Answer:
a) F = 1.70 10⁻⁹N, F = 1.47 10⁻⁸ N,
b) * the electronegative repulsion, from the repulsion by quantum effects
Explanation:
a) The atraicione force comes from the electric force given by Coulomb's law,
F = [tex]k \frac{ q_1 q_2}{r^2}[/tex]
divalent atoms
In this case q = 2q₀ where qo is the charge of the electron -1,6 10⁻¹⁹ C and the separation is given
F = k q² / r²
F = [tex]2 \ 10^9 \ \frac{2 (1.6 \ 10^{-19} )^2}{ (0.52 10^{-9} )^2 }[/tex]
F = 1.70 10⁻⁹N
monovalent atoms
in this case the load is q = q₀
F = 2 \ 10^9 \ \frac{ (1.6 \ 10^{-19} )^2}{ (0.125 10^{-9} )^2 }
F = 1.47 10⁻⁸ N
b) repulsive forces come from various sources
* the electronegative repulsion of positive nuclei
* the electrostatic repulsion of the electrons when it comes to bringing the electron clouds closer together
* from the repulsion of electron clouds, by quantum effects
A student club is designing a trebuchet for launching a pumpkin into projectile motion. Based on an analysis of their design, they predict that the trajectory of the launched pumpkin will be parabolic and described by the equation y(x) =ax^2+bx where a=−8.0×10^−3m^−1, b=1.0(unitless), x is the horizontal position along the pumpkin trajectory and y is the vertical position along the trajectory. The students decide to continue their analysis to predict at what position the pumpkin will reach its maximum height and the value of the maximum height.
Required:
What is the derivative of the vertical position of the pumpkin trajectory with respect to its horizontal position?
Answer:
[tex]\frac{dy}{dx}=2ax+b[/tex]
Explanation:
We are given that
[tex]y(x)=ax^2+bx[/tex]
[tex]a=-8.0\times 10^{-3}m^{-1}[/tex]
[tex]b=1.0[/tex]
x=Horizontal position along the trajectory
y=Vertical position along the trajectory
We have to find the derivative of the vertical position of the pumpkin trajectory with respect to its horizontal position.
Differentiate the equation w.r.t x
[tex]\frac{dy}{dx}=2ax+b[/tex]
Hence, the derivative of the vertical position of the pumpkin trajectory with respect to its horizontal position is given by
[tex]\frac{dy}{dx}=2ax+b[/tex]
Two students hear the same sound and their eardrums receive the same power from the sound wave. The sound intensity at the eardrums of the first student is 0.58 W/m^2, while at the eardrums of the second student the sound intensity is 1.18 times greater. If the diameter of the second student’s eardrum is 1.1 cm, how much acoustic power, in microwatts, is striking each of his (and the other student’s) eardrums?
Answer:
W = 65.04 μW
Explanation:
For this exercise we will use the definition of sound intensity which is the power per unit area
I = W / A
In this case, they indicate that for the intensity of the first student is I1 = 0.58 W / m², and the intensity of the second student is
I₂ = 1.18 I₁
I₂ = 1.18 0.58
I₂ = 0.6844 W / m²
Ask the power
W = I A
the diameter of the size is 1.1 cm and the radius is r = 0.55 cm = 0.55 10⁻² m, the approximate eardrum area as a circle
A = π r²
we substitute
W = I π r²
let's calculate
W = 0.6844 π (0.55 10⁻²)²
W = 6.504 10⁻⁵ W
they ask us to reduce to microwatts
W = 6.504 10⁻⁵ W (106 μW / 1W)
W = 65.04 μW
the power is constant, what changes is the intensity that depends on the sensor area, therefore the two students review the same power
Acoustic power, in microwatt W is 65.04 μW
We know that;
I = W / a
In case 1;
I1 = 0.58 W / m²
So, I₂ = 1.18 I₁
I₂ = 1.18 × 0.58
I₂ = 0.6844 W / m²
Diameter of the size = 1.1 cm
So,
Radius r = 1.1/2 = 0.55 cm
Area A = πr²W = Iπr²
W = 0.6844(3.4)(0.55)²
W = 6.504 × 10⁻⁵ W
Acoustic power, in microwatt W = 65.04 μW
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Des Linden won the Boston marathon in 2018, becoming the first American woman to win since 1985. The harsh conditions (heavy rain and a headwind) led to a winning time of 2 hrs 39 minutes and 54 seconds. If she weighed 99 pounds, her average heart rate during the race was 170 beats per minute and her heart pumped 2.5 mLs of blood per kg body mass per beat, how much blood did her heart pump during the race?
Answer:
3050.6 Litre .
Explanation:
Total time of heart beat = Total time of race = 2 hrs , 39 minutes and 54 seconds
= 2 x 60 + 39 + 54/60 min
= 120 + 39 + .9 min
= 159.9 min
rate of heart beat = 170 per min
Total no of heart beat during race = 170 x 159.9
volume of blood per kg per beat = 2.5 mL per kg of weight
body weight = 99 pounds = .4535 x 99 kg = 44.89 kg
volume of blood per beat = 2.5 mL x 44.89 mL
= 112.225 mL .
Total required volume of blood = 112.225 x 170 x 159.9 mL
= 3050612 mL
= 3050.6 L.
Determine the volume of an object that has a mass of 455.6 g and a density of 19.3 g/cm3.
Answer:
V = 23.6062 cm³
General Formulas and Concepts:
Math
Pre-Algebra
Order of Operations: BPEMDAS
Brackets Parenthesis Exponents Multiplication Division Addition Subtraction Left to RightChemistry
Gas Laws
Density = mass / volumeExplanation:
Step 1: Define
Mass m = 455.6 g
Density D = 19.3 g/cm³
Step 2: Solve for V
Set up: 19.3 g/cm³ = 455.6 g / VIsolate V: V = 23.6062 cm³The pressure in car tires is often measured in pounds per square inch (lb/in.2lb/in.2), with the recommended pressure being in the range of 25 to 45 lb/in.2lb/in.2. Suppose a tire has a pressure of 25.5 lb/in.2lb/in.2 . Convert 25.5 lb/in.2lb/in.2 to its equivalent in atmospheres. Express the pressure numerically in atmospheres.
Answer:
25.5 pounds per square inch are equivalent to 1.735 atmospheres.
Explanation:
An atmosphere equals 14.695 pounds per square inch. We find the equivalent of given pressure in atmospheres by means of simple rule of three:
[tex]x = 25.5\,\frac{lb}{in^{2}} \times \frac{1\,atm}{14.695\,\frac{lb}{in^{2}} }[/tex]
[tex]x = 1.735\,atm[/tex]
25.5 pounds per square inch are equivalent to 1.735 atmospheres.
Chameleons catch insects with their tongues, which they can rapidly extend to great lengths. In a typical strike, the chameleon's tongue accelerates at a remarkable 290 m/s2 for 20 ms, then travels at constant speed for another 30 ms. During this total time of 50 ms, 1/20 of a second, how far does the tongue reach?
Answer:
0.232m
Explanation:
We are given that
Acceleration,[tex]a=290m/s^2[/tex]
[tex]t_1=20ms=20\times 10^{-3}s=0.02s[/tex]
[tex]1ms=10^{-3}s[/tex]
[tex]t_2=30ms=30\times 10^{-3}s=0.03s[/tex]
We have to find the distance reached by tongue.
We know that
[tex]s=ut+\frac{1}{2}at^2[/tex]
Initial speed, u=0
Using the formula
[tex]x_1=0+\frac{1}{2}(290)(0.02)^2[/tex]
[tex]x_1=0.058m[/tex]
[tex]v=u+at[/tex]
Using the formula
Speed after 0.02s
[tex]v=0+(290)(0.02)[/tex]
[tex]v=5.8m/s[/tex]
Distance=vt
By using the formula
Distance traveled in 0.03 s
[tex]x_2=5.8(0.03)=0.174m[/tex]
Total distance traveled in 50 ms
[tex]\Delta x=x_1+x_2=0.058+0.174[/tex]
=0.232m
A 0.45 kg football traveling at a speed of 22 m/s is caught by a stationary receiver. If the football comes to a rest in the receiver's arms, the magnitude of the impulse imparted to the receiver by the ball is
Answer:
0.9Ns
Explanation:
Impulse formula is expressed as;
Impulse = Ft = m(v-u)
Impulse = m(v-u)
m is the mass of football = 0.45kg
v is the final velocity = 22m/s
u is the initial velocity = 0m/s
Impulse = 0.45(22-0)
Impulse = 0.45 * 22
Impulse = 0.9Ns
Hence the magnitude of the impulse imparted to the receiver by the ball is 0.9Ns
3. Most horoscopes are off by two astrological signs due to the past 2000 years of the Earth's wobble.
what is surface tension
Answer:
the tension of the surface film of a liquid caused by the attraction of the particles in the surface layer by the bulk of the liquid, which tends to minimize surface area.
Explanation:
A 25,000 kg jet is sitting still on the runway (O
m/s). If the jet goes to take off and reaches 80
m/s in 10 seconds, how much force was required
to get it there?
Answer:
The force required to get it there is 200000 N.
Explanation:
The force can be calculated by the second Newton's law:
[tex] F = ma [/tex]
Where:
m: is the mass = 25000 kg
a: is the acceleration
The acceleration is given by:
[tex] a = \frac{v}{t} [/tex]
Where:
v: is the velocity = 80 m/s
t: is the time = 10 s
[tex] a = \frac{v}{t} = \frac{80 m/s}{10 s} = 8 m/s^{2} [/tex]
Hence, the force is:
[tex] F = ma = 25000 kg*8 m/s^{2} = 200000 N [/tex]
Therefore, the force required to get it there is 200000 N.
I hope it helps you!
You are working in the movie industry. The director of a western film has come up with an impressive stunt and has asked you to assist with its setup. A cowboy sitting on a tree limb is to drop vertically from rest onto his galloping horse as it passes under the limb. The horse gallops at a constant rate of 13.0 m/s along a straight line and the vertical distance between the limb and the level of the saddle is 3.22 m. (Define the point at which the cowboy's bottom and the saddle meet as (x, y) = (0,0).)
Required:
You must advise the director as to the position of the horse along the line of its travel when the cowboy should begin his drop. What advice do you provide the director?
Answer:
x = 10.53 m
Explanation:
Let's analyze this problem a bit, the time that the cowboy must take to fall must be the time that the horse takes to arrive
Let's start by looking for the cowboy's time, which starts from rest and the point where the chair is is y = 0
y =y₀ + v₀ t - ½ g t²
0 = y₀ - ½ g t²
t = [tex]\sqrt{ \frac{2 y_o}{g} }[/tex]
we calculate
t = √(2 3.22 / 9.8)
t = 0.81 s
the horse goes at a constant speed
x = [tex]v_{c}[/tex] t
x = 13 0.81
x = 10.53 m
this is the distance where the horse should be when in cowboy it is left Cartesian
The position of the horse from the cowboy must be 10.53 m apart.
The time taken by the cowboy to fall must be equal to the time that the horse takes to arrive at the point where the cowboy falls.
We can calculate the time taken by the cowboy to reach the saddle by applying the second equation of motion
h = ut + (1/2)gt²
here h =3.22m , u = 0m/s as he starts from rest
[tex]t=\sqrt[]{\frac{2h}{g} }\\ t=\sqrt{\frac{2*3.22}{9.8} } [/tex]
t = 0.81 s
the horse goes at a constant speed
x = vt , given that v = 13 m/s
x = 13 × 0.81
x = 10.53 m is the distance where the horse should be when the cowboy jumps.
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Julie blows a bubble. At first, the pressure of the gas in the bubble is 4kPa. The bubble floats into the air and expands. When it gets to the top of a tree the bubble has a pressure of 2kPa and a volume of 3.5m³. Assuming a constant temperature, what was the volume in m³ of the bubble when it was first blown?
Answer:
V₁ = 1.75 m³
Explanation:
Assuming the gas to be an ideal gas. At constant temperature, the relationship between the volume and temperature of an ideal gas is given by Boyle's Law as follows:
[tex]P_{1}V_{1} = P_{2}V_{2}[/tex]
where,
P₁ = Initial Pressure of the Gas = 4 KPa
V₁ = Initial Volume of the Gas = ?
P₂ = Final Pressure of the Gas = 2 KPa
V₂ = Final Volume of the Gas = 3.5 m³
Therefore,
[tex](4\ KPa)V_{1} = (2\ KPa)(3.5\ m^{3})\\\\V_{1}=\frac{2\ KPa}{4\ KPa}(3.5\ m^{3})\\\\[/tex]
V₁ = 1.75 m³
A thin nonconducting spherical shell of radius R1 carries a total charge q1 that is uniformly distributed on its surface. A second, larger thin non-conducting spherical shell of radius R2 that is concentric with the first carries a charge q2 that is uniformly distributed on its surface.
Required:
a. Use Gauss's law to find the electric field in the regions r < R1, R1 < r < R2, and r > R2.
b. What should the ratio of the charges q1/q2 and their relative signs be for the electric field to be zero for r > R2?
Answer:
Explanation:
a )
in the regions r < R₁
charge q inside sphere of radius R₁ = 0
Applying gauss's law for electric field E at distance r <R₁
electric flux through Gaussian surface of radius r = 4π r² E
4π r² E = q / ε₀ = 0 / ε₀
E = 0
Applying gauss's law for electric field E at distance R₁ < r < R₂ .
charge q inside sphere of radius R₁ = q₁
Applying gauss's law for electric field E at distance R₁ < r < R₂
electric flux through Gaussian surface of radius r = 4π r² E
4π r² E = q₁ / ε₀
E = q₁ / 4πε₀
in the regions r> R₂
charge q inside sphere of radius R₂ = (q₁ + q₂)
Applying gauss's law for electric field E at distance r > R₂
electric flux through Gaussian surface of radius r = 4π r² E
4π r² E = (q₁ + q₂) / ε₀
E = (q₁ + q₂) /4π ε₀
b )
For electric flux to be zero at r > R₂
(q₁ + q₂) /4π ε₀ = 0
q₁ + q₂ = 0
q₁ / q₂ = - 1 .
(a) The value electric field in the regions r < R1, R1 < r < R2, and r > R2 will be [tex]\frac{ (q_1 + q_2)}{4\pi\varepsilon_0}[/tex]
(b)The ratio of the charges q1/q2 will be -1.
What is gauss law?The charge contained divided by the permittivity equals the total electric flux out of a closed surface, according to Gauss Law.
The electric flux in a given area is calculated by multiplying the electric field by the area of the surface projected in a plane parallel to the field.
Let's analyse regions r < R₁
charge q inside sphere will be zero.
According to gauss's law for electric field E at distance r <R₁
electric flux for the gaussian surface = 4π r² E
[tex]4\pi r^2E=\frac{Q}{\epsilon_0} =0[/tex]
By the applications of gauss's law for electric field E at distance R₁ < r < R₂ .
q is charge inside sphere having radius R₁ = q₁
By the applications of gauss's law for electric field E at distance R₁ < r < R₂
electric flux obtained due to Gaussian surface of radius r = 4π r² E
[tex]4\pi r^2 E =\frac{q}{\epsilon_0} \\\\\rm E=\frac{q_1}{\epsilon_0}[/tex]
For the regions r> R₂
q is the charge inside sphere of radius hyaving R₂ = (q₁ + q₂)
By the application of gauss's law for electric field E which is at distance r > R₂
Electric flux through Gaussian surface of radius r = 4π r² E
[tex]4\pi r^2 E =\frac{q}{\epsilon_0} \\\\\rm E=\frac{q_1+q_2}{\epsilon_0}[/tex]
Hence the value electric field in the regions r < R1, R1 < r < R2, and r > R2 will be [tex]\frac{ (q_1 + q_2)}{4\pi\varepsilon_0}[/tex].
(b) To finding the ratio of charges the value of electric flux must be zero at r > R₂
[tex]\frac{(q_1+ q_2) }{4\pi\epsilon_0} =0\\\\q_1 + q_2= 0\\\\\frac{q_1}{q_2} =-1[/tex]
Hence the ratio of the charges q1/q2 will be -1.
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If a tiger wood hits a 0.050 kg golf all giving it a speed of 75m/s what impulse does he impart to the ball
Answer:
impulse = change in linear momentum
3.75kgm/s
Explanation:
initial momentum is 0 because at the start golf ball is at rest
and our final momentum 0.050kg × 75m/s = 3.75kgm/s
momentum final - momentum initial
3.75kgm/s - 0
3.75kgm/s
Match the type of heat transfer with its description
Answer:
1. Convection
2. Radiation
3. Conduction
Hope this helps!
Explanation:
Please help I need finish by 10:57
Answer:
to answer it just get the distance traveled (m) and divide it by the time it took
Explanation:
so for the first one just do 100 divided by 47 which would equal 2.12 m/s
Lolliguncula brevis squid use a form of jet propulsion to swim -- they eject water out of jets that can point in different directions, allowing them to change direction quickly. When swimming at a speed of 0.15 m/s or greater, they can accelerate at 1.2 m/s^2.
Required:
a. Determine the time interval needed for a squid to increase its speed from 0.15m/s to 0.45m/s
b. What other questions can you answer using the data?
Answer:
a) the required time interval is 0.25 s
b) the distance traveled while accelerating is 0.075 m
Explanation:
Given that;
Initial speed of the squid u = 0.15 m/s
final speed of the squid v = 0.45 m/s
acceleration a = 1.2 m/s²
Time period T = ?
using equation of motion
v = u + at
solve for t
at = v - u
t = v - u / a
we substitute
t = ( 0.45 - 0.15 ) / 1.2
t = 0.3 / 1.2
t = 0.25 s
Therefore, the required time interval is 0.25 s
b)
the other possible question is; Determine the distance the squid has traveled while accelerating.
using the following expression
v² = u² + 2as
we solve for distance s
2as = v²- u²
s = v²- u² / 2a
we substitute
s = ((0.45)²- 0.15)²) / 2×1.2
s = (0.2025 - 0.0225) / 2.4
s = 0.18 / 2.4
s = 0.075 m
Therefore, the distance traveled while accelerating is 0.075 m
A 10 kg remote control plane is flying at a height of 111 m. How much
potential energy does it have?
Answer:
10.88kJ
Explanation:
Given data
mass= 10kg
heigth= 111m
Applying
PE= mgh
assume g= 9.81m/s^2
substitute
PE= 10*9.81*111
PE=10889.1 Joules
PE=10.881kJ
Hence the potential energy is 10.88kJ
A 3500kg truck moving rightward with a speed of 7km/hr collides head on with a 1200kg car moving leftward with a speed of 11 km/hr. The two vehicles stick together and move with the same velocity after the collision. Determine the post-collision velocity of the car and truck.
Answer:
2.40 Km/hr.
Explanation:
From the question given above, the following data were obtained:
Mass of truck (m₁) = 3500 kg
Velocity of truck (u₁) = 7 Km/hr
Mass of car (m₂) = 1200 kg
Velocity of car (u₂) = 11 km/hr
Post collision velocity (v) =?
The post collision velocity of the truck and car can be obtained as illustrated below:
m₁u₁ – m₂u₂ = v(m₁ + m₂)
(3500 × 7) – (1200 × 11) = v(3500 + 1200)
24500 – 13200 = v × 4700
11300 = v × 4700
Divide both side by 4700
v = 11300 / 4700
v = 2.40 Km/hr.
Therefore, the post collision velocity of the truck and car is 2.40 Km/hr.
Is the formula for velocity the same as speed or different?
Answer:
always same
Explanation:
velocity and speed are same upto some extend but velocity is vector while speed is scalar quantity
Help! Il give brainlest to who answers first
Answer:
1. The density of the cube is 1.03 g/mL.
2. Dish soap
Explanation:
1. Determination of the density of the cube.
From the question given above, the following data were obtained:
Mass (m) of cube = 21.7 g
Volume (V) of cube = 21 mL
Density (D) of cube =?
The density of a substance is simply defined as the mass of the substance per unit volume of the substance. Thus, density can be expressed mathematically as:
Density (D) = mass (m) / volume (V)
D = m / V
With the above formula, we can obtain the density of the cube as follow:
Mass (m) of cube = 21.7 g
Volume (V) of cube = 21 mL
Density (D) of cube =?
D = m / V
D = 21.7 / 21
D = 1.03 g/mL
Thus, the density of the cube is 1.03 g/mL.
2. Determination of the layer of density the cube will settle in.
From the question given above,
Subtance >>>>>>>> Density
Vegetable oil >>>>> 0.91 g/mL
Grape juice >>>>>> 0.97 m/L
Water >>>>>>>>>>> 1 g/mL
Dish soap >>>>>>>> 1.03 g/mL
Maple syrup >>>>>> 1.37 g/mL
Comparing the density of the cube (i.e 1.03 g/mL) with those in the table able, we can conclude that the cube will settle in the DISH SOAP layer since they both have the same density.
Arial
How do Seasons Help us Predict the
Weather ?
Answer:
if its winter
Explanation:
then we know it will me cold
Explain how the extension of a spring is determined
Answer:
For a given spring the extension is directly proportional to the force applied For example if the force is doubled, the extension doubles When an elastic object is stretched beyond its limit of proportionality the object does not return to its original length when the force is removed
Explanation:
Conduct research to identify claims made about the effects that certain frequencies of EMR have when absorbed by matter. Find two examples from published sources. Write brief descriptions of your two examples.
Answer:
The effects that certain frequencies of EMR have when absorbed by matter is explained below in complete detail.
Explanation:
Electromagnetic radiation of distinct frequencies associates with material adversely. ... Gamma rays, though commonly of somewhat greater frequency than X rays have the equivalent creation. When the power of gamma rays is consumed in material, its influence is practically indistinguishable from the outcome generated by X rays.
Explain some similarities and differences between:
Magnetism and electrostatic forces
Answer:
Electric field points in the direction of the force experienced by a positive charge. Magnetic field points in the direction of the force experienced by a north pole.
Quick Check:
12-17-20
Conservation of Energy
In the use of an energy resource like coal, do you get all
the energy back in one form(100%)? (circle one)
Yes
No
Answer:
Quick Check:
12-17-20
Conservation of Energy
In the use of an energy resource like coal, do you get all
the energy back in one form(100%)? (circle one)
Yes
No
Explanation:
Not sure
PLEASE HELP
How much work is required to raise a 1200 kg rock 15 m?
Answer:
18000
Explanation:
the is to raise how much work is required is 18000
Work done to raise an object of mass 1200 kg to the height of 15 meters will be equal to 176,400 J.
What is Work?In physics, the word "work" involves the measurement of energy transfer that takes place when an item is moved over a range by an externally applied, at least a portion of which is applied within the direction of the displacement.
The length of the path is multiplied by the element of a force acting all along the path to calculate work if the force is constant. The work W is theoretically equivalent towards the force f times the length d, or W = fd, to portray this concept.
As per the data given in the question,
Mass of the object, m = 1200 kg
Displacement, d = 15 meters
Then, use the equation of work done,
W = fd
W = (m×a)d
Substitute the values in above equation,
W = 1200 × 9.8 × 15
W = 176,400 J.
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A motorboat traveling on a straight course slows
uniformly from 65 km/h to 35 km/h in a distance
of 45 m.
Part A
What is the magnitude of the boat's acceleration?
Answer:
2.572 m/s²
Explanation:
Convert the given initial velocity and final velocity rates to m/s:
65 km/h → 18.0556 m/s35 km/h → 9.72222 m/sThe motorboat's displacement is 45 m during this time.
We are trying to find the acceleration of the boat.
We have the variables v₀, v, a, and Δx. Find the constant acceleration equation that contains all four of these variables.
v² = v₀² + 2aΔxSubstitute the known values into the equation.
(9.72222)² = (18.0556)² + 2a(45) 94.52156173 = 326.0046914 + 90a-231.4831296 = 90aa = -2.572The magnitude of the boat's acceleration is |-2.572| = 2.572 m/s².
First aid for a strain or a sprain
A) the FITT formula.
B) ineffective.
C) the RICE formula. .
D) the ABC formula
Answer:
The correct answer is - C) the RICE formula.
Explanation:
The first aid for a patient with a sprain or strain mainly includes the RICE formula which stands for the rest, ice, compression, and elevation. In case of a sprain or strain, one is advised to rest, use ice on the location of the sprain, compression on the part of the sprain, and elevate the body part for less pain, less swelling or reducing chances of making it worst.
The other three formulas given are used in different conditions. The FITT formula used in the exercise planning, and the ABC formula used for the first aid of the heart attack or blockage of the airway.
Which units are used to measure both velocity and speed? choose three options.
Explanation:
third and fifth option along with the first one are right
