therefore, the sign of [tex]g[/tex] should be negative. The question states that the magnitude of [tex]g\![/tex] is [tex]10\; \rm m \cdot s^{-2}[/tex]. Hence, the signed value of [tex]\! g[/tex] should be [tex]\left(-10\; \rm m \cdot s^{-2}\right)[/tex].
Similarly, the initial velocity of the ball thrown downwards should also be negative: [tex]\left(-8.0\; \rm m \cdot s^{-1}\right)[/tex].
On the other hand, the initial velocity of the ball thrown upwards should be positive: [tex]\left(12\; \rm m \cdot s^{-1}\right)[/tex].
Let [tex]v_0[/tex] and [tex]h_0[/tex] denote the initial velocity and height of one such ball. The following SUVAT equation gives the height of that ball at time [tex]t[/tex]:
[tex]\displaystyle h(t) = \frac{1}{2}\, g \cdot {t}^{2} + v_0 \cdot t + h_0[/tex].
For both balls, [tex]g = \left(-10\; \rm m \cdot s^{-2}\right)[/tex].
For the ball thrown downwards:
Initial velocity: [tex]v_0 = \left(-8.0\; \rm m \cdot s^{-1}\right)[/tex].Initial height: [tex]h_0 = 40\; \rm m[/tex].[tex]\displaystyle h(t) = -5\, t^{2} + (-8.0)\, t + 40[/tex] (where [tex]h[/tex] is in meters and [tex]t[/tex] is in seconds.)
Similarly, for the ball thrown upwards:
Initial velocity: [tex]v_0 = \left(12\; \rm m \cdot s^{-1}\right)[/tex].Initial height: [tex]h_0 = 0\; \rm m[/tex].[tex]\displaystyle h(t) = -5\, t^{2} + 12\, t[/tex] (where [tex]h[/tex] is in meters and [tex]t[/tex] is in seconds.)
Equate the two expressions and solve for [tex]t[/tex]:
[tex]-5\, t^{2} + (-8.0)\, t + 40 = -5\, t^{2} + 12\, t[/tex].
[tex]t = 2[/tex].
Therefore, the collision takes place [tex]2\, \rm s[/tex] after launch.
Substitute [tex]t = 2[/tex] into either of the two original expressions to find the height of collision:
[tex]h = -5\times 2^{2} + 12 \times 2 = 4\; \rm m[/tex].
In other words, the two balls collide when their height was [tex]4\; \rm m[/tex].
The time the two balls collide is 0.4 seconds while the height at which they collide is 4m
The given parameters are :
Initial Velocity U = 8m/s
Height H = 40m
For the second ball, the initial velocity = 12m/s
a.) For the first ball, the height attained at the point of collision will be
h = ut + 1/2gt^2
h = 8t + 1/2 x 10t^2 ........ (1)
For the second ball, the height attained at the point of collision will be
h = 12t - 1/2 x 10t^2 .........(2)
Since the height will be the same for the two balls, equate the two equations
8t + 10t^2 = 12t - 10t^2
Collect the like term
8t - 12t = -5t^2 - 5t^2
-4t = -10^2
10t = 4
t = 4/10
t = 0.4s
b.) Substitute time t in any of the equation to find the height
h = 12(0.4) - 0.5 x 10(0.4)^2
h = 4.8 - 0.8
h = 4m
Therefore, the time the two balls collide is 0.4 seconds while the height at which they collide is 4m
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Plz help this is so confusing
Answer:
5 Km/h
Explanation:
From the question given above, the following data were obtained:
Distance travelled = 10 Km
Time = 2 hours
Speed =?
Speed is simply defined as the distance travelled per unit time. Mathematically, it can be represented as:
Speed = distance travelled /time.
With the above formula, we can obtain the speed at which the duck is travelling as follow:
Distance travelled = 10 Km
Time = 2 hours
Speed =?
Speed = distance travelled /time.
Speed = 10 / 2
Speed = 5 Km/h
Thus, the duck is travelling at a speed of 5 Km/h
Waves in the ocean are tearing apart the shoreline. Which of the following two Earth Systems are interacting with each other.
Answer:
the Indian Ocean on 26 December 2004. This event claimed 227,898 dead and missing from 14 countries. The difference in mortality rates between these tsunamis reflects, in part, the benefits of understanding how tsunami waves are generated and move, and educating citizens to make scientifically
sound and potentially life-saving decisions.
A tsunami is a series of rapidly propagating, shallow-water ocean waves that develops when a submarine earthquake, landslide, or volcanic eruption displaces a large volume of water. Powerful earthquakes, with magnitudes of 9 or greater, caused both the 2004 and 2011 tsunamis. The earthquakes resulted from the movement of large tectonic plates. The 11 March 2011 earthquake occurred at 32 km (20
mi.) deep in Earth’s crust about 130 km (81 mi.) east of the city of Sendai. This location is on the boundary between two tectonic plates—the Pacific plate to the east and North American plate to the west. This
boundary fractured, releasing energy that was transmitted through the rocks and elevated portions of the
ocean floor. This drastic movement transmitted energy to the overlying ocean water, which generated
tsunami waves that radiated outward. The waves washed over the nearby coastlines and were felt around
the globe within hours (Figure 1.1).
Explanation:
Answer:
I believe Geosphere (lithosphere) and Hydrosphere
Explanation:
I hope it's right if not please notify me.
Which action will leave the dump trucks inertia unchanged?? PLEASE ANSWER FAST!!!
A. add gas
B. increase force applied to engine
Answer:
B.
Explanation:
Starting from the front door of your ranch house, you walk 50.0 m due east to your windmill, and then you turn around and slowly walk 30.0 m west to a bench where you sit and watch the sunrise. It takes you 27.0 s to walk from your house to the windmill and then 47.0 s to walk from the windmill to the bench. For the entire trip from the front door to the bench, what are your :
a. average velocity
b. average speed
Answer:
Explanation:
Total displacement for entire trip = final position - initial position
= 50 m - 30 m = 20 m
Total time = 27 + 47 = 74 s
Average velocity = Total displacement / total time
= 20 / 74 = .27 m /s
Total distance covered in entire trip = 50 + 30 = 80 m
Total time = 74 s
Average speed = Total distance covered / total time
= 80 / 74 = 1.08 m /s .
what are ribosomes?
I'm tired. But I have insomnia. Big ugh moment. <.<.
Answer:
Ribosomes are organelles the make protein for the cell.
In the laboratory, a ball is dropped onto a force-sensing platform several times, each time hitting a different surface (foam, feathers, clay, etc.). The momentum of the ball changes by the same amount in each trial; in each trial, the average scale reading is F, and the time of collision t are measured. What quantities would need to be graphed to exhibit a straight-line relationship
Answer:
Graphing the momentum against the change in moment yields a linear relationship.
Explanation:
This is an impulse experiment,
I = ∫ F .dt
where the force and time of the collision are measured, therefore if we assume an average force the integral reduces to
I = F t
Furthermore, the momentum is equal to the change in moment of the ball, this change in moment can be found using the energy relations measuring the height of the ball and calculating its speed, in the two intervals for the descent and for the exit, possibly the heights are different so the moment change is different from zero.
Starting point. Higher
Em₀ = U = mgh
Lower end point, just before hitting the scale
[tex]Em_{f}[/tex] = K = ½ m v²
in the path in the air there is no friction
Em₀ = Em_{f}
m g h = ½ m v²
v = [tex]\sqrt{2gh}[/tex]
this height is different for the descent and ascent of the ball, so we have two moments
Δp = [tex]p_{f}[/tex] - p₀
Δp = m (v_{f} -v₀)
therefore we have the relationship
I = Δp
Graphing the momentum against the change in moment yields a linear relationship.
In the equation for the gravitational force between two objects, which quantity must be squared?
•mi
•m2
•G
•d
Answer:
d
Explanation:
The quantity that must be squared in the equation of gravitational force is distance d.
According to the universal gravitational law, the square of the distance between two objects is inversely proportional to the force of gravity.
Therefore, the quantity to be squared is dThe formula is given as:
Fg = [tex]\frac{G m_{1} m_{2} }{d^{2} }[/tex]
So d is the quantity that must be squared
Learning task 2: Using the information you gathered from Learning Task 1, make a concept web of the contributions of the following scientist in the DEVELOPMENT OF MAGNETIC THEORY
A. Andre- Marie Ampere
B. Michael Faraday
C. Heinrich Herts
D. James Clerk Maxwell
E. Hans Christian Oersted
Answer:
The contributions of the following scientist in the DEVELOPMENT OF MAGNETIC THEORY
James Clerk Maxwell Hans Christian OerstedExplanation:
George Green was the first personality to formulate a mathematical principle of magnetism and electricity and his system created the framework for the work of different scientists such as William Thomson, James Clerk Maxwell, and others. Magnetism is the power exercised by magnets when they drag or deflect each other. Magnetism is produced by the movement of electric charges.
The contributions of James Clerk Maxwell and Hans Christian Oersted, et al in the DEVELOPMENT OF MAGNETIC THEORY are as follows:
They discovered that the speed at which electromagnetic waves traveled was similar to that of lightThey proved that there was a proportional connection between electricity and magnetismAccording to the given question, we are asked to show the contributions which the aforementioned scientists had in the development of the magnetic theory.
As a result of this, we can see that James Maxwell first developed this theory in the nineteenth century and the theory was modified by other scientists who made the framework for the electrical system and magnetism.
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John runs 3 km north then walks 2 km south. What is his total distance traveled and displacement?
Answer:
the total distance is 5km and the displacement is 1km
Explanation:
The total distance would be the addition of John running both ways so 3 km, 2 km.
However since he only walked back from a distance of 3 km to 2 km, he would be displaced 1 km because displacement is more like the position from the original point.
Think about 2 km as a positive value for the first part of the question and a negative value for the second part.
A receiver catches a football on the 50.0 yard line and is tackled 5.42 seconds later on the 12 yard line. What
was the runner's average speed?
Answer:
7.01yard/sec
Explanation:
Given parameters:
Initial position = 50yard
Final position = 12yard
Time = 5.42s
Unknown:
Average speed of runner = ?
Solution:
To solve this problem;
Speed = [tex]\frac{distance}{time}[/tex]
Distance covered = Initial position - final position = 50 - 12 = 38yards
So;
Speed = [tex]\frac{38}{5.42}[/tex] = 7.01yard/sec
Precisely 1.00 s after the speeder passes, the police officer steps on the accelerator; if the police car's acceleration is 2.70 m/s2 , how much time passes after the police car is passed by a speeder and before the police car overtakes the speeder (assumed moving at constant speed)
Answer:
t= 16.75 s
Explanation:
We will solve this exercise using the kinematic expressions
corridor that goes at constant speed, suppose that its speed is v₁ = 20 m/s, it does not appear in the statement, we start counting the time when it passes the policeman.
x₁ = v₁ t
The policeman starts from rest, so his initial velocity is zero and he has an acceleration a = 2.70 m /s², to use the same time counter we take into account that the policeman left at = 1.00 s after passing the corridor
x₂ = v₀ (t-t₀) + ½ a (t-t₀)²
x₂ = ½ a (t-1)²
at the point where the two meet, the position must be the same
x₁ = x₂
v₁ t = ½ a (t-1)²
(t-1)² = [tex]\frac{2 v_1 t}{a}[/tex]
t² - 2t + 1 - \frac{2 v_1 t}{a} +1 = 0
t² - 2(1 + [tex]\frac{v_1}{a}[/tex]) t +1
let's we solve the second degree equation
t² - 2 ( 1 + [tex]\frac{20}{2.7}[/tex]) t + 1=0
t² - 16.81 t +1=0
t = [ 16.81 ± [tex]\sqrt{ 16.81^2 - 4 )}[/tex] ] /2
t = [16.81 ± 16.695]/2
t₁= 16.75 s
t2= 0.06 s
Time t₂ is less than the reaction time of humans, so the correct answer is the first time
t= 16.75 s
Some giant ocean waves have a wavelength of 25 m and travel at 6.5 m/s with a frequency of 0.26 HZ. What is the period of such a wave ?
Answer:
3.85s
Explanation:
Given parameters:
Wavelength = 25m
Velocity = 6.5m/s
Frequency = 0.26Hz
Unknown:
Period of the wave = ?
Solution:
The period of a wave is the inverse of the frequency of the wave.
Period = [tex]\frac{1}{frequency}[/tex]
Period = [tex]\frac{1}{0.26}[/tex] = 3.85s
There are two different isotopes; X and Y, both contain the same number of radioactive substances. If sample X
has a longer half-life than Y, compare their rate of radioactive decay.
O A. Rate does not depend on half-life
B. Both of their rates are equal
O C. X has a smaller rate than Y
O D. X has a greater rate than Y
Answer:
Half life refers to the time for 1/2 of the radioactive atoms to decay.
Suppose that X has a half life of 10 days and Y has a half life of 20 days
If both start out with 1000 radioactive atoms then after 20 days
X would have 250 radioactive atoms and Y would have 500 atoms
The rate of decay is greater for the shorter half life:
In the example given X must have the smaller rate of decay because it has a longer half life.
if an electric is not grounded, it is best to reach out and touch it to provide the ground
Answer:
No. Touching a live electric current is never a good idea.
Answer:
false you would electrocute yourself
Explanation:
!!!!!!!!!!! LOGICAL !!!!!!!!!
A 12-kg object is moving rightward with a constant velocity of 4 m/s. How much net force is required to keep the object moving with
the same speed and in the same direction?
Explain how momentum is determined and conserved.
ASAP!!
Explanation:
Momentum is conserved in the collision. Momentum is conserved for any interaction between two objects occurring in an isolated system.
An 88 kg person steps into a car of mass 2002 kg, causing it to sink 5.36 cm on itssprings. Assuming no damping, with what fre-quency will the car and passenger vibrate onthe springs? Answer in units of Hz. The acceleration of gravity is 9.81 m/s^2.
Answer:
The required frequency = 0.442 Hz
Explanation:
Frequency [tex]f = ( \dfrac{1}{2 \pi}) \omega[/tex]
where;
[tex]\omega = \sqrt{\dfrac{k}{m} }[/tex]
Then;
[tex]f = \Bigg ( \dfrac{1}{2 \pi} \Bigg ) \Bigg( \sqrt{\dfrac{k}{m} } \Bigg )[/tex]
However;
[tex]k = \dfrac{F}{x}[/tex] and;
mass [tex]m = m_{car } + m_{person}[/tex]
[tex]f = \Bigg ( \dfrac{1}{2 \pi} \Bigg ) \Bigg( \sqrt{\dfrac{\dfrac{F}{x}}{m_{car}+m_{person}} } \Bigg )[/tex]
[tex]f = \Bigg ( \dfrac{1}{2 \pi} \Bigg ) \Bigg( \sqrt{\dfrac{{F}}{x(m_{car}+m_{person})} } \Bigg )[/tex]
where;
[tex]F = m_{person}g[/tex]
Then;
[tex]f = \Bigg ( \dfrac{1}{2 \pi} \Bigg ) \Bigg( \sqrt{\dfrac{ {m_{person}g }}{x(m_{car}+m_{person})} } \Bigg )[/tex]
replacing the values;
[tex]f = \Bigg ( \dfrac{1}{2 \pi} \Bigg ) \Bigg( \sqrt{\dfrac{ {(88 \ kg)* (9.81 \ m/s^2) }}{(5.36 \times 10^{-2} \ m) (2002 \ kg +88 \ kg)} } \Bigg )[/tex]
[tex]\mathbf{f = 0.442 \ Hz}[/tex]
Q1. A man wants to install a surveillance mirror in his shop, which mirror should he use?(1)
a) Convex mirror
b) Concave mirror
c) Plane mirror
d) Both (a) and (b)
answer is convex mirror
Explanation:
A
Because convex mirror will provide maximum view
How do pulleys help move objects?
Pulleys are powerful simple machines. They can change the direction of a force, which can make it much easier for us to move something. If we want to lift an object that weighs 10 kilograms one meter high, we can lift it straight up or we can use a pulley, so we can pull down on one end to lift the object up.
Answer:
Pulleys are powerful simple machines. They can change the direction of power, which can make it much easier for us to move something. If we want to lift an object that weighs 10 kilograms one meter high, we can lift it straight up or use a pulley, so we can pull one end down and lift the object.
Explanation:
describe measurement in our daily life
A small rock is thrown vertically upward with a speed of 17.0m/s from the edge of the roof of a 26.0m tall building. The rock doesn't hit the building on its way back down and lands in the street below. Air resistance can be neglected.
Part A
What is the speed of the rock just before it hits the street?
Express your answer with the appropriate units.
Part B
How much time elapses from when the rock is thrown until it hits the street?
Express your answer with the appropriate units.
Answer:
A) v = 28.3 m/s
B) t = 4.64 s
Explanation:
A)
Assuming no other forces acting on the rock, since the accelerarion due to gravity close to the surface to the Earth can be taken as constant, we can use one of the kinematic equations in order to get first the maximum height (over the roof level) that the ball reaches:[tex]v_{f}^{2} - v_{o}^{2} = 2* g* \Delta h (1)[/tex]
Taking into account that at this point, the speed of the rock is just zero, this means vf=0 in (1), so replacing by the givens and solving for Δh, we get:[tex]\Delta h = \frac{-v_{o} ^{2}}{2*g} = \frac{-(17.0m/s)^{2} }{2*(-9.8m/s2)} = 14.8 m (2)[/tex]
So, we can use now the same equation, taking into account that the initial speed is zero (when it starts falling from the maximum height) and that the total vertical displacement is the distance between the roof level and the ground (26.0 m) plus the maximum height that we have just found in (2) , 14.8m:Δh = 26.0 m + 14. 8 m = 40.8 m (3)Replacing now in (1), we can solve for vf, as follows:[tex]v_{f} =\sqrt{2*g*\Delta h} = \sqrt{2*9.8m/s2*40.8m} = 28.3 m/s (4)[/tex]
B)
In order to find the total elapsed from when the rock is thrown until it hits the street, we can divide this time in two parts:1) Time elapsed from the the rock is thrown, until it reaches to its maximum height, when vf =02) Time elapsed from this point until it hits the street, with vo=0.For the first part, we can simply use the definition of acceleration (g in this case), making vf =0, as follows:[tex]v_{f} = v_{o} + a*\Delta t = v_{o} - g*\Delta t = 0 (5)[/tex]
Replacing by the givens in (5) and solving for Δt, we get:[tex]\Delta t = \frac{v_{o}}{g} = \frac{17.0m/s}{9.8m/s2} = 1.74 s (6)[/tex]
For the second part, since we know the total vertical displacement from (3), and that vo = 0 since it starts to fall, we can use the kinematic equation for displacement, as follows:[tex]\Delta h = \frac{1}{2} * g * t^{2} (7)[/tex]
Replacing by the givens and solving for t in (7), we get:[tex]t_{fall} =\sqrt{\frac{2*\Delta h}{g}} = \sqrt{\frac{2*40.8m}{9.8m/s2} } = 2.9 s (8)[/tex]
So, total time is just the sum of (6) and (8):t = 2.9 s + 1.74 s = 4.64 s