The question is incomplete, the complete question is;
When an object with an electric charge of −7.0μC is 5.0cm from an object with an electric charge of 4.0μC, the force between them has a strength of 100.7N. Calculate the strength of the force between the two objects if they are 1.7cm apart. Round your answer to 2 significant digits
Answer:
865.1 N
Explanation:
F1 = Kq1q2/r1^2 ---------1
F2 = Kq1q2/r2^2 -------2
We have that;
r1 = 5cm
r2 =1.7 cm
F1 = 100.7 N
Comparing equations 1 and 2
F2 = F1r1^2/r2^2
F2 = 100.7N[(5cm)^2/(1.7cm)^2]
F2= 865.1 N
If a person weighs 140 lb'on Earth, their mass in kilograms is
Answer:
70 kg
Explanation:
divide it by 2
Hope this helped!
Answer:
63.502932 Kilograms
Explanation:
If there is "waste" energy, does the Law of Conservation of Energy still apply?
Explanation:
Yes, the law of conservation of energy still applies even if there is waste energy.
The waste energy are the transformation products of energy from one form to another.
According to the law of conservation of energy "energy is neither created nor destroyed by transformed from one form to another in a system".
But of then times, energy is lost as heat or sound within a system.
If we take into account these waste energy, we can see that energy is indeed conserved. The sum total of the energy generated and those produced will be the same if we factor in other forms in which the energy has been transformed into.The mass of 60 paper clips is 18.0 grams. What is the mass of one paper clip?
Answer:
3.333333333333333333333333333333333333333
Explanation:
3.3333333333333333333333333333333333
A 500 kg wrecking ball is knocking down a wall. When it is pulled back to its highest point, it is at a height of 6.2 m. When it hits the wall, it is moving at 3.1 m/s. How high is the wrecking ball when it hits the wall? (Show your work and follow all of the steps of the GUESS method. Check your answer after you submit the form - it's in the feedback for this question.) |
A sprinter starts from rest and accelerated at a rate of 0.16 m/s over a distance of 50.0 meters. How fast is the athletes traveling at the end of the 50.0 meters?
Answer:
40m/s
Explanation:
v²=u²+2as
v²=0²+2(16)(50)
v²=160v=40m/s
As waves crash into rock along the shoreline, particles of sand, shell, and other materials in the ocean water loosen tiny bits of sediment from the rock. As the waves recede, they carry the sediment away. In this scenario, which process represents weathering, and which process represents erosion?
Answer:
WEATHERING is represented by the scenario (As waves crash into rock along the shoreline, particles of sand, shell, and other materials in the ocean water loosen tiny bits of sediment from the rock).
Erosion is represented by the scenario (As the waves recede, they carry the sediment away).
Explanation:
A wave is a disturbance which travels through a medium and transfers energy from one point to another. When wind blows over a water body like the ocean, ocean waves are formed. As the generated energy from the wind is transported through the water by the waves, the can hit against rocks on the shores leading to its break down with time. WEATHERING occurs when tiny bit of sediments from rocks are loosened due to the impact of ocean waves.
Erosion can be described as the wearing away of the earth's surface due to the impact of wind, rainfall ( water) or waves. There are different types of erosion which is classified according it's cause of formation.
Wave erosion occurs when sediments such as sand, shell and other materials are carried to the shoreline by ocean waves. This erodes the shore over time as the sediments act like sandpapers.
An atom undergoes nuclear decay, but its atomic number is not changed.
What type of nuclear decay did the atom undergo?
A. Gamma decay
B. Beta decay
C. Nuclear fission
D. Alpha decay
Answer:
A. Gamma decay
Explanation:
A form of nuclear decay in which the atomic number is unchanged is a gamma decay.
The atom has undergone a gamma decay.
In a gamma decay, no changes occur to the mass and atomic number of the substance.
Gamma rays have zero atomic and mass numbers. When they cause decay, they cause no change to the mass and atomic numbers. They simply produce gamma rays during such reactions and these rays are very energetic.If there is "waste" energy, does the Law of Conservation of Energy still apply? please don't type something random if so i'll just report it.
Explanation:
Yes, the law of conservation of energy still applies even if there is waste energy.
The waste energy are the transformation products of energy from one form to another.
According to the law of conservation of energy "energy is neither created nor destroyed by transformed from one form to another in a system".
But of then times, energy is lost as heat or sound within a system.
If we take into account these waste energy, we can see that energy is indeed conserved. The sum total of the energy generated and those produced will be the same if we factor in other forms in which the energy has been transformed into.A solid sphere of radius R = 5 cm is made of non-conducting material and carries a total negative charge Q = -12 C. The charge is uniformly distributed throughout the interior of the sphere.
What is the magnitude of the electric potential V at a distance r = 30 cm from the center of the sphere, given that the potential is zero at r = [infinity] ?
Answer:
V= -3.6*10⁻¹¹ V
Explanation:
Since the charge is uniformly distributed, outside the sphere, the electric field is radial (due to symmetry), so applying Gauss' Law to a spherical surface at r= 30 cm, we can write the following expression:[tex]E* A = \frac{Q}{\epsilon_{0} } (1)[/tex]
At r= 0.3 m the spherical surface can be written as follows:[tex]A = 4*\pi *r^{2} = 4*\pi *(0.3m)^{2} (2)[/tex]
Replacing (2) in (1) and solving for E, we have:[tex]E = \frac{Q}{4*\pi *\epsilon_{0}*r^{2} } = \frac{(9e9N*m2/C2)*(-12C)}{(0.3m)^{2} y} (3)[/tex]
Since V is the work done on the charge by the field, per unit charge, in this case, V is simply:V = E. r (4)Replacing (3) in (4), we get:[tex]V =E*r = E*(0.3m) = \frac{(9e9N*m2/C2)*(-12C)}{(0.3m)} = -3.6e11 V (5)[/tex]
V = -3.6*10¹¹ Volts.The electrical potential module will be [tex]-3.6*10^-^1^1 V[/tex]
We can arrive at this answer as follows:
To answer this, we owe Gauss's law. This is because the charge is evenly distributed across the sphere. This will be done as follows:[tex]E*A=\frac{Q}{^E0} \\\\\\A=4*\pi*r^2[/tex]
Solving these equations will have:[tex]E=\frac{Q}{4*\pi*^E0*r^2} \\E= \frac{(9e9N*m2/c2)*(-12C)}{(0.3m)^2y}[/tex]
As we can see, the electric potential is carried out on the field charge. In this case, using the previous equations, we can calculate the value of V as follows:[tex]V=E*r\\V=E*0.3m= \frac{(9e9N*m^2/C2)*(-12C)}{0.3m} \\V= -3.6*10^-^1^1 V.[/tex]
More information about Gauss' law at the link:
https://brainly.com/question/14705081
A stone is dropped from the top of a high cliff with zero initial velocity. In which system is the net momentum zero as the stone falls freely
Answer:
A system that includes the stone and the earth.
Explanation:
If the system of being dropped from the height of the cliff consists of just the stone alone, then it means that its momentum will certainly undergo changes as it falls freely. However, If the system is now expanded to include not only the stone but also the Earth, then it implies that the momentum of the stone which is in the downward direction will be equal and opposite to the momentum of the Earth in the upwards direction towards the stone. Therefore, the momentum will cancel out and net momentum will be zero.
A system of stone and earth can result to a net zero momentum.
Conservation of linear momentum
The principle of conservation of linear momentum states that the sum of the initial momentum is equal to the sum of final momentum.
[tex]m_1u_1 + m_2 u_2 = m_1v_1 + m_2 v_2[/tex]
A system that consists a linear system of stone and earth can result to a net zero momentum.
Thus, a system of stone and earth can result to a net zero momentum.
Learn more about conservation of momentum here: https://brainly.com/question/7538238
Which of the physical variables listed below will change when you change the area of the capacitor plates (while keeping the battery connected).
a. Capacitance
b. Charge on the plates
c. Voltage across the plates
d. Net electric field between the plates
e. Energy stored in the capacitor
Answer:
a. Capacitance
b. Charge on the plates
e. Energy stored in the capacitor
Explanation:
Let A be the area of the capacitor plate
The capacitance of a capacitor is given as;
[tex]C = \frac{Q}{V} = \frac{\epsilon _0 A}{d} \\\\[/tex]
where;
V is the potential difference between the plates
The charge on the plates is given as;
[tex]Q = \frac{V\epsilon _0 A}{d}[/tex]
The energy stored in the capacitor is given as;
[tex]E = \frac{1}{2} CV^2\\\\E = \frac{1}{2} (\frac{\epsilon _0 A}{d} )V^2[/tex]
Thus, the physical variables listed that will change include;
a. Capacitance
b. Charge on the plates
e. Energy stored in the capacitor
When the bowling ball has fallen halfway down the building (height = 20 m), it has a speed of 19.8 m/s.
How much potential energy does the bowling ball have?
How much kinetic energy does the bowling ball have?
How much total energy (potential + kinetic) does the bowling ball have?
Of the bowling ball’s total energy, is more in the form of potential or kinetic energy?
Answer:
I think the answer is 19.8 potential energy
Explanation:
NONE.
A neutral metal bob is hanging on the bottom of a pendulum that is 15 cm long. A charged balloon is held near the metal bob and the pendulum is pulled up to a vertical angle of 20-deg. If the mass of the metal bob is 0.025kg, what is the charge on the balloon.
Answer:
Explanation:
See the figure attached
F is electrostatic force .
T cos20 = mg
T sin20 = F
Tan20 = F / mg
F = mg tan 20 = .025 x 9.8 tan20
= .09 N
Distance between bob and balloon
= 15 sin20 = 5.1 cm = .051 m
If q be the charge on balloon
F = 9 x 10⁹ x q² / .051²
= 3460 x 10⁹ q² = .09
q² = 26 x 10⁻⁶ x 10⁻⁹
q = 16.12 x 10⁻⁸ C .
Two metal bricks are held off the edge of a balcony from the same height above the ground. The bricks are the same size but one is made of Titanium (density of 4.5 g/cm%) and one is made of Lead (density of 11.3 g/cm3) so the Lead is about twice as heavy as the Titanium. The time it takes the bricks to reach the ground will be:________.
a. less but not necessarily half as long for the heavier brick
b. about half as long for the lighter brick
c. less but not necessarily half as long for the lighter brick
d. about half as long for the heavier brick
e. about the same time for both bricks
Answer:
e.
Explanation:
Assuming that the air resistance is neglectable, both bricks are only accelerated by gravity, which produces a constant acceleration on both bricks, which is the same, according Newton's 2nd Law, as we can see below:[tex]F_{g} = m*g = m*a (1)[/tex]⇒a = g = 9.8m/s² (pointing downward)Since acceleration is constant, if both fall from the same height, we can apply the following kinematic equation:[tex]\Delta y = v_{o} * t - \frac{1}{2} *g*t^{2} (2)[/tex]
Since both bricks are held off the edge, the initial speed is zero, so (2) reduces to the following equation:[tex]h =\frac{1}{2} *g*t^{2} (3)[/tex]
Since h (the height of the balcony) is the same, we conclude that both bricks hit ground at exactly the same time.If the air resistance is not negligible, due both bricks have zero initial speed, and have the same shape, they will be affected by the drag force in similar way, so they will reach the ground at approximately the same time.A long, straight wire carries a current of 5.20 A. An electron is traveling in the vicinity of the wire. At the instant when the electron is 4.40 cm from the wire and traveling at a speed of 6.20 * 104 m>s directly toward the wire, what are the magnitude and direction (relative to the direction of the current) of the force that the magnetic field of the current exerts on the electron
Answer:
Explanation:
Magnetic field due to current at a distance of 4.4 cm
B = 10⁻⁷ x 2 x 5.2 / 4.4 x 10⁻² [ B = 10⁻⁷ x 2i / r = ]
= 2.36 x 10⁻⁵ T.
Force on moving electron = Bqv , B is magnetic field , q is charge and v is velocity of charge .
Force = 2.36 x 10⁻⁵ x 1.6 x 10⁻¹⁹ x 6.2 x 10⁴
= 23.41 x 10⁻²⁰ N .
This force will be perpendicular to the direction of current .
A heavy book is launched horizontally out a window from the first floor, a height, h, above the ground, with initial velocity, v0, and it hits the ground a horizontal distance X1 away from the window. Another book is similarly launched (same initial velocity) from the second floor window, a height 2h above the ground. Where does the second book land relative to the first book
Answer:
x₂ / x₁ = √2
Explanation:
To solve this exercise we can use the projectile launch ratios, let's find the time it takes for the second book to reach the ground
y = y₀ + [tex]v_{oy}[/tex] t - ½ g t²
as the book is thrown horizontally v_{oy} = 0, when it reaches the ground its height is zero y= 0
0 = y₀ - ½ g t²
t = [tex]\sqrt{ \frac{2y_o}{ g} }[/tex]
t = \sqrt{ \frac{2 \ 2h}{ g} }
with this time we calculate the horizontal distance traveled
x = v₀ t
x₂ = v₀ [tex]\sqrt{ \frac{4h}{g} }[/tex]
now let's calculate the time it takes him to get to the floor when he leaves from the first floor
t =\sqrt{ \frac{2y_o}{ g} }
the horizontal distance traveled is
x₁ = v₀ [tex]\sqrt{ \frac{2h}{g} }[/tex]
therefore the difference in distance between the two runs is
Δx = x₂-x₁
Δx = v₀ \sqrt{ \frac{4h}{g} } - v₀ \sqrt{ \frac{2h}{g} }
Δx = v₀ \sqrt{ \frac{2h}{g} } √2
Δx =√2 x₁
the relationship between the two distances is
x₂ / x₁ = √2
A cylindrical tank of radius R, filled to the top with a liquid, has a small hole in the side, of radius r, at distance d below the surface. Find an expression for the volume flow rate through the hole. A 4.0-mm-diameter hole is 1.0 m below the surface of a 2.0-m-diameter tank of water. What is the rate, in mm/min, at which the water level will initially drop if the water is not replenished?
Answer:
a)[tex]V=\pi *r^2 * \sqrt{2gd}[/tex]
b)[tex]dh / dt = 0.2658 mm / min[/tex]
Explanation:
From the question we are told that
Diameter of hole [tex]d_h=4mm=>0.004m[/tex]
Depth of hole [tex]D=0mm=>0.001m[/tex]
Diameter of tank [tex]d_t=2mm=>0.002m[/tex]
Generally the equation for pressure is mathematically given as
[tex]Pressure P= \rho*g*d[/tex]
[tex]P= 1/2*\rho *v^2[/tex]
Where
[tex]v = \sqrt {2gd}[/tex]
[tex]V = Area*v[/tex]
[tex]V=\pi *r^2 * \sqrt{2gd}[/tex]
Generally the level at which the water level will initially drop if the water is not replenished is mathematically given by
[tex]dh / dt = (r/R)^2 *sqrt{2gd}\\dh / dt = (2/2000)^2 *sqrt(2*9.81*1) \\dh / dt = 4.429*10^-3 mm/s \\[/tex]
Therefore the level at which the water level will initially drop if the water is not replenished
[tex]dh / dt = 0.2658 mm / min[/tex]
The rate, in mm/min, at which the water level will initially drop will be 1.0625 mm/min.
Given data:
The diameter of hole is, d = 4.0 mm = 0.004 m.
The depth of hole is, h = 1.0 m.
The diameter of tank is, d' = 2.0 m.
The given problem is based on the flow rate, which is defined as the flow of liquid through a given section per unit time.
Let us first obtain the equation of pressure as,
[tex]P=\dfrac{1}{2} \times \rho \times v^{2}[/tex]
Here, v is the velocity of efflux and its value is,
[tex]v=\sqrt{2gh} \\\\v^{2}=2gh[/tex]
And the level at which the water level will initially drop if the water is not replenished is mathematically given by,
[tex]\dfrac{dH}{dt}=(r/R)^{2} \times v[/tex]
Here,
r is the radius of hole.
R is the radius of tank.
Solving as,
[tex]\dfrac{dH}{dt}=((d/2) /(d'/2))^{2} \times \sqrt{2gh} \\\\\dfrac{dH}{dt}=((0.004/2) /(2/2))^{2} \times \sqrt{2 \times 9.8 \times 1}\\\\\dfrac{dH}{dt}=1.77 \times 10^{-5} \;\rm m/s\\\\\dfrac{dH}{dt}=1.77 \times 10^{-5} \times 6 \times 10^{4} \;\rm mm/min\\\\\dfrac{dH}{dt}=1.0625 \;\rm mm/min[/tex]
Thus, we can conclude that the rate, in mm/min, at which the water level will initially drop will be 1.0625 mm/min.
Learn more about the flow rate here:
https://brainly.com/question/11816739
When a rattlesnake strikes, its head accelerates from rest to a speed of 22 m/s in 0.48 seconds. Assume for simplicity that the only moving part of the snake is its head of mass 170 g. How much (average) power does the rattlesnake need to accelerate its head that fast? Answer in units of W.
Answer:
P = 85.72 W
Explanation:
Given that,
Initial speed, u = 0
Final speed, v = 22 m/s
Time, t = 0.48 s
Mass, m = 170 g = 0.17 kg
Let a be the acceleration of the rattlesnake.
[tex]a=\dfrac{v-u}{t}\\\\a=\dfrac{22-0}{0.48}\\\\a=45.84\ m/s^2[/tex]
Let x is the displacement of a rattlesnake. It can be given by :
[tex]x=ut+\dfrac{1}{2}at^2\\\\x=0+\dfrac{1}{2}\times 45.84\times (0.48)^2\\\\x=5.28\ m[/tex]
The power of the rattlesnake is given by :
[tex]P=\dfrac{W}{t}\\\\P=\dfrac{m\times a\times x}{t}\\\\P=\dfrac{0.17\times 45.84\times 5.28}{0.48}\\\\P=85.72\ W[/tex]
So, the power of the rattlesnake is 85.72 W.
In which number are the zeros not significant?
100.0
O 0.0003
O 4.00005
O 1.0004
Answer:
0.0003
Explanation:
In the rules of Sig Figs, all zeros before with decimals are not sigificant. I.E. 0.00000000000000009. Despite how many 0's there are, only the 9 is significant. Zeros before a number is not significant. In 100, only the one is signficant in 100. with a dot at the end, the one and the two zeros are significant. hope this helps.
Answers:
the second option
Explanation:
Which of the following is a vector quantity?
speed
distance
acceleration
When researchers replicate a study, they are seeking to __________.
A.
prove that the hypothesis upon which the study was founded is untestable
B.
develop a new hypothesis
C.
change the study to provide new results
D.
support or reject the hypothesis upon which the study was founded
Please select the best answer from the choices provided
A
B
C
D
Answer:
D
Explanation:
right edge 2022
One disadvantage to experimental research is that experimental conditions do not always reflect reality.
Please select the best answer from the choices provided
T
F
Answer:
It's true I took the test on Edge.
Explanation:
Answer:
True
Explanation:
Got it right on edg
In the graph, which two regions show the particle undergoing zero acceleration and negative acceleration respectively?
A.
BC shows zero acceleration, and AB shows negative acceleration.
OB.
AB shows zero acceleration, and CD shows negative acceleration.
O C.
BC shows zero acceleration, and CD shows negative acceleration.
D.
AB shows zero acceleration, and BC shows negative acceleration.
Answer:
c
Explanation:
Answer:
c
Explanation:
what is momentum of a train that is 60,000 kg that is moving at velocity of 17m/s?
explain your answer
What kind of scattering (Rayleigh, Mie, or non-selective) would you expect to be most important when radiation of the specified wavelength encounters the following natural or anthropogenic particles?
Wavelength O2molecules Smoke particles Cloud droplets Rain droplets
(size 10-10m) (size 0.3 mm) (size 20 mm) (size 3 mm)
200 nm
0.6 mm
10 mm
1.0 mm
1.0 km
Answer:
hello your question is not properly arranged attached below is the arranged table and solution
answer : attached table below
Explanation:
Given data:
02 molecules size = 10^-10m
smoke particles size = 0.3 mm
cloud droplets size = 20 mm
Rain droplets size = 3 mm
Attached below is a table showing the kind of scattering that is expected to occur at various wave lengths
Note : For Rayleigh scattering the wave particle is smaller than the wave length while for Non-selective scattering the wave particle is greater than the wavelength.
and For Mie scattering the wavelength is the same as the wavelength.
A vertical piston-cylinder device contains a gas at a pressure of 100 kPa. The piston has a mass of 10 kg and a diameter for 14 cm. Pressure of the gas is to be increased by placing some weights on the piston. Determine the local atmospheric pressure and the mass of the weights that will doublethe pressure of the gas inside the cylinder.
Answer:
the local atmospheric pressure is 93.63 kPa
the mass of the weights is 156.9 kg
Explanation:
Given that;
Initial pressure of gas = 100 kPa
mass of piston = 10 kg and diameter = 14 cm = 0.14 m
g = 9.81 m/s²
Now,
P_gas = P_atm + P_piston
100 = P_atm + P_piston --------- let this equation 1
P_piston = M_piston × g / A = (10 × 9.81) / π/4×d²
P_piston = 98.1 / (π/4×( 0.14 )²)
P_piston = 98.1 / 0.01539 = 6374,269 Pa = 6.37 kPa
now, from equation 1
100 = P_atm + P_piston
we substitute
100 = P_atm + 6.37
P_atm = 100 - 6.37
P_atm = 93.63 kPa
Therefore, the local atmospheric pressure is 93.63 kPa
Now for pressure of the gas in the cylinder ⇒ 2×initial pressure
Pgas_2 = 2 × 100 = 200 kPa
Pgas_2 = P_atm + P_piston + P_weight
Pgas_2 = P_gas + P_weight
we substitute
200 kPa = 100 kPa + P_weight
P_weight = 200 kPa - 100 kPa
P_weight = 100 kPa = 100,000 Pa
Also;
P_weight = M×g / A
100,000 Pa = ( M × 9.81 ) / (π/4 × (0.14)²)
100,000 × 0.01539 = M × 9.81
1539 = M × 9.81
M = 1539 / 9.81
M = 156.9 kg
Therefore, the mass of the weights is 156.9 kg
what is the direction of the third force that would cause the box to remain stationary on the ramp ?
An arrow pointing from the bottom of the ramp to the top, I assume it would be friction.
What Coulombs discovered almost 300
years ago
Answer:
ummm hehe this is my time to shine
Explanation:
MERICIA!!!!!!!!!!!!!!!!!!!!!!!
A student releases a small cart at the top of an incline with height H above the floor. The cart experiences very little friction. The student is attempting to cause the cart to go around a vertical loop of radius R without the cart losing contact with the track at the top. The student suggests that the heigt H should equal 2R so that the release height and maximum height of th eloop are the same. However, the student finds that it requires noticably higher hieght than 2R for the cart to go around the loop. Explain why H must be noticably greater than 2R to complete the loop. (Hint: In order for the cart to go around the loop it must have a nonzero velocity at the top of the loop.) answer
Answer:
Explanation:
In the whole process , potential energy of the cart is converted into kinetic energy . At the top of the vertical loop , the whole of potential energy is regained and kinetic energy becomes zero if we release the cart from a height of 2R because difference of height between lowest and highest point of motion is 2R . In that case kinetic energy at top = 0 , velocity v = 0
At the top , weight mg is acting which is providing centripetal force . So cart must have some velocity at the top . If it be v
mv²/R = mg
v = √ gR .
For that purpose , the cart must be released from a height greater than 2R .
The extra height beyond 2R will make the velocity at the top non-zero.
Which is larger: 65 mph (miles per hour) or 120 kph (kilometers per hour)? As a percentage, how much faster is one than the other?
To Find :
Which is larger: 65 mph (miles per hour) or 120 kph (kilometers per hour).
Solution :
We know, 1 mph = 1.61 kph
So, 65 mph = 1.61 × 65 kph
65 mph = 104.65 kph
Since, 65 mph is 104.65 kph which is smaller than 120 kph.
Therefore, 120 kph is faster than 65 mph by ( 120 - 104.65 ) = 15.35 kph.
