Answer:
r ’= 4 r
Explanation:
Electric potential energy is
U = [tex]k \frac{q_1q_2}{r_{12}}[/tex]k q1q2 / r12
in this exercise
q₁ = q₂ = q
U = k q² / r
for when the charge change
U ’= k q’² / r’
indicate that
q ’= 2q
U ’= U
we substitute
U = k (2q) ² / r ’
U = 4 k q² / r ’
we substitute
[tex]k \ \frac{q^2}{r} = 4 k \ \frac{q^2}{r'}[/tex]k q² / r = 4 k q² / r ’
r ’= 4 r
the acceleration due to gravity jupiter is 25m/s Square . what does it mean
Answer:
The acceleration due to gravity of Jupiter is 25 m/s^2. means that, Any object dropped near Jupiter's Surface will accelerate downward (towards the Jupiter's surface) at the rate of 25 m/s^2 due to the gravity of Jupiter
Explanation:
hope it's help
Answer:
Explanation:
If any object is dropped from a height above the Jupiter's surface the object will fall towards jupiter's surface with a constant acceleration of 25m/s^2.
In which situation should a parent be proactive and act to assume responsibility?
Answer: Patsy is eager to learn how to bake a cake but does not know how to do it.
Explanation: i picked this and it is correct, you’re welcome:)
e. Your father bought you a pair of shoes. When you wore the shoes, you realized there was a problem. The shoes were too long Why might such a problem arise and how can it be mitigated?
The problem arose due to a difference in length. This was due to father not knowing the exact length of shoe used by the son. And this can be mitigated by the use of shoe fillers.
The length of an object implies how long the object is. And it is one of the fundamental unit of quantities measured in SI unit of meters.
Considering the given question, it can be observed that the father do not know the exact length of shoe that would fit the son appropriately. Thus the realized problem of the pair of shoes too long arose due to difference in length of the pair of shoes and the son's leg. This variation would not have occurred if the exact length of pair of shoes has been bought.
To mitigate this little problem, shoe fillers can be used.
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3. A microscope is focused on a black dot. When a 1.30 cm -thick piece of plastic is placed over the dot, the microscope objective has to be raised 0.410 cm to bring the dot back into focus. What is the index of refraction of the plastic
The index of refraction of the plastic is approximately 1.461
The known values in the question are;
The thickness of the piece of plastic placed on the dot = 1.30 cm
The height to which the microscope objective is raised to bring the dot back to focus = 0.410 cm
The unknown values in the question are;
The index of refraction
Strategy;
Calculate the refractive index by making use of the apparent height and real height method for the black dot under the thick piece of plastic
[tex]\mathbf{ Refractive \ index, n = \dfrac{Real \ depth}{Apparent \ depth}}[/tex]
The real depth of the dot below the piece of plastic, d₁ = 1.30 cm
The apparent depth of the dot, d₂ = The actual depth - The height to which the microscope is raised
Therefore;
The apparent depth of the dot, d₂ = 1.30 cm - 0.410 cm = 0.89 cm
[tex]The \ refractive \ index, \ n = \dfrac{d_1}{d_2}[/tex]
Therefore, n = 1.30/0.89 ≈ 1.461
The refractive index of the plastic block, n ≈ 1.461
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calculate the pressure of water having density 1000 kilo per metre square at a depth of 20 m inside the water
Answer:
the pressure of the water at the given depth is 196,200 N/m².
Explanation:
Given;
density of the water, ρ = 1000 kg/m³
depth of the water, h = 20 m
acceleration due to gravity, g = 9.81 m/s²
The pressure at the given depth of the water is calculated as;
P = ρgh
P = 1000 x 9.81 x 20
P = 196,200 N/m²
Therefore, the pressure of the water at the given depth is 196,200 N/m².
According to Newton's law of universal gravitation, the force F between two bodies of constant mass m and M is given by the formula F = G m M d 2 , where G is the gravitational constant and d is the distance between the bodies. a. Suppose that are constants. Find the rate of change of force F with respect to distance d .
Answer:
One can write F = K d^-2 where K = G M m
So dF/dd = -2 K d^-3 = -2 K / d^3 (As d increases F decreases - it is opposite to the direction of F)
State the law of conservation of momentum
Explanation:
Conservation of momentum, general law of physics according to which the quantity called momentum that characterizes motion never changes in an isolated collection of objects; that is, the total momentum of a system remains constant
Two friends, Al and Jo, have a combined mass of 194 kg. At the ice skating rink, they stand close together on skates, at rest and facing each other. Using their arms, they push on each other for 1 second and move off in opposite directions. Al moves off with a speed of 7.9 m/sec in one direction and Jo moves off with a speed of 6.7 m/sec in the other. You can assume friction is negligible.
What is Al's mass? 110.58 What is Jo's mass? If you assume the force is constant during the 1 second they are pushing on each other, what is the magnitude of the force of Al on Jo? If you assume the force is constant during the 1 second they are pushing on each other, what is the magnitude of the force of Jo on Al?
Answer:
The mass of Al is 89.027 kilograms.
The mass of Jo is 104.973 kilograms.
The magnitude of the force of Jo on Al is 596.481 newtons.
Explanation:
Given the absence of external forces, this situation can be described will by Principle of Linear Momentum Conservation and Impact Theorem on each skater:
Al:
[tex]m_{1}\cdot (v_{1, f}-v_{1, o}) = -F \cdot \Delta t[/tex] (1)
Jo:
[tex]m_{2}\cdot (v_{2,f}-v_{2,o}) = F\cdot \Delta t[/tex] (2)
Total mass:
[tex]m_{1} + m_{2} = 194\,kg[/tex]
Where:
[tex]m_{1}[/tex], [tex]m_{2}[/tex] - Masses of the skaters, in kilograms.
[tex]v_{1,o}[/tex], [tex]v_{1,f}[/tex] - Initial and final velocities of Al, in meters per second.
[tex]v_{2,o}[/tex], [tex]v_{2,f}[/tex] - Initial and final velocities of Jo, in meters per second.
[tex]F[/tex] - Impact force between skaters, in newtons.
[tex]\Delta t[/tex] - Impact time, in seconds.
If we know that [tex]v_{1,o} = 0\,\frac{m}{s}[/tex], [tex]v_{1,f} = -7.9\,\frac{m}{s}[/tex], [tex]\Delta t = 1\,s[/tex], [tex]v_{2,o} = 0\,\frac{m}{s}[/tex] and [tex]v_{2,f} = 6.7\,\frac{m}{s}[/tex], then the masses of the skaters are, respectively:
[tex](194-m_{2})\cdot (-7.9) = -F[/tex] (1b)
[tex]m_{2} \cdot 6.7 = F[/tex] (2b)
(2b) in (1b):
[tex](194-m_{2})\cdot (-7.9) = -m_{2}\cdot 6.7[/tex]
[tex]-1532.6 +7.9\cdot m_{2} = -6.7\cdot m_{2}[/tex]
[tex]14.6\cdot m_{2} = 1532.6[/tex]
[tex]m_{2} = 104.973\,kg[/tex]
[tex]m_{1} = 194\,kg - 104.973\,kg[/tex]
[tex]m_{1} = 89.027\,kg[/tex]
And the magnitude of the force is:
[tex]F = 6.7\cdot m_{2}[/tex]
[tex]F = 596.481\,N[/tex]
The mass of Al is 89.027 kilograms.
The mass of Jo is 104.973 kilograms.
The magnitude of the force of Jo on Al is 596.481 newtons.
What is not one of the main uses of springs?
A. Car suspension
B. Bike suspension
C. The seasons
D. Clock making
What is the submarine's maximum safe depth in sea water? The pressure inside the submarine is maintained at 1.0 atm
Answer:
The submarine's maximum safe depth in sea water is 801.678 m.
Explanation:
P=Po+(rho)*g*h
Max Pressure = Initial Pressure + (Water Density)(Gravity)(Max Depth)
Area of Window = Pi*(Diameter/2)^2 = Pi*(.4m/2)^2 = 0.125664 m^2
Max Pressure= (1.0*10^6 N)/(0.125664 m^2)= 7.95775-E6 Pa
Initial Pressure= 1atm= 101.3kPa= 101300Pa
Water Density (rho) = 1000kg/m^3
Gravity= 9.8m/s^2
So rearranging for h= (P-Po)/((rho)*g)
h=((7.95775-E6Pa)-(101300Pa))/((1000kg/m^3)(9.8m/s^2))= 801.678 m
Which of the following quantities is measured by the area under the velocity time graph? (a) Magnitude of velocity (b) Magnitude of acceleration (c) Magnitude of displacement (d) Average Speed
Answer:
c.
magnitude of displacement
Keisha writes that if an object has any external forces acting on it, then the object can be in dynamic equilibrium but not
static equilibrium
Which statement best describes Keisha's error?
An object that is not moving is always in static equilibrium.
O An object that is moving must be in dynamic equilibrium.
An object in either state of equilibrium must have no forces acting on it.
An object in either state of equilibrium must have no net force acting on it.
Answer:
An object in either state of equilibrium must have no net force acting on it.
Explanation:
Answer: An object in either state of equilibrium must have no net force acting on it.
Explanation:
a baseball is given an initial velocity with magnitude v at the angle beta above the surface of an incline which in turn inclined at angle teta above horizontal calculate the distance measured along incline from the launch point to where the baseball strike the incline
Explanation:
The maximum height of an object, given the initial launch angle and initial velocity is found with:h=v2isin2θi2g h = v i 2 sin 2 θ i 2 g .
Determine the magnitude as well as direction of the electric field at point A, shown in the above figure. Given the value of k = 8.99 × 1012N/C. where, d= 11 cm Q= 12.5 C
Answer:
The electric field is 9.3 x 10^12 N/C and the direction is away from the charge.
Explanation:
charge, Q = 12.5 C
distance, d = 11 cm = 0.11 m
Let the electric field is E.
[tex]E =\frac{K Q}{d^2}\\\\E = \frac{9\times 10^9\times 12.5}{0.11\times 0.11}\\\\E = 9.3\times 10^{12} N/C[/tex]
The direction of electric filed is away from the charge.
Large cockroaches can run as fast as 1.50 m/s in short bursts. Suppose you turn on the light in a cheap motel and
see one scurrying directly away from you at a constant 1.50 m/s. If you start 0.90 m behind the cockroach with
an initial speed of 0.80 m/s toward it, what minimum constant acceleration would you need to catch up with it
when it has traveled 1.20 m, just short of safety onder a counter?
Answer:
The time that you need to use 1.2/1.5 because this is how long it took the cockroach to travel the 1.2 meters to the counter. That is therefore how long you have to catch up to it.
Explanation:
Consider newtonian mechanics here.
Dynamic equation is
The time we have to use 1.2/1.5 this how long it took the cockroach to travel the 1.2 meters to the counter.
we'll consider newtonian mechanics here.
so the dynamic equations is S = ut + 0.5at^2
we know u=0.8
S=1.2+0.9
t=1.2/1.5
find a.
Condensation is the process of ____________________.
a. planetesimals accumulating to form protoplanets.
b. planets gaining atmospheres from the collisions of comets.
c. clumps of matter adding material a small bit at a time.
d. clumps of matter sticking to other clumps.
e. clouds formed from volcanic eruptions.
A capacitor is connected to an ac generator that has a frequency of 3.2 kHz and produces a rms voltage of 2.0 V. The rms current in the capacitor is 28 mA. When the same capacitor is connected to a second ac generator that has a frequency of 4.7 kHz, the rms current in the capacitor is 70 mA. What rms voltage does the second generator produce
Answer:
The rms voltage of new generator is 3.4 V.
Explanation:
f = 3200 Hz
rms voltage, V = 2 V
rms current, i = 28 mA
Now
f' = 4700 Hz
rms current, i' = 70 mA
let the new rms voltage is V'.
[tex]i = \frac{V}{Xc} = V \times 2\pi fC....(1)\\\\i' = V' \times 2 \pi f' C..... (2)\\\\\frac{i}{i'} =\frac{V f}{V' f'}\\\\\frac{28}{70}=\frac{2\times 3200}{V'\times 4700}\\\\V' = 3.4 V[/tex]
A student graphs power (p) on the vertical axis and time (t) on the horizontal axis. The graph appears to be a hyperbola.
a) What should the student graph on each axis to test whether the relationship is actually
hyperbolic?
b) If the relationship is actually hyperbolic, what is the general equation for the relationship between power and time?
Answer: it would be daddy
Explanation:
Because I’m daddy
Snell's Law: Light goes from material having a refractive index of n 1 into a material with refractive index n 2. If the refracted light is bent away from the normal, what can you conclude about the indices of refraction
Answer:
a) the light is close to normal therefore the reference incidence of medium 1 is less than medium n2 where the ray is transmitted.
b) The ray is far from normal in this case the refractive index of medium 1 is greater than index of medium 2
Explanation:
The expression for the angle of refraction is
n₁ sin θ₁ = n₂ sin θ₂
refractive index n₁ is for incident light and n₂ is for transmitted light.
We have two cases
a) the light is close to normal therefore the reference incidence of medium 1 is less than medium n2 where the ray is transmitted.
b) The ray is far from normal in this case the refractive index of medium 1 is greater than index of medium 2
Sound is an example of a:
Select one:
O a. rolling waves
b. longitudinal wave
O c. traverse waves
O d. surface wave
Ez Physics question will mark brainliest.
Answer:
The answer is B. longitude wave
Given that two vectors A = 5i-7j-3k, B = -4i+4j-8k find A×B
[tex]\textbf{A}×\textbf{B}= 68\hat{\textbf{i}} + 52\hat{\textbf{j}} - 8\hat{\textbf{k}}[/tex]
Explanation:
Given:
[tex]\textbf{A} = 5\hat{\textbf{i}} - 7\hat{\textbf{j}} - 3\hat{\textbf{k}}[/tex]
[tex]\textbf{B} = -4\hat{\textbf{i}} + 4\hat{\textbf{j}} - 8\hat{\textbf{k}}[/tex]
The cross product [tex]\textbf{A}×\textbf{B}[/tex] is given by
[tex]\textbf{A}×\textbf{B} = \left|\begin{array}{ccc}\hat{\textbf{i}} & \hat{\textbf{j}} & \hat{\textbf{k}} \\\:\:5 & -7 & -3 \\ -4 & \:\:4 & -8 \\ \end{array}\right|[/tex]
[tex]= \left|\begin{array}{cc}-7 & -3\\\:4 & -8\\ \end{array}\right|\:\hat{\textbf{i}}\:+\:\left|\begin{array}{cc}-3 & \:\:5\\-8 & -4\\ \end{array}\right|\:\hat{\textbf{j}}\:+\: \left|\begin{array}{cc}\:\:5 & -7\\-4 & \:\:4\\ \end{array}\right|\:\hat{\textbf{k}}[/tex]
[tex]= 68\hat{\textbf{i}} + 52\hat{\textbf{j}} - 8\hat{\textbf{k}}[/tex]
A simple model of the human eye ignores its lens entirely. Most of what the eye does to light happens at the outer surface of the transparent cornea. Assume that this surface has a radius of curvature of 6.50 mm and that the eyeball contains just one fluid, with a refractive index of 1.41. Determine the distance from the cornea where a very distant object will be imaged.
Answer:
the distance from the cornea where a very distant object will be imaged is 23.35 mm
Explanation:
Given the data in the question;
For a spherical refracting surface;
[tex]n_i[/tex]/[tex]d_0[/tex] + [tex]n_t[/tex]/[tex]d_i[/tex] = ( [tex]n_t[/tex] - [tex]n_i[/tex] )/R
where [tex]n_i[/tex] is the index of refraction of the light of ray in the incident medium
[tex]d_0[/tex] is the object distance
[tex]n_t[/tex] is the index of refraction of light ray in the refracted medium
[tex]d_i[/tex] is the image distance
R is the radius of curvature
Now, let [tex]d_0[/tex] = ∞, such that;
[tex]n_i[/tex]/∞ + [tex]n_t[/tex]/[tex]d_i[/tex] = ( [tex]n_t[/tex] - [tex]n_i[/tex] )/R
0 + [tex]n_t[/tex]/[tex]d_i[/tex] = ( [tex]n_t[/tex] - [tex]n_i[/tex] )/R
we make [tex]d_i[/tex] subject of the formula
[tex]n_t[/tex]R = [tex]d_i[/tex]( [tex]n_t[/tex] - [tex]n_i[/tex] )
[tex]d_i[/tex] = ( [tex]n_t[/tex] × R ) / ( [tex]n_t[/tex] - [tex]n_i[/tex] )
given that; R = 6.50 mm, [tex]n_t[/tex] = 1.41, we know that [tex]n_i[/tex] = 1.00
so we substitute
[tex]d_i[/tex] = (1.41 × 6.50 mm ) / ( 1.41 - 1.00 )
[tex]d_i[/tex] = 9.165 / 0.41
[tex]d_i[/tex] = 23.35 mm
Therefore, the distance from the cornea where a very distant object will be imaged is 23.35 mm
At room temperature, sound travels at a speed of about 344 m/s in air. You see a distant flash of lightning and hear the thunder arrive 7.5 seconds later. How many miles away was the lighting strike? (Assume the light takes essentially no time to reach you).
Answer:
1.6031 miles
Explanation:
Given the following data;
Speed = 344 m/s
Time = 7.5 seconds
To find how many miles away was the lighting strike;
Mathematically, the distance travelled by an object is calculated by using the formula;
Distance = speed * time
Distance = 344 * 7.5
Distance = 2580 meters
Next, we would have to convert the value of the distance travelled in meters to miles;
Conversion:
1609.344 metres = 1 mile
2580 meters = X mile
Cross-multiplying, we have;
X * 1609.344 = 2580
X = 2580/1609.344
X = 1.6031 miles
a concave mirror has a radius of curvature of 60cm. How close to the mirror should an object be placed so that the rays travel parallel to each other after reflection
Answer:
Answer:30 cm
Answer:30 cmExplanation:
Answer:30 cmExplanation:Given=ROC= 60cm
Answer:30 cmExplanation:Given=ROC= 60cmObject be placed so that the rays that came from the object to them mirror are reflected from the mirror, and, then travel parallel to each other= 30cm at focus.
What frequency is received by a person watching an oncoming ambulance moving at 110 km/h and emitting a steady 800-Hz sound from its siren? The speed of sound on this day is 345 m/s. Group of answer choices
Answer:
check photo for solve
Explanation:
find the velocity of the object for all relevent times find the position of the object for all relevent times a softball is popped up vertically velocity of 32 m/s
Answer:
whats the formula
Explanation:
Which of the following groups is the largest ?
population
community
ecosystem
biome
Answer:
B. Community
Took science classes for 6 years now
The spectral lines of two stars in a particular eclipsing binary system shift back and forth with a period of 6 months. The lines of both stars shift by equal amounts, and the amount of the Doppler shift indicates that each star has an orbital speed of 64,000 m/s. What are the masses of the two stars
Answer:
the masses of the two stars are; m₁ = m₂ = 4.92 × 10³⁰ kg
Explanation:
Given the data in the question;
Time period = 6 months = 1.577 × 10⁷ s
orbital speed v = 64000 m/s
since its a circular orbit,
v = 2πr / T
we solve for r
r = vT/ 2π
r = ( 64000 × 1.577 × 10⁷ ) / 2π
r = 1.6063 × 10¹¹ m = ( (1.6063 × 10¹¹) / (1.496 × 10¹¹) )AU = 1.0737 AU
Now, from Kepler's law
T² = r³ / ( m₁ + m₂ )
T = 6 months = 0.5 years
we substitute
(0.5)² = (1.0737)³ / ( m₁ + m₂ )
0.25 = 1.2378 / ( m₁ + m₂ )
( m₁ + m₂ ) = 1.2378 / 0.25
( m₁ + m₂ ) = 4.9512
m₁ = m₂ = 4.9512 / 2 = 2.4756 solar mass
we know that solar mass = 1.989 × 10³⁰ kg
so
m₁ = m₂ = 2.4756 × 1.989 × 10³⁰ kg
m₁ = m₂ = 4.92 × 10³⁰ kg
Therefore, the masses of the two stars are; m₁ = m₂ = 4.92 × 10³⁰ kg
A 30-year-old astronaut goes off on a long-term mission in a spacecraft that travels at speeds close to that of light. The mission lasts exactly 20 years as measured on Earth. Biologically speaking, at the end of the mission, the astronaut's age would be:_______.
a) exactly 50 years.
b) exactly 25 years.
c) exactly 30 years.
d) less than 50 years.
e) more than 50 years.
Answer:
I think D) less than 50 years
Biologically speaking, at the end of the mission, the astronaut's age would be less than 50 years. The correct option is d.
Who is an astronaut?An astronaut observes and performs the experiments based on the universe.
A 30-year-old astronaut goes off on a long-term mission in a spacecraft that travels at speeds close to that of light. The mission lasts exactly 20 years as measured on Earth.
Due to special relativity, between space and Earth, both moving with different speeds.
The total age will be less than 30 +20 =50 years. In space, he is moving with speed of light. So, time will move slowly. As measured with respect to Earth, exact time spent in space 20 years will be less on Earth.
So, biologically speaking, at the end of the mission, the astronaut's age would be less than 50 years.
Thus, the correct option is d.
Learn more about astronaut.
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Which of the following accurately describes circuits?
options:
A)
In a parallel circuit, there's only one path for the current to travel.
B)
In a series circuit, the amount of current passing through each part of the circuit may vary.
C)
In a series circuit, the current can flow through only one path from start to finish.
D)
In a parallel circuit, the same amount of current flows through each part of the circuit.
' C ' is the only correct statement.
