Answer:Nuclear fusion is the process by which two or more atomic nuclei join together, or “fuse,” to form a single heavier nucleus. During this process, matter is not conserved because some of the mass of the fusing nuclei is converted to energy, which is released.
Explanation:
Atoms of different elements can combine to make new substances. A molecule is formed when two or more atoms join together chemically. If atoms combine that are of two or more different elements, we call that a compound. ... When two hydrogen atoms combine with one oxygen atom, it becomes the compound water.
Answer:
nuclear fusion
Explanation:
The effective spring constant describing the potential energy of the HBr molecule is 410 N/m and that for the NO molecule is 1530 N/m.A) Calculate the minimum amplitude of vibration for the NO molecule.
B) Calculate the minimum amplitude of vibration for the HCl molecule.
Answer:
a. the minimum amplitude of vibration for the NO molecule A [tex]\simeq[/tex] 4.9378 pm
b. the minimum amplitude of vibration for the HCl molecule A [tex]\simeq[/tex] 10.9336 pm
Explanation:
Given that:
The effective spring constant describing the potential energy of the HBr molecule is 410 N/m
The effective spring constant describing the potential energy of the NO molecule is 1530 N/m
To calculate the minimum amplitude of vibration for the NO molecule, we use the formula:
[tex]\dfrac{1}{2}kA^2= \dfrac{1}{2}hf[/tex]
[tex]kA^2= hf[/tex]
[tex]A^2= \dfrac{hf}{k}[/tex]
[tex]A = \sqrt{\dfrac{hf}{k}}[/tex]
[tex]A = \sqrt{\dfrac{(6.626 \times 10^{-34} \ J. s ) ( 5.63 \times 10^{13} \ s^{-1})}{1530 \ N/m}}[/tex]
[tex]A = \sqrt{\dfrac{3.730438 \times 10^{-20} \ m}{1530 }[/tex]
[tex]A = \sqrt{2.43819477 \times 10^{-23}\ m[/tex]
[tex]A =4.93780799 \times 10^{-12} \ m[/tex]
A [tex]\simeq[/tex] 4.9378 pm
The effective spring constant describing the potential energy of the HCl molecule is 480 N/m
To calculate the minimum amplitude using the same formula above, we have:
[tex]A = \sqrt{\dfrac{hf}{k}}[/tex]
[tex]A = \sqrt{\dfrac{(6.626 \times 10^{-34} \ J. s ) ( 8.66 \times 10^{13} \ s^{-1})}{480 \ N/m}}[/tex]
[tex]A = \sqrt{\dfrac{5.738116\times 10^{-20} \ m}{480 }[/tex]
[tex]A = \sqrt{1.19544083\times 10^{-22}\ m[/tex]
[tex]A = 1.09336217 \times 10^{-11} \ m[/tex]
[tex]A = 10.9336217 \times 10^{-12} \ m[/tex]
A [tex]\simeq[/tex] 10.9336 pm
Find p1, the gauge pressure at the bottom of tube 1. (Gauge pressure is the pressure in excess of outside atmospheric pressure.) Express your answer in terms of quantities given in the problem introduction and g, the magnitude of the acceleration due to gravity.
Answer:
(a). The gauge pressure at the bottom of tube 1 is [tex]P_{1}=\rho g h_{1}[/tex]
(b). The speed of the fluid in the left end of the main pipe [tex]\sqrt{\dfrac{2g(h_{2}-h_{1})}{(1-(\gamma)^2)}}[/tex]
Explanation:
Given that,
Gauge pressure at bottom = p₁
Suppose, an arrangement with a horizontal pipe carrying fluid of density p . The fluid rises to heights h1 and h2 in the two open-ended tubes (see figure). The cross-sectional area of the pipe is A1 at the position of tube 1, and A2 at the position of tube 2.
Find the speed of the fluid in the left end of the main pipe.
(a). We need to calculate the gauge pressure at the bottom of tube 1
Using bernoulli equation
[tex]P_{1}=\rho g h_{1}[/tex]
(b). We need to calculate the speed of the fluid in the left end of the main pipe
Using bernoulli equation
Pressure for first pipe,
[tex]P_{1}=\rho gh_{1}[/tex].....(I)
Pressure for second pipe,
[tex]P_{2}=\rho gh_{2}[/tex].....(II)
From equation (I) and (II)
[tex]P_{2}-P_{1}=\dfrac{1}{2}\rho(v_{1}^2-v_{2}^2)[/tex]
Put the value of P₁ and P₂
[tex]\rho g h_{2}-\rho g h_{1}=\dfrac{1}{2}\rho(v_{1}^2-v_{2}^2)[/tex]
[tex]gh_{2}-gh_{1}=\dfrac{1}{2}(v_{1}^2-v_{2}^2)[/tex]
[tex]2g(h_{2}-h_{1})=v_{1}^2-v_{2}^2[/tex]....(III)
We know that,
The continuity equation
[tex]v_{1}A_{1}=v_{2}A_{2}[/tex]
[tex]v_{2}=v_{1}(\dfrac{A_{1}}{A_{2}})[/tex]
Put the value of v₂ in equation (III)
[tex]2g(h_{2}-h_{1})=v_{1}^2-(v_{1}(\dfrac{A_{1}}{A_{2}}))^2[/tex]
[tex]2g(h_{2}-h_{1})=v_{1}^2(1-(\dfrac{A_{1}}{A_{2}}))^2[/tex]
Here, [tex]\dfrac{A_{1}}{A_{2}}=\gamma[/tex]
So, [tex]2g(h_{2}-h_{1})=v_{1}^2(1-(\gamma)^2)[/tex]
[tex]v_{1}=\sqrt{\dfrac{2g(h_{2}-h_{1})}{(1-(\gamma)^2)}}[/tex]
Hence, (a). The gauge pressure at the bottom of tube 1 is [tex]P_{1}=\rho g h_{1}[/tex]
(b). The speed of the fluid in the left end of the main pipe [tex]\sqrt{\dfrac{2g(h_{2}-h_{1})}{(1-(\gamma)^2)}}[/tex]
Using the lensmaker's formula (equation (5) of your lab manual), calculate the index of refraction of the acrylic lens. You should use the f_are you calculated in part (1) above instead of the value for the focal length of the concave lens that you measured. Remember that the focal length of a concave lens is negative so in this case, f = -f_are.
Answer:
n = 1 + R / f
Explanation:
The equation of the constructor is optical is
1 / f = 1 / p + 1 / q
where f is the focal length, p and q are the distance to the object and image, respectively
The exercise tells us that it is a concave lens with focal length fo, in these lenses the focal length is negative. The relationship to calculate the focal length is
1 / f = (n -n₀) (1 /R₁ - 1 /R₂)
where is n₀ the refractive index of the medium that surrounds the lens in this case it is air with n₀ = 1, you do not indicate the type of lens, but the most used lens is the concave plane, in this case R₂ = ∞, so which 1 / R₂ = 0, let's substitute
1 / f = (n-1) / R₁
n - 1 = R₁ / f
let's calculate
n = 1- R₁ / f
remember that the radius of curvature is negative, so the equation is
n = 1 + R / f
At t = 0, one toy car is set rolling on a straight track with initial position 13.0 cm, initial velocity -3.6 cm/s, and constant acceleration 2.20 cm/s2. At the same moment, another toy car is set rolling on an adjacent track with initial position 11.5 cm, initial velocity 5.40 cm/s, and constant zero acceleration. (a) At what time, if any, do the two cars have equal speeds? (Enter NA if the cars never have equal speeds.) s (b) What are their speeds at that time? (Enter NA if the cars never have equal speeds.) cm/s (c) At what time(s), if any, do the cars pass each other? (If there is only one time, enter NA in the second blank. If there are two times, enter the smaller time first. If they never pass, enter NA in both blanks.) s s (d) What are their locations at that time? (If there is only one position, enter NA in the second blank. If there are two positions, enter the smaller position first. If they never pass, enter NA in both blanks.) cm cm (e) Explain the difference between question (a) and question (c) as clearly as possible.
Answer:
that's too much to read
List two scientific questions that would be best explained using a model.
Answer:
Is a flame hottest when it is blue? Is it cold today?
Explanation:
Verification questions. These are basic data collecting questions. They are useful in building knowledge.
which of the following is not a method of caculating percentage of body fat
Body mass index, known as BMI, is not a method of calculating body fat percentage, but it is a technique used to check nutritional status and to see if a person is within normal range with respect to weight and height.. This technique is measured by the formula: BMI = Weight (Kg) / (Height (m)) ².
Ergo, the answer is BMI!!!!!!!!!!!
Hope this helped you!
A clock battery wears out after moving 10,900 C of charge through the clock at a rate of 0.450 mA. (a) How long (in s) did the clock run
Answer:
2.42×10⁷ s
Explanation:
From the question above,
Applying,
Q = It.................... Equation 1
Where Q = quantity of charge in coulombs, I = electric current in Ampere, t = time in seconds
make t the subject of the equation
t = Q/I................ Equation 2
Given: Q = 10900 C, I = 0.450 mA = 4.5×10⁻⁴ A
Substitute these values into equation 2
t = 10900/(4.5×10⁻⁴)
t = 2.42×10⁷ s
Hence the clock runs for 2.42×10⁷ s
give me an example of orderliness In nature
Answer:
Here are some examples of orderliness In nature
1. The proposal that the order of nature showed evidence of having its own human-like "intelligence" goes back to the origins of Greek natural philosophy and science, and its attention to the orderliness of nature, often with special reference to the revolving of the heavens.
2. For Stillman, the orderliness of Astaire-like dance is an actual cure for destructive emotions.
3. Tells about the orderliness of the crowds, and of the dispatch with which the trains were being filled and emptied.
A NASA spacecraft measures the rate of at which atmospheric pressure on Mars decreases with altitude. The result at a certain altitude is:
Complete Question
A NASA spacecraft measures the rate R of at which atmospheric pressure on Mars decreases with altitude. The result at a certain altitude is: [tex]R = 0.0498 \ kPAkm^{-1}[/tex] Convert R to [tex]kJ*m^{-4}[/tex]
Answer:
The value is [tex]R = 0.0498 *10^{-3} \frac{kJ}{m^4}[/tex]
Explanation:
From the question we are told that
The altitude is [tex]R = 0.0498 \ kPAkm^{-1}[/tex]
Generally
[tex]1 k PA = 1000 PA[/tex]
So
[tex]R = 0.0498 \frac{1000PA}{ km}[/tex]
Also
1 km = 1000 m
So
[tex]R = 0.0498 \frac{1000PA}{ 1000m}[/tex]
=> [tex]R = 0.0498 \frac{1 PA}{ 1 m}[/tex]
Now PA is Pascal which is mathematically represented as
[tex]PA = \frac{N}{m^2 }[/tex]
So
[tex]R = 0.0498 \frac{\frac{N}{m^2} }{m}[/tex]
[tex]R = 0.0498 \frac{N}{m^3}[/tex]
Looking the unit we are arrive at we see that it contains J which is mathematically represented as
[tex]J = N * m[/tex]
So
[tex]R = 0.0498 \frac{ N \frac{m}{m} }{m^3}[/tex]
=> [tex]R = 0.0498 \frac{\frac{J}{m} }{m^3}[/tex]
=> [tex]R = 0.0498 \frac{J}{m^4}[/tex]
Generally
[tex]1 J \to 1.0*10^{-3} kJ[/tex]
[tex]0.0498 J \to x kJ[/tex]
=> [tex]x = \frac{0.0498 * 1.0*10^{-3}}{1}[/tex]
=> [tex]0.0498 *10^{-3} kJ[/tex]
So
[tex]R = 0.0498 *10^{-3} \frac{kJ}{m^4}[/tex]
43 Points For Answering & +22 for Brainliest
Question 1
A student conducts an experiment to test how the temperature of a ball affects its bounce height. The same ball is used for each test, and the ball is dropped from the same height each time. What is the dependent variable?
A.The type of ball
B.The temperature of the ball
C.The drop height of the ball
D.The bounce height of the ball
Question 2
A student conducts an experiment to test how the temperature affects the amount of sugar that can dissolve in water. In the experiment, she uses 100 millilitres of water in each trial and stirs for five minutes each time. What is the independent variable?
A.The amount of water
B.The temperature of the water
C.The amount of sugar
D The time stirred
Question 3
Which of the following is a way for scientists to limit the amount of errors in their experimentation?
A.Using controls
B.Only completing an experiment once
C.Using equipment to measure the experiment that has been damaged
D.There is no need to record data from an experiment
Answer:
Question 1: D because the height the ball bounces depends on all the other factors in the experiment.
Question 2: B because the the temperature of the water is not affected by the other variables.
Question 3: A because the more that they can control in the experiment, the more accurate the results will be.
Hopefully this helps :)
1:D 2:B 3:A
just took the test
Which particle is most likely to interact with your hand? Select one: a. Alpha particle b. Beta particle c. Gamma particle d. Neutrino
Answer:
The correct option is a
Explanation:
The alpha particle has the lowest penetrating power of the trio of alpha, beta and gamma particles and can be stopped by a sheet of paper and hence cannot penetrate a human skin. Beta particle has a higher penetrating power than alpha particle (some of it penetrates the human skin and some do not) while the gamma particle has the highest penetrating power (with all of it penetrating the human skin).
From the above description, it can be deduced that the alpha particle will stay and interact with the hand (because of its low penetrating power) as the remaining particles move through the skin.
A hockey puck moves 26 meters northward, then 12 meters southward, and finally 6 meters
northward
For this motion, what is the distance moved?
What is the magnitude and direction of the displacement?
Answer:
The distance moved is 44 metres.
The magnitude of displacement is 20 metres with northward direction.
Answer:
44 m.
North 20 m.
Explanation:
Distance moved = 26 + 12 + 6
= 44 m.
Magnitude of the displacement = 26 - 12 + 6
= 20m
Direction is Northward.
A beam of helium-3 atoms (m = 3.016 u) is incident on a target of nitrogen-14 atoms (m = 14.003 u) at rest. During the collision, a proton from the helium-3 nucleus passes to the nitrogen nucleus, so that following the collision there are two atoms: an atom of "heavy hydrogen" (deuterium, m = 2.014 u) and an atom of oxygen-15 (m = 15.003 u). The incident helium atoms are moving at a velocity of 6.346 x 10° m/s. After the collision, the deuterium atoms are observed to be moving forward (in the same direction as the initial helium atoms) with a velocity of 1.531 x 107 m/s.A) What is the final velocity of the oxygen-15 atoms? B) Compare the total kinetic energies before and after the collision.
Answer:
Explanation:
We shall apply law of conservation of momentum to solve the problem .
Helium-3 collides with nitrogen-14 at rest . After the collision the newly formed deuterium atom and oxygen-15 atom moves .
momentum before the collision
= 3.016 x 6.346 x 10⁶ + 14.003 x 0 = 19.14 x 10⁶ unit
momentum after collision
2.014 x 1.531 x 10⁷ + 15.003 V
3.083 x 10⁷ + 15.003 V units
Applying the law of conservation of momentum ,
19.14 x 10⁶ = 3.083 x 10⁷ + 15.003 V
1.914 x 10⁷ = 3.083 x 10⁷ + 15.003 V
15.003 V = - 1.169 x 10⁷
V = .077917 x 10⁷
= 7.79 x 10⁵ m /s
= .0779 x 10⁷ m /s
mass of helium atom = 3.016 u = 3.016 x 1.67 x 10⁻²⁷ kg
velocity = 6.346 x 10⁶ m /s
kinetic energy = 1 /2 x 3.016 x 1.67 x 10⁻²⁷ x (6.346 x 10⁶ )²
= 101.42 x 10⁻¹⁵ J
kinetic energy of nitrogen atoms = 0
Total energy before collision = 101.42 x 10⁻¹⁵ J
Similarly kinetic energy after collision
= 1 /2 x [ 2.014 x 1.531² + 15.003 x .0779² ] x 1.67 x 10⁻²⁷ x 10¹⁴
= .835 x [ 4.72 + .09 ] x 10⁻¹³ J
= 4.016 x 10⁻¹³ J
= 401.6 x 10⁻¹⁵ J
value of kinetic energy is increased .
Distance is the length of a path followed by a particle. The displacement of a particle is defined as its change in position in some time interval.
a. True
b. False
Answer:
a. True
Explanation:
Distance is described with only magnitude. It is defined as the total path covered by an object, in other words it is the length of a path followed by a particle.
Displacement is described with both magnitude and direction. It is distance traveled in a specified direction or change in position in some time interval.
Therefore, the correct option is " a. True"
A hollow conductor carries a net charge of ++3QQ. A small charge of −−2QQ is placed inside the cavity in such a way that it is isolated from the conductor. How much charge is on the outer surface of the conductor?
Answer:
The value is [tex]q_o = Q[/tex]
Explanation:
From the question we are told that
The net charge is [tex]Q = + 3Q[/tex]
The charge place on the inside of the cavity is [tex]q = -2Q[/tex]
Since we are told from the question that small charge placed inside the cavity is isolated from the conductor
Then it implies that the electric flux is Zero
Which mean that the charge place within the conductor + the charge on the inner region of the conductor = 0
i.e
[tex]q + q_i = 0[/tex]
=> [tex]-2Q + q_i = 0[/tex]
=> [tex]q_i = 2Q[/tex]
Now the net charge on the conductor is mathematically represented as
[tex]Q = q_o + q_i[/tex]
Here [tex]q_o[/tex] is the charge on the outer surface
So
[tex]3Q = q_o + 2Q[/tex]
=> [tex]q_o = Q[/tex]
Visible light passes through a diffraction grating that has 900 slits per centimeter, and the interference pattern is observed on a screen that is 2.40 m from the grating. Part A In the first-order spectrum, maxima for two different wavelengths are separated on the screen by 2.92 mm . What is the difference between these wavelengths? Express your answer in meters.
Answer:
13.51 nm
Explanation:
To solve this problem, we are going to use angle approximation that sin θ ≈ tan θ ≈ θ where our θ is in radians
y/L=tan θ ≈ θ
and ∆θ ≈∆y/L
Where ∆y= wavelength distance= 2.92 mm =0.00292m
L=screen distance= 2.40 m
=0.00292m/2.40m
=0.001217 rad
The grating spacing is d = (90000 lines/m)^−1
=1.11 × 10−5 m.
the small-angle
approx. Using difraction formula with m = 1 gives:
mλ = d sin θ ≈ dθ →
∆λ ≈ d∆θ = =1.11 × 10^-5 m×0.001217 rad
=0.000000001351m
= 13.51 nm
hor and Hulk are fighting eachother, back and forth in a straight line in the master's colosseum. Hulk punches thor into a wall 30 meters away from the start. Thor then hit's hulk with a hammer, hurling Hulk 45 meters back from the wall. This all happens in the course of 3 seconds. What is the speed of their fight across the arena?
a. 25 m/s
b. 10 m/s forward from Hulk and Thor's starting point.
c. 5 m/s backwards from hulk's starting point
d. 5 m/s
Answer:
A
Explanation:
cant be c because its asking for speed and speed doesnt have direction
In the drawing, what is the vector sum of forces A→+B→+C→ if each grid square is 7.00 N on a side? If the resultant is eastward, enter a positive value and if the resultant is westward, enter a negative value.
Answer:
resultant force = 14 N ( East direction)
Explanation:
A = [tex]\sqrt{(4*7)^2 + (4*7)^2}[/tex]
A = 39.6 N
B = 4 * 7
B = 28 N
C = 2 * 7
C = 14 N
∑ y forces = Ay - B = (4*7) - 28 = 0
∑ x forces = Ax - C = (4*7) - 14 = 14 N
so the resultant force = 14 N ( East direction)
Answer:
resultant = 14 N to the right
Explanation:
A→+B→+C→
adding all forces acting on x
and
adding all forces acting on y
A = sqrt(4*7)^2 + (4*7)^2 = 39.6
B = 4 * 7 = 28
C = 2 * 7 = 14
forces acting on x = (4*7) - 28 = 0
forces acting on y = (4*7) - 14 = 14 N
so the resultant = 14 N to the right
en un experimento se retiro a un pardela de su nido se le llevo a 5150 km de distancia y luego fue liberada su regreso a su nido fue 13.5 dias despues de haberse soltado si el origen es el nido y extendemos el eje +x al punto de liberacion ¿cual fue la velocidad media del ave?a)en el vuelo de regreso b) desde que se retirodel nido hasta que regreso
Answer:
a) v = - 4.4168 m / s , b) V = 0
Explanation:
The average speed is the variation of the displacement in time used.
v = Δx /Δt = (x₂-x₁) / (t₂-t₁)
let's apply this equation to our case
Let's reduce the magnitudes of the system Yes
Δx = 5150 km (1000 m / 1km) = 5.15 106 m
Δt = 13.5 day (86,400 s / 1 day) = 1,166 10 6 s
a) return trip
the vector is negative because it points long towards the center of the system
v = - 5.15 106 / 1.166 106
v = - 4.4168 m / s
the negative sign indicates that he is coming back, to the lair
b) In a complete trip the distance is zero, because it is a vector, consequently the mean fickleness is also zero
V(j_ = 0
ame
2. A train moving at 15 m/s slows down, and eventually stops after 5
seconds. What is the acceleration of the train?
G: Vi =
Vf=
U:
E: Formula
S: Substitute
S: Solve
Answer:
Acceleration, a = -3 m/s²
Explanation:
Given that,
Initial speed, u = 15 m/s
Final speed, v = 0
Time, t = 5 s
We need to find the acceleration of the train. It is equal to the change in velocity divided by time taken. So,
[tex]a=\dfrac{v-u}{t}\\\\a=\dfrac{0-15\ m/s}{5\ s}\\\\a=-3\ m/s^2[/tex]
So, the acceleration of the train is 3 m/s² and it is deaccelerating.
In lifting a heavy weight from the floor, one should use the power of the __________ in order to avoid straining the lower back.
Answer:
Hip and knee extensors
Explanation:
These are gluteus muscles and hamstring muscles the there are the major movers for your body and are very important in pelvic alignment and lower back support during weight lifting
A truck took 4 hours to complete a journey. At the first 1 h 45
min, it travelled at an average speed of 70 km/h. For the rest of
the journey, it travelled at an average speed of 80 km/h. What
was the total distance of the journey?
Answer:302.5
Explanation:
The peak value of an alternating current in a 1500-W device is 6.4 A. What is the rms voltage across it
Answer:
6787.5 V
Explanation:
From the question,
P = IV..................... Equation 1
Where P = Power, I = rms current, V = rms voltage.
make V the subject of the equation
V = P/I................. Equation 2
Given: P = 1500 W, I = 6.4/√2 = 4.525 A
Substitute these values into equation 2
V = 1500(4.525)
V = 6787.5 V
Hence the rms voltage = 6787.5 V
Determine the scalar components Ra and Rb of the force R along the nonrectangular axes a and b. Also determine the orthogonal projection Pa of R onto axi
Answer: Hello your question is incomplete attached below is the complete question
Ra = 1132 N
Rb = 522.6 N
Pa = 679.7 N
Explanation:
To determine the scalar components Ra
[tex]\frac{Ra}{Sin 120^o} = \frac{750}{sin 35^o}[/tex]
therefore : Ra = [tex]\frac{sin120^o * 750}{sin 35^o}[/tex] = 1132 N
To determine the scalar component Rb
[tex]\frac{Rb}{sin 25^o} = \frac{750}{sin 35^o}[/tex]
therefore : Rb = [tex]\frac{sin 25^o * 750}{sin 35^o}[/tex] = 522.6 N
To determine the orthogonal projection Pa of R onto
Pa = 750 cos[tex]25^o[/tex] = 679.7 N
Two small metal cubes with masses 2.0 g and 4.0 g are tied together by a 4.7-cm-long massless string and are at rest on a frictionless surface. Each is charged to +2.5 μC .
What is the energy of this system?
Express your answer using two significant figures.
What is the tension in the string?
Express your answer using two significant figures.
The string is cut. What is the speed of each sphere when they are far apart?
Hint: There are two conserved quantities. Make use of both.
Express your answers using two significant figures. Enter your answers numerically separated by a comma.
(v2g,v4g)=_____m/s
Answer:
1.2J
26 N
(-28, 14) m/s
Explanation:
energy
U = kQq / d = 8.99*10^9 * (2.5*10^-6C)² / 0.047m
U = 0.0562/0.047
U = 1.20 J
to two significant figures
tension
T = kQq / d²
T = U / d
T = 1.2 / 0.047
T = 25.53 N = 26 N to 2 sf
Momentum is conserved, and the initial momentum is zero:
0 = 0.0020 * V2 + 0.0040 * V4
so
V2 = -2 * V4
Energy is also conserved:
½ * 0.0020 * (-2V4)² + ½ * 0.0040 * (V4)² = 1.2 J
-½ * 0.0080 * V4² + ½ * 0.0040 * V4² = 1.2 J
-0.0040V4² + 0.002V4² = 1.2 J
0.0060V4² = 1.2 J
V4² = 1.2/0.0060
V4² = 200
V4 = √200
V4 = 14 m/s
and since V2 = -2 * V4
V2 = -28 m/s
(V2, V4) = (-28, 14)
The energy of this system is 1.2J
The tension in the string is 26 N
The speed of each sphere when they are far apart is (-28, 14) m/s
Calculation of energy, tension, and speed:The energy should be
U = kQq / d
= 8.99*10^9 * (2.5*10^-6C)² / 0.047m
U = 0.0562/0.047
U = 1.20 J
The tension should be
T = kQq / d²
T = U / d
T = 1.2 / 0.047
T = 25.53 N
= 26 N
The speed should be
Since Momentum should be conserved, and the initial momentum is zero:
So,
0 = 0.0020 * V2 + 0.0040 * V4
Now
V2 = -2 * V4
Due to this, Energy is also conserved:
½ * 0.0020 * (-2V4)² + ½ * 0.0040 * (V4)² = 1.2 J
-½ * 0.0080 * V4² + ½ * 0.0040 * V4² = 1.2 J
-0.0040V4² + 0.002V4² = 1.2 J
0.0060V4² = 1.2 J
V4² = 1.2/0.0060
V4² = 200
V4 = √200
V4 = 14 m/s
and now V2 = -2 * V4
V2 = -28 m/s
(V2, V4) = (-28, 14)
Learn more about energy here: https://brainly.com/question/15182235
Why is it harder pushing a car than pushing a bike?
Answer:
When inertia increases, it's because the mass increased, which increases the normal force, which ultimately increases friction.
i need help Mr or ms tutor
Explanation:
Height is the x-axis, and gravitational potential energy is the y-axis. As the height increases, the gravitational potential energy increases linearly.
The siren of a fire engine that is driving northward at 31.0 m/s emits a sound of frequency 2020 Hz. A truck in front of this fire engine is moving northward at 19.0 m/s.
a) What is the frequency of the siren's sound that the fire engine's driver hears reflected from the back of the truck?
b) What wavelength would this driver measure for these reflected sound waves?
Answer:
A. Using
Fl= ( v+vl/v+vz)fz
= (340+19/340+31) x 2020
= 1954.7Hz
Then to find the frequency of sound when reflected from the truck such that the driver becomes the listener
we use
F"= ( v+vz/v+vl) fz
= 340+31/340+19 x 2020
2087.5Hz
B to find the wavelength of sound we use
Wavelength= V+vl/ F"
= 340+31/2087.5= 0.18m
When the Apollo 11 lunar module Eagle lands on the moon it comes to a stop 10m above the surface of the moon.The last 10m it freely falls to the surface of the Moon. i) How long does it take for the Eagle to touch down
Answer:
3.5s
Explanation:
Using equation of motion number 2
S= ut + 1/2 gt²
Substituting
10= 0.5+1.62t²
t= 3.5seconds
You drive your car in a straight line at 15 m/s for 10 kilometers, then at 25 m/s for another 10 kilometers.
a. What is your average speed?
b. Choose the best explanation from among the following:
1) More time is spent at 15 m/s than at 25 m/s.
2) The average of 15 m/s and 25 m/s is 20 m/s.
3) Less time is spent at 15 m/s than at 25 m/s.
Answer:
A) Average speed = 18.75 m/s
B) More time is spent at 15 m/s than at 25 m/s.
Explanation:
Let the first distance be d1 and the second distance be d2.
We are given;
d1 = 10 km = 10000 m
d2 = 10 km = 10000 m
Speed; v1 = 15 m/s
Speed; v2 = 25 m/s
Now, the formula for distance is; Distance = speed x time
Thus:
d1 = v1 x t1
t1 = d1/v1 = 10000/15 = 666.67 seconds
Also,
d2 = v2 x t2
t2 = d2/v2 = 10000/25 = 400 seconds
Average speed = total distance/total time = (10000 + 10000)/(666.67 + 400) = 18.75 m/s
From earlier, since t1 = 666.67 seconds and t2 = 400 seconds, then;
More time at 15 m/s than at 25 m/s.