Answer:
The launched angle θ = 30 degrees, the initial velocity Vo = 20 m/s, the initial horizontal velocity Vox= ?, the initial vertical velocity Voy = ?, the time of flight t = ? the maximum height h = ?
Vox = Vo * (cos of 30 degrees)
Voy = Vo * (sin of 30 degrees)
t = 2 * (Voy / g)
h = Voy * 0.5 t - 1/2 g * (0.5t)^2
I have given the equations for you to use, just plug – in the values and then solve in a step by step manner.
Answer:
approximately 15.68 meters.
Explanation:
Here is how;
First, let's calculate the time of flight for the t-shirt. We can use the vertical motion equation:
y = y0 + v0y * t - 0.5 * g * t^2
where:
y is the vertical displacement (27.7 m)
y0 is the initial vertical position (0 m)
v0y is the vertical component of the initial velocity (v0 * sin(theta))
g is the acceleration due to gravity (9.8 m/s^2)
t is the time of flight
Plugging in the values:
27.7 = 0 + (25.8 * sin(63.6°)) * t - 0.5 * 9.8 * t^2
Simplifying the equation, we get a quadratic equation:
4.9t^2 - (25.8 * sin(63.6°))t + 27.7 = 0
Solving this quadratic equation will give us the time of flight, t. Using the quadratic formula, we find that:
t ≈ 1.23 s
Now, let's find the horizontal displacement of the t-shirt using the horizontal motion equation:
x = x0 + v0x * t
where:
x is the horizontal displacement
x0 is the initial horizontal position (0 m)
v0x is the horizontal component of the initial velocity (v0 * cos(theta))
t is the time of flight
Plugging in the values:
x = 0 + (25.8 * cos(63.6°)) * 1.23
Calculating this:
x ≈ 14.92 m
The t-shirt falls short of reaching the person by the horizontal distance of:
Shortfall = 30.6 m - 14.92 m
Calculating this:
Shortfall ≈ 15.68 m
Therefore, the t-shirt will be approximately 15.68 meters short of reaching the person.
A 1.5kg block slides along a frictionless surface at 1.3m/s . A second block, sliding at a faster 4.3m/s , collides with the first from behind and sticks to it. The final velocity of the combined blocks is 2.0m/s . What was the mass of the second block?
Answer:
The mass of the second block=0.457 kg
Explanation:
We are given that
m1=1.5 kg
v1=1.3m/s
v2=4.3 m/s
V=2.0 m/s
We have to find the mass of the second block.
[tex]m_1v_1+m_2v_2=(m_1+m_2)V[/tex]
Let m2=m
Substitute the values
[tex]1.5(1.3)+m(4.3)=(1.5+m)(2)[/tex]
[tex]1.95+4.3m=3+2m[/tex]
[tex]4.3m-2m=3-1.95[/tex]
[tex]2.3m=1.05[/tex]
[tex]m=\frac{1.05}{2.3}[/tex]
[tex]m=0.457 kg[/tex]
Hence, the mass of the second block=0.457 kg
Two loudspeakers, 5.5 m apart and facing each other, play identical sounds of the same frequency. You stand halfway between them, where there is a maximum of sound intensity. Moving from this point toward one of the speakers, you encounter a minimum of sound intensity when you have moved 0.25 m . Assume the speed of sound is 340 m/s.
Required:
a. What is the frequency of the sound?
b. If the frequency is then increased while you remain 0.21 m from the center, what is the first frequency for which that location will be a maximum of sound intensity?
c.
Solution :
Let [tex]$d_1=\frac{5.5}{2}[/tex]
= 2.75 m
[tex]d_2 = 0.21 \ m[/tex]
And [tex]$d=|d_1-d_2|$[/tex]
[tex]$d=(d_1+d_2) - (d_1-d_2)$[/tex]
[tex]$d=(2.75+0.21) - (2.75-0.21)$[/tex]
[tex]$d = 2.96-2.54$[/tex]
[tex]d = 0.42 \ m[/tex]
a). At minimum,
[tex]$d=\frac{\lambda}{2}$[/tex]
[tex]$\lambda = 2d$[/tex]
= 2 x 0.42
= 0.84 m
Frequency, [tex]$\nu = \frac{v}{\lambda}$[/tex]
[tex]$=\frac{340}{0.84}$[/tex]
= 404.76 Hz
Therefore, the frequency of he sound, [tex]$\nu$[/tex] = 404.76 Hz
b). At maximum, λ = d = 0.42 m
Therefore, the frequency, [tex]$\nu = \frac{v}{\lambda}[/tex]
[tex]$=\frac{350}{0.42}$[/tex]
= 809.52 Hz
A surveyor measures the distance across a straight river by the following method: Starting directly across from a tree on the opposite bank, he walks x = 106 m along the riverbank to establish a baseline. Then he sights across to the tree. The angle from his baseline to the tree is = 32.8°. How wide is the river?
Answer:
x = 68.3 m
Explanation:
tan 32.8 = x / 106
The potential difference between the plates of a capacitor is 234 V. Midway between the plates, a proton and an electron are released. The electron is released from rest. The proton is projected perpendicularly toward the negative plate with an initial speed. The proton strikes the negative plate at the same instant the electron strikes the positive plate. Ignore the attraction between the two particles, and find the initial speed of the proton.
I have tried looking at the cramster.com solution manual and do not like the way it is explained. Simply put, I cannot follow what is going on and I am looking for someone who can explain it in plain man's terms and help me understand and get the correct answer. I am willing to give MAX karma points to anyone who can help me through this. Thank you kindly.
Answer:
The speed of proton is 2.1 x 10^5 m/s .
Explanation:
potential difference, V = 234 V
let the initial speed of the proton is v.
The kinetic energy of proton is
KE = q V
[tex]0.5 mv^2 = e V \\\\0.5\times 1.67\times 10^{-27} v^2 = 1.6\times 10^{-19} \times 234\\\\v=2.1\times 10^5 m/s[/tex]
Difference between uniform motion and non uniform motion
Answer:
When an object covers equal distance in an equal interval of time, it is uniform motion but when an object covers unequal distance in an equal interval of time, it is called non uniform motion.
Two train 75 km apart approach each other on parallel tracks, each moving at
15km/h. A bird flies back and forth between the trains at 20km/h until the trains pass
each other. How far does the bird fly?
Answer:
The correct solution is "37.5 km".
Explanation:
Given:
Distance between the trains,
d = 75 km
Speed of each train,
= 15 km/h
The relative speed will be:
= [tex]15 + (-15)[/tex]
= [tex]30 \ km/h[/tex]
The speed of the bird,
V = 15 km/h
Now,
The time taken to meet will be:
[tex]t=\frac{Distance}{Relative \ speed}[/tex]
[tex]=\frac{75}{30}[/tex]
[tex]=2.5 \ h[/tex]
hence,
The distance travelled by the bird in 2.5 h will be:
⇒ [tex]D = V t[/tex]
[tex]=15\times 2.5[/tex]
[tex]=37.5 \ km[/tex]
Consider two closely spaced and oppositely charged parallel metal plates. The plates are square with sides of length L and carry charges Q and -Q on their facing surfaces. What is the magnitude of the electric field in the region between the plates
Answer:
E_ {total} = [tex]\frac{Q }{L^2 \epsilon_o}[/tex]
Explanation:
In this exercise you are asked to calculate the electric field between two plates, the electric field is a vector
E_ {total} = E₁ + E₂
E_ {total} = 2 E
where E₁ and E₂ are the fields of each plate, we have used that for the positively charged plate the field is outgoing and for the negatively charged plate the field is incoming, therefore in the space between the plates for a test charge the two fields point in the same direction
to calculate the field created by a plate let's use Gauss's law
Ф = ∫ E . dA = q_{int} /ε₀
As a Gaussian surface we use a cylinder with the base parallel to the plate, therefore the direction of the electric field and the normal to the surface are parallel, therefore the scalar product is reduced to the algebraic product.
E 2A = q_{int} / ε₀
where the 2 is due to the surface has two faces
indicate that the surface has a uniform charge for which we can define a surface density
σ = q_{int} / A
q_{int} = σ A
we substitute
E 2A = σ A /ε₀
E = σ / 2ε₀
therefore the total field is
E_ {total} = σ /ε₀
let's substitute the density for the charge of the whole plate
σ= Q / L²
E_ {total} = [tex]\frac{Q }{L^2 \epsilon_o}[/tex]
A plastic dowel has a Young's Modulus of 1.50 ✕ 1010 N/m2. Assume the dowel will break if more than 1.50 ✕ 108 N/m2 is exerted. What is the maximum force (in kN) that can be applied to the dowel assuming a diameter of 2.30 cm?
A.
52.3 kN
B.
62.3 kN
C.
72.3 kN
D.
42.3 N
Answer:
cobina
Explanation:
me 2
A 2kg ball is rolled along the floor for 0.8 m at a constant speed of 6 m/s. What is the work done by gravity?
A, 0
B, 16 J
C, 72 J
D, 450 J
E, 90 J
=F×s×cosa=2×g×0,8×cos90°= 0
The work done by gravity on a ball of 2 kg which is moving with a constant speed of 6 meter per second is zero. Thus, the correct option is A.
What is Work?Work is the energy transfer to or from an object through the application of force along with the displacement. For a constant force aligned with the direction of motion, the work done is equal to the product of the force strength which is applied and the distance traveled by the object.
Work = Force × Displacement
Force = Mass × Acceleration
Acceleration of the ball is zero as it is moving with a constant speed. Therefore, the work done by the gravity is zero.
Therefore, the correct option is A.
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Ashley, a psychology major, remarks that she has become interested in the study of intelligence. In other words, Ashley is interested in?
Group of answer choices.
a) the capacity to learn from experience, solve problems, and to adapt to new situations.
b) how behavior changes as a result of experience.
c) the factors directing behavior toward a goal.
d) the ability to generate novel
Answer:
a) the capacity to understand the world, think rationally, and use resources effectively.
Explanation:
Psychology can be defined as the scientific study of both the consciousness and unconsciousness of the human mind such as feelings, emotions and thoughts, so as to understand how it functions and affect human behaviors in contextual terms.
This ultimately implies that, psychology focuses on studying behaviors and the mind that controls it.
In this scenario, Ashley who is a psychology major, stated that she's interested in the study of intelligence.
Intelligence can be defined as a measure of the ability of an individual to think, learn, proffer solutions to day-to-day life problems and effectively make informed decisions.
In other words, Ashley is interested in the capacity of humans to understand the world, think rationally, and use resources effectively to produce goods and services that meet the unending requirements, needs or wants of the people (consumers or end users) living around the world.
A smokestack of height H = 50 m emits a pollutant in a 3 m/s wind. The plume is carried downwind by advection (wind speed U = 3 m/s) and is simultaneously dispersing vertically with a turbulent diffusion coefficient D. The vertical diffusion causes the plume to widen vertically over time, with halfâwidth (distance from centerline to edge) increasing as:
half width = 2 â2Dt
The plume reaches the ground some distance L downwind of the base of the smokestack (see sketch in book on page 203)
a. If L = 2 km, estimate the value of the turbulent diffusion coefficient D.
b. Under the same wind speed and turbulence conditions, what would be the value of L if the smokestack were twice as high?
Answer:
a) 0.46875
b) 8 km
Explanation:
Smokestack height ( H ) = 50 m
speed of pollutant / wind speed = 3 m/s
Half width = 2 [tex]\sqrt{2Dt }[/tex] = 50 m ---- ( 1 )
a) If L = 2 km
value of turbulent diffusion coefficient D
back to equation 1
50 = 2 √ 2 * D * ( 2000/3 )
2500 = 4 * 2 * D * ( 2000/3 )
D = 2500 / ( 8 * ( 2000/3 ) )
= 0.46875
where : time to travel ( t ) = Distance / speed = 2000 / 3
b) when the smoke stack = 50 * 2 = 100 m
L = 800 m = 8 km
attached below is the detailed solution
Identify the factors that affect the intensity of radiation detected from a radioactive source. Select one or more: The color of the source Type of emission from the source Distance of the detector from the source Type of materials between the source and the detector
The intensity of radiation is the defined as amount of energy per surface angle which can be used to determine the amount of energy emitting from a source that will hit another surface.
The factors that affect the intensity of radiation are
Type of emission from the source :This can be alpha, gamma, beta or electromagnetic rays etc
Distance of the detector from the source: The shorter the distance between the source and the detector, the more the effect and vice versa for the longer the distance.
Type of materials between the source and the detector: The type of material between the source and the detector will tell how absorbing and penetrating the radiation is.
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1 A thing ring has a mass of 6kg and a radius of 20cm. calculate the rotational inertia.
Answer:
2400kgm²
Explanation:
Rotational inertia=mass x radius²
Michelson and Morley concluded from the results of their experiment that Group of answer choices the experiment was successful in not detecting a shift in the interference pattern. the experiment was a failure since they detected a shift in the interference pattern. the experiment was a failure since there was no detectable shift in the interference pattern. the experiment was successful in detecting a shift in the interference pattern.
Answer:
The results of the experiment indicated a shift consistent with zero, and certainly less than a twentieth of the shift expected if the Earth's velocity in orbit around the sun was the same as its velocity through the ether.
Explanation:
Is there any absolute rest or motion? Describe the types of motion with one example of each type
Is it true that as we gain mass the force of gravity on us decreases
Answer:
No. As we gain mass the force of gravity on us does not decrease
A friend lends you the eyepiece of his microscope to use on your own microscope. He claims that since his eyepiece has the same diameter as yours but twice the focal length, the resolving power of your microscope will be doubled. Is his claim valid? Explain.
Answer:
The resolving power remains same.
Explanation:
The resolving power of the lens is directly proportional to the diameter of the lens not on the focal length.
As the diameter is same but the focal length is doubled so the resolving power remains same.
The resistance of a thermistor over a limited range of temperature is given by the equation R= ( c/T-203 )where c is a constant and T is the absolute temperature. What will be the temperature on the Celsius scale of the thermistor at absolute temperature of T = 300K?
Temperature in kelvin scale=300k
We know
[tex]\boxed{\sf 0°C=273K}[/tex]
[tex] \\ \Large\sf\longmapsto 300K[/tex]
[tex] \\ \Large\sf\longmapsto 300-273[/tex]
[tex] \\ \Large\sf\longmapsto 27°C[/tex]
Explain how newton's first law of motion follows from second law?
Answer:
Newton's First Law of Motion states that a body will stay at rest or continue its path with constant velocity unless an external force acts upon it. Newton's Second Law of Motion states that the net force that acts upon a body is equal to the mass of the body multiplied by the acceleration due to the net force.
Partial tides _______. Question 7 options: represent various components of local tides that are resolved mathematically are predicted individually are added together to predict the height and timing of astronomical tides All of the above are correct. Only a and c are correct.
Complete Question
Partial tides __________.
Question 7 options:
a. represent various components of local tides that are resolved mathematically
b. are added together to predict the height and timing of astronomical tides
c. consist of 4 components due to the influence of celestial bodies
d. consist of up to 60 components due to astronomical and non-astronomical factors
e. All of the above except c are correct.
Answer:
Option E
Explanation:
Generally
Partial tides represent various components of local tides that are resolved mathematically
Partial tides are added together to predict the height and timing of astronomical tides
Partial tides consist of up to 60 components due to astronomical and non-astronomical factors
But Partial tides do not consist of 4 components due to the influence of celestial bodies
Therefore
All of the above except c are correct.
Option E
A space ship has four thrusters positioned on the top and bottom, and left and right as shown below. The thrusters can be operated independently or together to help the ship navigate in all directions.
Initially, the Space Probe is floating towards the East, as shown below, with a velocity, v. The pilot then turns on thruster #2.
Select one:
a.
Space ship will have a velocity to the West and will be speeding up.
b.
Space ship will have a velocity to the East and will be speeding up.
c.
Space ship will have a velocity to the East and will be slowing down.
d.
Space ship will have a velocity to the West and will be slowing down.
e.
Ship experiences no change in motion.
Answer:
The correct answer is - c. Spaceship will have a velocity to the East and will be slowing down.
Explanation:
In this case, if turned on thruster #2 then it will exert force on the west side as thruster 2 is on the east side and it can be understood by Newton's third law that says each action has the same but opposite reaction.
As the spaceship engine applies force on the east side then according to the law the exhauster gas applies on towards west direction. It will try to decrease the velocity of the spaceship however, the direction of floating still be east side initally.
Which of the following is not true about Triton, the large moon of Neptune? It is more reflective than Earth's Moon. It is larger than Earth's Moon. It is in a retrograde orbit. It has a thin atmosphere. It has nitrogen geysers.
Answer:
Triton is the largest of Neptune's 13 moons. It is unusual because it is the only large moon in our solar system that orbits in the opposite direction of its planet's rotation―a retrograde orbit. ... Like our own moon, Triton is locked in synchronous rotation with Neptune―one side faces the planet at all times.
A physical pendulum in the form of a planar object moves in simple harmonic motion with a frequency of 0.680 Hz. The pendulum has a mass of 2.00 kg, and the pivot is located 0.340 m from the center of mass. Determine the moment of inertia of the pendulum about the pivot point.
Answer:
Therefore, the moment of inertia is:
[tex]I=0.37 \: kgm^{2} [/tex]
Explanation:
The period of an oscillation equation of a solid pendulum is given by:
[tex]T=2\pi \sqrt{\frac{I}{Mgd}}[/tex] (1)
Where:
I is the moment of inertiaM is the mass of the pendulumd is the distance from the center of mass to the pivotg is the gravityLet's solve the equation (1) for I
[tex]T=2\pi \sqrt{\frac{I}{Mgd}}[/tex]
[tex]I=Mgd(\frac{T}{2\pi})^{2}[/tex]
Before find I, we need to remember that
[tex]T = \frac{1}{f}=\frac{1}{0.680}=1.47\: s[/tex]
Now, the moment of inertia will be:
[tex]I=2*9.81*0.340(\frac{1.47}{2\pi})^{2}[/tex]
Therefore, the moment of inertia is:
[tex]I=0.37 \: kgm^{2} [/tex]
I hope it helps you!
A visible violet light emits light with a wavelength of 4.00 × 10-7 m.
Calculate the frequency of the violet light.
A)6.30 × 10 -1 Hz
B)7.50 × 10 14 Hz
C)6.30 × 10 24 Hz
D)7.50 × 10 1 Hz
Answer:
The correct option is B. 7.5 * 10¹⁴ Hz
Explanation:
Frequency = (speed) / (wavelength)
= (3 x 10⁸ m/s) / (4 x 10⁻⁷ m)
= (3/4 x 10¹⁵) ( m / m - s )
= (0.75 x 10¹⁵) /sec
= 7.5 x 10¹⁴ Hz
= 750,000 GHz
Answer:
Mark Brainliest please
answer is
Explanation:
For any wave,
Frequency = (speed) / (wavelength)
= (3 x 10⁸ m/s) / (4 x 10⁻⁷ m)
= (3/4 x 10¹⁵) ( m / m - s )
= (0.75 x 10¹⁵) /sec
= 7.5 x 10¹⁴ Hz
= 750,000 GH
Convert the unit of 0.00023 kilograms into grams. (Answer in scientific notation)
Answer:
2.3 × [tex]10^{-1}[/tex]
Explanation:
1 kg = 1000 g.
0.00023 kg x 1000 g = 0.23 grams
Answer:
0.23×10⁴
Explanation:
kilogram to gram ÷ 1000
0.00023kg ÷ 1000
=0.23g
scientific notation=0.23×10⁴
1 Poin Question 4 A 85-kg man stands in an elevator that has a downward acceleration of 2 m/s2. The force exerted by him on the floor is about: (Assume g = 9.8 m/s2) А ON B 663 N C) 833 N D) 1003 N
Answer:
D) 1003 N
Explanation:
Given the following data;
Mass of man = 85 kg
Acceleration of elevator = 2 m/s²
Acceleration due to gravity, g = 9.8 m/s²
To find the force exerted by the man on the floor;
Force = mg + ma
Two divers, G and H, are at depths 20 m and 40 m respectively
below the water surface in lake. The pressure on G is P, while
the pressure on H is P2 if the atmospheric pressure is equivalent
to 10 m of water, then the value of P2/P1 is.
A. 1.67.
B. 2.00.
C. 0.50.
D. 0.60.
Answer:
B
Explanation:
P1/P1 = 40/20
=2
A) In terms of electrolysis, it’s been said from multiple sources online that “Using water's density and relative atomic populations, it is estimated by a mass balance that approximately 2.38 gallons of water are consumed as a feedstock to produce 1 kg of hydrogen gas (14.13 liters), assuming no losses.”
B) However, 1 Gallon of water is said to contain approximately 4,707 liters of hydrogen.
How can both statements be correct under normal atmospheric conditions, since even with 80% efficiency of current PEM electrolyzers the first statement (A) is nowhere near the +4,000 liters of the second approximation (B)?
Answer:
hhhhhhhjjjkkllkcftkbgfjknhglncg
Two 51 g blocks are held 30 cm above a table. As shown in the figure, one of them is just touching a 30-long spring. The blocks are released at the same time. The block on the left hits the table at exactly the same instant as the block on the right first comes to an instantaneous rest. What is the spring constant?
The concept of this question can be well understood by listing out the parameters given.
The mass of the block = 51 g = 51 × 10⁻³ kgThe distance of the block from the table = 30 cmLength of the spring = 30 cmThe purpose is to determine the spring constant.
Let us assume that the two blocks are Block A and Block B.
At point A on block A, the initial velocity on the block is zero
i.e. u = 0
We want to determine the time it requires for Block A to reach the table. The can be achieved by using the second equation of motion which can be expressed by using the formula.
[tex]\mathsf{S = ut + \dfrac{1}{2}gt^2}[/tex]
From the above formula,
The distance (S) = 30 cm; we need to convert the unit to meter (m).
Since 1 cm = 0.01 mThen, 30cm = 0.3 mThe acceleration (g) due to gravity = 9.8 m/s²
∴
inputting the values into the equation above, we have;
[tex]\mathsf{0.3 = (0)t + \dfrac{1}{2}*(9.80)*(t^2)}[/tex]
[tex]\mathsf{0.3 = \dfrac{1}{2}*(9.80)*(t^2)}[/tex]
[tex]\mathsf{0.3 =4.9*(t^2)}[/tex]
By dividing both sides by 4.9, we have:
[tex]\mathsf{t^2 = \dfrac{0.3}{4.9}}[/tex]
[tex]\mathsf{t^2 = 0.0612}[/tex]
[tex]\mathsf{t = \sqrt{0.0612}}[/tex]
[tex]\mathbf{t =0.247 \ seconds}[/tex]
However, block B comes to an instantaneous rest on point C. This is achieved by the dropping of the block on the spring. During this process, the spring is compressed and it bounces back to oscillate in that manner. The required time needed to get to this point C is half the period, this will eventually lead to the bouncing back of the block with another half of the period, thereby completing a movement of one period.
By applying the equation of the time period of a simple harmonic motion.
[tex]\mathbf{T = 2 \pi \sqrt{\dfrac{m}{k}}}[/tex]
where the relation between time (t) and period (T) is:
[tex]\mathsf{t = \dfrac{T}{2}}[/tex]
T = 2t
T = 2(0.247)
T = 0.494 seconds
[tex]\mathbf{T = 2 \pi \sqrt{\dfrac{m}{k}}}[/tex]
By making the spring constant (k) the subject of the formula:
[tex]\mathbf{\dfrac{T}{2 \pi } = \sqrt{ \dfrac{m}{k}}}[/tex]
[tex]\Big(\dfrac{T}{2 \pi }\Big)^2 = { \dfrac{m}{k}[/tex]
[tex]\dfrac{T^2}{(2 \pi)^2 }= { \dfrac{m}{k}[/tex]
[tex]\mathsf{ T^2 *k = 2 \pi^2*m} \\ \\ \mathsf{ k = \dfrac{2 \pi^2*m}{T^2}}[/tex]
[tex]\mathsf{ k =\Big( \dfrac{(2 \pi)^2*(51 \times 10^{-3})}{(0.494)^2} \Big) N/m}[/tex]
[tex]\mathbf{ k =8.25 \ N/m}[/tex]
Therefore, we can conclude that the spring constant between the two 51 g blocks held at a distance 30 cm above a table as a result of instantaneous rest caused by the compression of the spring is 8.25 N/m.
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11. From this lab we can conclude that a) the heat transferred when objects are rubbed together creates an energy that can cause objects to move towards or away from each other. b) objects such as balloons and sweaters have a natural affinity towards each other. They will attract each other whether they are rubbed together or not. c) charges exert forces on other charges. do) charges do not exert forces on other charges.
Answer:
c) charges exert forces on other charges.
Explanation:
When two different materials are rubbed together, there is a transfer of electrons from one material to the other material so this causes one object to become positively charged and the other object is negatively charged so they will attract each other not repel each other. Charges exert forces on other charges i.e. opposite charges attract each other whereas similar charges repel each other so in both cases force are exerted on one another.
