Answer:
[tex]q_1=9.9998\mu C[/tex] and [tex]q_2=0.0002\mu C[/tex]
Or
[tex]q_1=0.00016\mu C[/tex] and [tex]q_2=9.99984\mu C[/tex]
Explanation:
We are given that
Force between two charges=0.036 N=[tex]36\times 10^{-3}N[/tex]
Distance between two charges, r=2cm=[tex]2\times 10^{-2}[/tex]m
1m=100cm
Sum of two charges=[tex]10\mu C[/tex]
Let one charge=[tex]q_1=q\mu C=q\times 10^{-6}C[/tex]
[tex]q_2=(10-q)\times 10^{-6} C[/tex]
We know that
Electric force between two charges
[tex]F=\frac{kq_1q_2}{r^2}[/tex]
Where [tex]k=\frac{1}{4\pi \epsilon_0}=9\times 10^{9}[/tex]
Using the formula
[tex]36\times 10^{-3}=9\times 10^{9}\times \frac{q\times 10^{-6}\times(10-q)\times 10^{-6}}{(2\times 10^{-2})^2}[/tex]
[tex]\frac{144\times 10^{-7}}{9\times 10^{9}\times 10^{-12}}=q(10-q)[/tex]
[tex]0.0016=10q-q^2[/tex]
[tex]q^2-10q+0.0016=0[/tex]
[tex]10000q^2-100000q+16=0[/tex]
[tex]q=\frac{100000\pm\sqrt{(100000)^2-4\times 10000\times 16}}{2\times 10000}[/tex]
Using the formula
[tex]x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]
[tex]q=9.999[/tex] and [tex]q=0.00016[/tex]
[tex]q_2=10-9.9998=0.0002[/tex]
[tex]q_2=10-0.00016=9.99984[/tex]
Hence, two charges are
[tex]q_1=9.9998\mu C[/tex] and [tex]q_2=0.0002\mu C[/tex]
Or
[tex]q_1=0.00016\mu C[/tex] and [tex]q_2=9.99984\mu C[/tex]
How many planets on the solar system?
Answer:
8
Explanation:
tsijtsiztuztuistizrizturzurz
Answer:
8
Explanation:
Mercury, Venus, earth , Mars, jupiter, saturn , Uranus,Neptune
In a certain region of space the electric potential increases uniformly from east to west and does not vary in any other direction. The electric field:Group of answer choicespoints east and varies with positionpoints east and does not vary with positionpoints west and varies with positionpoints west and does not vary with positionpoints north and does not vary with position
Answer:
Explanation:
The relation between electric field and potential difference is as follows
E = - dV / dr
That means if dV is positive , E is negative . In other words , if potential increases , E is negative or in opposite direction in which potential increases .
Here the electric potential increases uniformly from east to west , that means electric field is from west to east . Since potential is uniformly increasing that means
dV / dr = constant
E = constant
Electric field is constant .
So the option which is correct is
" points east and does not vary with position " .
Plzzz answer this question correctly
Answer:
changing the direction in which a force is exerted
Which two statements help explain why digital storage of data is so reliable?
A. Memory chips are sturdy.
U B. Digital data usually deteriorate over time.
C. It is usually possible to recover data from a memory chip even
when the device containing it is broken.
D. Digital data are easier to copy than analog data are, making them
more accessible to thieves.
Answer:
A. Memory chips are sturdy.
C. It is usually possible to recover data from a memory chip even when the device containing it is broken.
Explanation:
Digital storage of data refers to the process which typically involves saving computer files or documents on magnetic storage devices usually having flash memory. Some examples of digital storage devices are hard drives, memory stick or cards, optical discs, cloud storage, etc.
A reliable storage ensures that computer files or documents are easily accessible and could be retrieved in the event of a loss.
The two statements which help explain why digital storage of data is so reliable are;
A. Memory chips are sturdy: they are designed in such a way that they are compact and firm.
C. It is usually possible to recover data from a memory chip even when the device containing it is broken.
Answer:
A and C
Explanation:
got it right on a p e x
Use Newton's laws of motion to explain why it is important that baseballs and softballs each have a small acceptable range of masses.
Explanation:
new non law neutron means neutral then it's important that baseball and softball features small respectable range of masses soft it means that when a ball hits anything hard it comes back by the Newton Law if the baseball is big and the small boy small and then if the contract with each other they ignore triple so when a ball hits the wall if the comeback because of the Mutants and when a big ball if we throw it to the wall it doesn't come that it comes back but in a very low way because it contains less neutrons in it if it is helpful please share with me
A cyclist cover 6km in 20minutes. His speed is
Answer:
The speed of a cyclist is 0.3 km/min.
Explanation:
Given
The distance d = 6km Time t = 20 minutesTo determine
We need to determine the speed of a cyclist.
In order to determine the speed of a cyclist, all we need to do is to divide the distance covered by a cyclist by the time taken to cover the distance.
We know the formula involving speed, time, and distance
[tex]s=\frac{d}{t}[/tex]
where
s = speedd = distance coveredt = time takensubstitute d = 6, and t = 20 in the formula
[tex]s=\frac{d}{t}[/tex]
[tex]s=\frac{6}{20}[/tex]
Cancel the common factor 2
[tex]s=\frac{3}{10}[/tex]
[tex]s=0.3[/tex] km/min
Thus, the speed of a cyclist is 0.3 km/min.
A 1 m3tank containing air at 10oC and 350 kPa is connected through a valve to another tank containing 3 kg of air at 35oC and 150 kPa. Now the valve is opened, and the entire system is allowed to reach thermal equilibrium with the surroundings, which are at 20oC. Determine the volume of the second tank and the final equilibrium pressure of air.
Answer:
- the volume of the second tank is 1.77 m³
- the final equilibrium pressure of air is 221.88 kPa ≈ 222 kPa
Explanation:
Given that;
[tex]V_{A}[/tex] = 1 m³
[tex]T_{A}[/tex] = 10°C = 283 K
[tex]P_{A}[/tex] = 350 kPa
[tex]m_{B}[/tex] = 3 kg
[tex]T_{B}[/tex] = 35°C = 308 K
[tex]P_{B}[/tex] = 150 kPa
Now, lets apply the ideal gas equation;
[tex]P_{B}[/tex] [tex]V_{B}[/tex] = [tex]m_{B}[/tex]R[tex]T_{B}[/tex]
[tex]V_{B}[/tex] = [tex]m_{B}[/tex]R[tex]T_{B}[/tex] / [tex]P_{B}[/tex]
The gas constant of air R = 0.287 kPa⋅m³/kg⋅K
we substitute
[tex]V_{B}[/tex] = ( 3 × 0.287 × 308) / 150
[tex]V_{B}[/tex] = 265.188 / 150
[tex]V_{B}[/tex] = 1.77 m³
Therefore, the volume of the second tank is 1.77 m³
Also, [tex]m_{A}[/tex] = [tex]P_{A}[/tex][tex]V_{A}[/tex] / R[tex]T_{A}[/tex] = (350 × 1)/(0.287 × 283) = 350 / 81.221
[tex]m_{A}[/tex] = 4.309 kg
Total mass, [tex]m_{f}[/tex] = [tex]m_{A}[/tex] + [tex]m_{B}[/tex] = 4.309 + 3 = 7.309 kg
Total volume [tex]V_{f}[/tex] = [tex]V_{A}[/tex] + [tex]V_{B}[/tex] = 1 + 1.77 = 2.77 m³
Now, from ideal gas equation;
[tex]P_{f}[/tex] = [tex]m_{f}[/tex]R[tex]T_{f}[/tex] / [tex]V_{f}[/tex]
given that; final temperature [tex]T_{f}[/tex] = 20°C = 293 K
we substitute
[tex]P_{f}[/tex] = ( 7.309 × 0.287 × 293) / 2.77
[tex]P_{f}[/tex] = 614.6211119 / 2.77
[tex]P_{f}[/tex] = 221.88 kPa ≈ 222 kPa
Therefore, the final equilibrium pressure of air is 221.88 kPa ≈ 222 kPa
Astronauts aboard the ISS move at about 8000 m/s, relative to us when we look upward.How long does an astronaut need to stay aboard the space station to be a full second youngerthan people on the ground? Please show and explain how you would set-up the problem,before you actually try to solve it. If you cannot solve it exactly, please try to offer an estimate.(5 pts)
Answer:
#_time = 7.5 10⁴ s
Explanation:
In order for the astronaut to be younger than the people on earth, it follows that the speed of light has a constant speed in vacuum (c = 3 108 m / s), therefore with the expressions of special relativity we have.
t = [tex]\frac{t_p}{ \sqrt{1- (v/c)^2} }[/tex]
where t_p is the person's own time in an immobile reference frame,
[tex]t_{p} = t \sqrt{1 - (\frac{v}{c})^2 }[/tex]
let's calculate
we assume that the speed of the space station is constant
[tex]t_p = 1 \sqrt{1 - \frac{8 \ 10^3}{3 \ 10^8} }[/tex]
[tex]t_p = 1 \sqrt{1- 2.6666 \ 10^{-5}}[/tex]
t_ = 0.99998666657 s
therefore the time change is
Δt = t - t_p
Δt = 1 - 0.9998666657
Δt = 1.3333 10⁻⁵ s
this is the delay in each second, therefore we can use a direct rule of proportions. If Δt was delayed every second, how much second (#_time) is needed for a total delay of Δt = 1 s
#_time = 1 / Δt
#_time =[tex]\frac{1}{1.3333 \ 10^{-5}}[/tex]
#_time = 7.5 10⁴ s
20. For each improvement in glider design, engineers follow
O A. the written instructions that are provided in the hang glider build kit.
O B. an iterative process of testing, modifying, retesting, and modifying again.
O C. a complicated process of checks and balances while obtaining financing.
O D. a mathematical process, rejecting designs that don't follow blueprint dimensions.
Turn In
A point charge, Q1 = -4.2 μC, is located at the origin. A rod of length L = 0.35 m is located along the x-axis with the near side a distance d = 0.45 m from the origin. A charge Q2 = 10.4 μC is uniformly spread over the length of the rod.Part (a) Consider a thin slice of the rod, of thickness dx, located a distance x away from the origin. What is the direction of the force on the charge located at the origin due to the charge on this thin slice of the rod? Part (b) Write an expression for the magnitude of the force on the point charge, |dF|, due to the thin slice of the rod. Give your answer in terms of the variables Q1, Q2, L, x, dx, and the Coulomb constant, k. Part (c) Integrate the force from each slice over the length of the rod, and write an expression for the magnitude of the electric force on the charge at the origin. Part (d) Calculate the magnitude of the force |F|, in newtons, that the rod exerts on the point charge at the origin.
Answer:
a) attractiva, b) dF = [tex]k \frac{Q_1 \ dQ_2}{dx}[/tex], c) F = [tex]k Q_1 \frac{Q_2}{d \ (d+L)}[/tex], d) F = -1.09 N
Explanation:
a) q1 is negative and the charge of the bar is positive therefore the force is attractive
b) For this exercise we use Coulomb's law, where we assume a card dQ₂ at a distance x
dF = [tex]k \frac{Q_1 \ dQ_2}{dx}[/tex]
where k is a constant, Q₁ the charge at the origin, x the distance
c) To find the total force we must integrate from the beginning of the bar at x = d to the end point of the bar x = d + L
∫ dF = [tex]k \ Q_1 \int\limits^{d+L}_d {\frac{1}{x^2} } \, dQ_2[/tex]
as they indicate that the load on the bar is uniformly distributed, we use the concept of linear density
λ = dQ₂ / dx
DQ₂ = λ dx
we substitute
F = [tex]k \ Q_1 \lambda \int\limits^{d+L}_d \, \frac{dx}{x^2}[/tex]
F = k Q1 λ ([tex]-\frac{1}{x}[/tex])
we evaluate the integral
F = k Q₁ λ [tex](- \frac{1}{d+L} + \frac{1}{d} )[/tex]
F = k Q₁ λ [tex]( \frac{L}{d \ (d+L)})[/tex]
we change the linear density by its value
λ = Q2 / L
F = [tex]k Q_1 \frac{Q_2}{d \ (d+L)}[/tex]
d) we calculate the magnitude of F
F =9 10⁹ (-4.2 10⁻⁶) [tex]\frac{10.4 10x^{-6} }{0.45 ( 0.45 +0.35)}[/tex]
F = -1.09 N
the sign indicates that the force is attractive
Answer:
a)Toward the rod
b)|dF| = k|Q1|Q2(dx/L)/x^2
c)|F| = k|Q1|Q2/(d(d+L))
d)Plug in for answer c and solve
Explanation:
A)
Q1 is negative and Q2 is positive so it is an attractive force to where the rod is located.
B)
The formula for Force due to electric charges is F=kQ1Q2/r^2
In this case, Q2 is distrusted through the length of the rod as opposed to a single point charge. As such Q2 is actually Q2*dx/L as dx is a small portion of the full length, L.
The radius between Q1 and Q2 depends on the section of the rod taken so r will be the variable x distance from Q1.
The force is only from a small portion of the rod so more accurately, we are finding |dF| as opposed to the full force, F, caused by the whole rod.
The final formula is |dF| = k|Q1|Q2(dx/L)/x^2
C)
Integrating with respect to the only changing variable, x, which spans the length of the rod, from radius = d to d+L we get this:
F = integral from d to d+L of k|Q1|Q2(dx/L)/x^2
factor out constants
F = kQ1Q2/L * integral d to d+L(1/x^2)dx
F = kQ1Q2/L * (-1/x)| from d to d+L
F = kQ1Q2/L * (-1/d+L - -1/d)
F = kQ1Q2/L * (-d/(d(d+L)) + (d+L)/(d(d+L))
F = kQ1Q2/L * (L)/(d(d+L))
F = kQ1Q2/(d(d+L))
D)
Plug in the given values into c and you have your answer.
Light of wavelength 425.0 nm in air falls at normal incidence on an oil film that is 850.0 nm thick. The oil is floating on a water layer 1500 nm thick. The refractive index of water is 1.33, and that of the oil is 1.40. The number of wavelengths of light that fit in the oil film is closest to:
Answer:
in oil film λ = 303.57 10⁻⁹ m
in the water film λ = 319.55 10⁻⁹ m
Explanation:
When electromagnetic radiation reaches a material, its propagation is by a process that we call absorption and reflection,
when light reaches a surface it has a mass much greater than the mass of the photons (m = 0), therefore there is an elastic collision where the frequency does not change, due to the speed of light in the material medium changes, therefore the only possibility is that the wavelength in the material changes, to maintain the relationship
v = λ f
in the void we have
c = λ₀ f
we divide the two expression
c / v = λ₀ / λ
the refractive index is
n = c / v
n = λ₀ /λ
λ = λ₀ / n
let's calculate
in oil film
λ = 425 10⁻⁹ / 1.40
λ = 303.57 10⁻⁹ m
in the water film
λ = 425 10⁻⁹ / 1.33
λ = 319.55 10⁻⁹
those wavelengths are in the ultraviolet
The new springs will be identical to the original springs, except the force constant will be 5655.00 N/m smaller. When James removes the original springs, he discovers that the length of each spring expands from 8.55 cm (its length when installed) to 12.00 cm (its length with no load placed on it). If the mass of the car body is 1355.00 kg, by how much will the body be lowered with the new springs installed, compared to its original height
Answer:
Explanation:
For original spring , compression in spring due to a load of 1355 kg is
x = 12 - 8.55 = 3.45 cm = .0345 m
spring constant = W / x
= 1355 x 9.8 / .0345
= 384898.55 N /m
Spring constant of new spring
k = 384898.55 - 5655 = 379243.55 N /m
New compression for new spring
= W / k
= 1355 x 9.8 / 379243.55
= .035 m
= 3.50 cm
Difference of compression = 3.50 - 3.45
= .05 cm .
In later case , car will be more lowered by .05 cm .
Q4. (a) An acre-foot is the volume of water that would cover 1 acre of flat land to a depth of 1
foot. How many gallons are in 1 acre-foot?
Answer:
326,000
Explanation:
One acre-foot equals about 326,000 gallons, or enough water to cover an acre of land, about the size of a football field, one foot deep. An average California household uses between one-half and one acre-foot of water per year for indoor and outdoor use.
What is the correct coefficient for 2H2 + O2 →2H2O
Explanation:
2forH2,1for02,and2forH20
A toy car can go 5 mph. How long would it take to go 12 miles?
In the human arm, the forearm and hand pivot about the elbow joint. Consider a simplified model in which the biceps muscle is attached to the forearm 3.80 cm from the elbow joint. Assume that the person's hand and forearm together weigh 15.0 N and that their center of gravity is 15.0 cm from the elbow (not quite halfway to the hand). The forearm is held horizontally at a right angle to the upper arm, with the biceps muscle exerting its force perpendicular to the forearm.
A. Find the force exerted by the biceps when the hand is empty.
B. Now the person holds a 80.0-N weight in his hand, with the forearm still horizontal. Assume that the center of gravity of this weight is 33.0 cm from the elbow. Find the force now exerted by the biceps.
C. Explain why the biceps muscle needs to be very strong.
D. Under the conditions of part B, find the magnitude of the force that the elbow joint exerts on the forearm.
E. Under the conditions of part B, find the direction of the force that the elbow joint exerts on the forearm.
F. While holding the 80.0-N weight, the person raises his forearm until it is at an angle of 53.0∘ above the horizontal. If the biceps muscle continues to exert its force perpendicular to the forearm, what is this force when the forearm is in this position?
G. Has the force increased or decreased from its value in part B? Explain why this is so, and test your answer by actually doing this with your own arm.
Answer:
Answer is explained in the explanation section below.
Explanation:
Part A)
From conserve moment of force, we have:
F1d1 = F2d2
F1 x (3.80 x [tex]10^{-2}[/tex] m) = 15N x (15 x [tex]10^{-2}[/tex] m)
F1 = [tex]\frac{15 . 15 . 10^{-2} }{3.80 . 10^{-2} }[/tex]
F1 = 59.2 N
Force exerted by the biceps when the hand is empty.
Part B)
The 80 N weight acts at 33 cm and 15 N at 15 cm, then the center of mass is:
x = [tex]\frac{m1x1 + m2x2}{m1+m2}[/tex]
x = [tex]\frac{\frac{80}{9.8} (33 .10^{-2}) + \frac{15}{9.8}(15.10^{-2} }{\frac{80}{9.8} + \frac{15}{9.8} }[/tex]
x = 30.16 x [tex]10^{-2}[/tex] m
Total Weight is:
F = 80N + 15N = 95N
From the conserve moment of force, we have:
F ( 3.8 x [tex]10^{-2}[/tex] ) = 95N (30.16 x [tex]10^{-2}[/tex])
F = 754 N
Part C:
From the above two examples solved, the force exerted by the biceps is higher than downward force, due to this muscle need to be very strong.
Part D)
The force exerted by elbow on the forearm is:
The force exerted by the elbow and biceps are in upward direction and total weight is in downward direction. So, the balancing force in vertical direction is:
F2 + 754N = 95N
F2 = 95N -754N
F2 = -659N
Negative sign shows the force is in downward direction.
Part E)
The bicep muscle acts perpendicular to the forearm, so it is lever arm stays the same. but those of the other two forces decreases as the arm is raised. There tension in the biceps muscle decreases.
Part F)
Angle = 53 degrees.
So,
Force = FcosФ
Force = 754 cos 53
Force = 453.76 N
Part G)
The value of force has gone downwards. It has decreased from that of part B.
What are regular and irregular reflection of light? plz help its
urgent..
Explanation:
Regular reflection: It is the reflection from a smooth surface such that the light rays are evenly parallel to each other and an image is formed. ... Irregular reflection: It is the diffused reflection from uneven surface such that the light rays are not parallel to each other and do not form an image.
How do you think that changing the mass of the pendulum bob will affect the period of the pendulum swing?
convert 0.0345mW
to MW
Answer:
3.45e-11MV
that is ur answer
To fully describe velocity you must have a _____
A. Magnitude and unit
B. Speed and unit
C. Average speed and position
D. Magnitude and direction
1
What kind of adaptation is a long neck on a tortoise? *
(10 Points)
O
A. Structural
B. Behavioral
a
C. Functional
a
D. Physiological
Answer:
The answer is ......... structural adaptation
Explanation:
because structural adaptations is a physical thinng in their body so its A please give me brainliest
To increase the gravitational force between the two objects above, I could
Your answer:
A. increase the mass of the objects and decrease the distance between the objects.
B. increase BOTH the distance and the mass between the two objects
C. decrease the mass of the objects and increase the distance between the objects.
D. decrease BOTH the mass and distance between the two objects.
Answer:
Option A
Explanation:
Cole drives to school from home, starting from rest and accelerating for 10 minutes as he travels 6.0 km to school.
1) What is Cole's acceleration?
2) What is his velocity when he reaches school?
Explanation:
this is the answer for your question. if you have any doubt.
you can send your doubt to:6369514784(what's app)
what happens when a wave passes through a medium ?
Answer:
When waves travel from one medium to another the frequency never changes. As waves travel into the denser medium, they slow down and wavelength decreases. Part of the wave travels faster for longer causing the wave to turn. The wave is slower but the wavelength is shorter meaning frequency remains the same.
Explanation:
Which of the following is the BEST explanation for why oceans have two different types of currents?
Answer:
sddww
Explanation:
szsswa
This is the build up of substance such as pesticides in an organism and occurs when an organism absorb a substance at a rate faster than that at which the substance is lost
Answer:
which the substance is lost by catabolism and excretion.
Explanation:
An aluminum wire having a cross-sectional area equal to 2.20 10-6 m2 carries a current of 4.50 A. The density of aluminum is 2.70 g/cm3. Assume each aluminum atom supplies one conduction electron per atom. Find the drift speed of the electrons in the wire.
Answer:
The drift speed of the electrons in the wire is 2.12x10⁻⁴ m/s.
Explanation:
We can find the drift speed by using the following equation:
[tex] v = \frac{I}{nqA} [/tex]
Where:
I: is the current = 4.50 A
n: is the number of electrons
q: is the modulus of the electron's charge = 1.6x10⁻¹⁹ C
A: is the cross-sectional area = 2.20x10⁻⁶ m²
We need to find the number of electrons:
[tex] n = \frac{6.022\cdot 10^{23} atoms}{1 mol}*\frac{1 mol}{26.982 g}*\frac{2.70 g}{1 cm^{3}}*\frac{(100 cm)^{3}}{1 m^{3}} = 6.03 \cdot 10^{28} atom/m^{3} [/tex]
Now, we can find the drift speed:
[tex]v = \frac{I}{nqA} = \frac{4.50 A}{6.03 \cdot 10^{28} atom/m^{3}*1.6 \cdot 10^{-19} C*2.20 \cdot 10^{-6} m^{2}} = 2.12 \cdot 10^{-4} m/s[/tex]
Therefore, the drift speed of the electrons in the wire is 2.12x10⁻⁴ m/s.
I hope it helps you!
Two students are on a balcony a distance h above the street. One student throws a ball vertically downward at a speed vi; at the same time, the other student throws a ball vertically upward at the same speed. Answer the following symbolically in terms of vi, g, h, and t. (Take upward to be the positive direction.)
(a) What is the time interval between when the first ball strikes the ground and the second ball strikes the ground?
?t = ______
(b) Find the velocity of each ball as it strikes the ground.
For the ball thrown upward vf = ______
For the ball thrown downward vf = ______
(c) How far apart are the balls at a time t after they are thrown and before they strike the ground?
d = _______
Answer:
Explanation:
a )
Time for first ball to reach top position
v = u - gt
0 = vi - gt
t = vi / g
Time to reach balcony while going downwards
= vi /g
Total time = 2 vi / g
Time to go down further to the ground = t₁
Total time = 2 vi / g + t₁
Time for the other ball to go to the ground = t₁
Time difference = ( 2 vi / g + t₁ ) - t₁
= 2vi / g .
( b )
v² = u² + 2gh
For both the throw ,
final displacement = h , initial velocity downwards = vi
( For the first ball also , when it go down while passing the balcony , it acquires the same velocity vi but its direction is downwards.)
vf² = vi² + 2gh
vf = √ ( vi² + 2gh )
(c )
displacement of first ball after time t
s₁ = - vi t + 1/2 g t² [ As initial velocity is upwards , vi is negative ]
displacement of second ball after time t
s₂ = vi t + 1/2 g t²
Difference = d = s₂ - s₁
= vi t + 1/2 g t² - ( - vi t + 1/2 g t² )
d = 2 vi t .
In which situation are waves transmitted?
O A. A patient wears a lead apron at the dentist's office when getting
teeth X-rays.
O B. A light in a swimming pool comes on after dark to prevent
accidents in the water.
O C. A person wears earplugs to prevent hearing damage when fueling
a jet plane at the airport.
O D. A reflective screen is put on a parked car's dashboard to keep the
car from heating up in sunlight.
Answer: B. A light in a swimming pool comes on after dark to prevent
accidents in the water.
A loaded wagon of mass 10,000 kg moving with a speed of 15 m/s strikes a stationary wagon of the same mass making a perfect inelastic collision. What will be the speed of coupled wagons after collision?
Answer:
7.5 m/s
Explanation:
Unfortunately, I don't have an explanation but I guessed the correct answer.