I'm not sure of this question, yet i know you may get angry but, just know your loved and if going through a hard time i'm so so sorry, i know how it feels it sucks. take care of yourself, drink water, listen to that one song that makes you dance in your room and go do what YOU love. your beautiful your perfect and dont listen to what anyone says. i love u, everyone loves u, i know its scary but it happens to everyone life isnt a peice of cake unfortunatly and it sucks but woahh your strong! if you need someone to talk to leave a comment below, happy hoildays everyone!
P.S, im looking into the answer of this question. I may not answer it right on time (since i'm not the smartest...) but i promise ill try my best! Also, i am not doing this for points im doing this to make others smile. I hate how no one checks up on anyone, lots of people are going through a stressful time (including me) and i just want to let everyone know that its okay not to be okay! I hope that you all find happiness in every possible situation, once again HAPPY HOLIDAYS!
A 3 kg mass is travelling in a circle of 0.1 m radius with a speed of 2 m/s. What is the centripetal acceleration?
a = v² / R = (2 m/s)² / (0.1 m) = 40 m/s²
• The net force acting on an object equals the applied force plus the force of friction.
True or False
Answer:
false
Explanation:
ke sath hi hai j clin oncol biol to u h
Uranium-235 undergoes fission, forming krypton-92, barium-141, and 3
neutrons. The mass of the uranium-235 is greater than the total mass of the
products. Which statement explains this difference in mass?
A. Some of the mass was transformed into neutrons during the
process.
O B. Mass was destroyed and disappeared during the process.
C. Some of the mass was transformed into gases during the
process.
D. Mass was transformed into energy during the process.
Answer:
C
Explanation:
Some of the mass
Answer:
D. Mass was transformed into energy during the process.
derivative equation of F = kx
Given that,
The equation is "F=kx"
To find,
The derivative equation.
Solution,
We are given with the given equation which is as follows :
F = kx ...(1)
Where, F is force, k is spring constant and x is distance covered by the spring
Differentiate equation (1) wrt x.
[tex]\dfrac{dF}{dx}=\dfrac{d}{dx}(kx)\\\\\dfrac{dF}{dx}=k[/tex]
As k is constant.
Hence, this is the required solution.
Find the total current through this circuit
Answer:
I=0.02[amp]
Explanation:
Since the resistors are connected in series, we can find the total resistance.
[tex]Rt = 100 + 300+50\\Rt = 450[ohm][/tex]
Now we can use ohm's law which tells us that the voltage is equal to the product of resistance by the current.
[tex]V =I*R\\I=V/R\\I=9/450\\I=0.02[amp][/tex]
A 100 g ball collides elastically with a 300 g ball that is at rest. If the 100 g ball was traveling
in the positive x-direction at 6.20 m/s before the
collision, what are the velocities of the two
balls after the collision?
Answer:
The magnitude of the velocities of the two balls after the collision is 3.1 m/s (each one).
Explanation:
We can find the velocity of the two balls after the collision by conservation of linear momentum and energy:
[tex] P_{1} = P_{2} [/tex]
[tex] m_{1}v_{1_{i}} + m_{2}v_{2_{i}} = m_{1}v_{1_{f}} + m_{2}v_{2_{f}} [/tex]
Where:
m₁: is the mass of the ball 1 = 100 g = 0.1 kg
m₂: is the mass of the ball 2 = 300 g = 0.3 kg
[tex]v_{1_{i}}[/tex]: is the initial velocity of the ball 1 = 6.20 m/s
[tex]v_{2_{i}}[/tex]: is the initial velocity of the ball 2 = 0 (it is at rest)
[tex]v_{1_{f}}[/tex]: is the final velocity of the ball 1 =?
[tex]v_{2_{f}}[/tex]: is the initial velocity of the ball 2 =?
[tex] m_{1}v_{1_{i}} = m_{1}v_{1_{f}} + m_{2}v_{2_{f}} [/tex]
[tex] v_{1_{f}} = v_{1_{i}} - \frac{m_{2}v_{2_{f}}}{m_{1}} [/tex] (1)
Now, by conservation of kinetic energy (since they collide elastically):
[tex] \frac{1}{2}m_{1}v_{1_{i}}^{2} = \frac{1}{2}m_{1}v_{1_{f}}^{2} + \frac{1}{2}m_{2}v_{2_{f}}^{2} [/tex]
[tex] m_{1}v_{1_{i}}^{2} = m_{1}v_{1_{f}}^{2} + m_{2}v_{2_{f}}^{2} [/tex] (2)
By entering equation (1) into (2) we have:
[tex] m_{1}v_{1_{i}}^{2} = m_{1}(v_{1_{i}} - \frac{m_{2}v_{2_{f}}}{m_{1}})^{2} + m_{2}v_{2_{f}}^{2} [/tex]
[tex] 0.1 kg*(6.20 m/s)^{2} = 0.1 kg*(6.2 m/s - \frac{0.3 kg*v_{2_{f}}}{0.1 kg})^{2} + 0.3 kg(v_{2_{f}})^{2} [/tex]
By solving the above equation for [tex]v_{2_{f}}[/tex]:
[tex]v_{2_{f}} = 3.1 m/s [/tex]
Now, [tex]v_{1_{f}}[/tex] can be calculated with equation (1):
[tex] v_{1_{f}} = 6.20 m/s - \frac{0.3 kg*3.1 m/s}{0.1 kg} = -3.1 m/s [/tex]
The minus sign of [tex] v_{1_{f}}[/tex] means that the ball 1 (100g) is moving in the negative x-direction after the collision.
Therefore, the magnitude of the velocities of the two balls after the collision is 3.1 m/s (each one).
I hope it helps you!
What happens to liquid nitrogen when it gets warmer?
Answer:
see below
Explanation:
It boils (or evaporates) and becomes gaseous nitrogen like water turns to water vapor when it gets warm enough
I'm getting conflicting information on brainly about this question. It's either fly-by or GPS satellites...
What constitute the most common way that images of space are captured?
A. Fly-by missions
B. Drones landing on planets
C. GPS satellites in Earth's orbit
D. Astronauts visiting the
international space center
Answer:
C
Explanation:
GPS satellites in Earth's orbit constitute the most common way that images of space are captured.
I hope this helps! And this is the real answer.
a train traveling at 48 m/s begins to slow down as it approaches a bend in the tracks. if it travels around the bend at a speed of 14 m/s and it takes 40 seconds to properly slow down what is the acceleration acting on the train during this time?
Answer:
The acceleration acting on the train during this
time of travel is 0.85m/s²
HOW TO CALCULATE ACCELERATION:
• The acceleration of a moving body can be calculated by using the formula below:
a = (v - u)/t
Where;
1. a = acceleration (m/s²)
2. v = final velocity (m/s) 3. u = initial velocity (m/s)
4. t = time (s)
According to this question, a = ?, v = 48m/s, u = 14m/s, t = 40s.
= (48 - 14)/40 a=
• a = 34/40
• a = 0.85m/s²
Therefore, the acceleration acting on the train during this time of travel is 0.85m/s².
Hope this helps!
Don't forget to mark me as Brainliest.
The illustration shows a roller coaster and indicates four different positions the car might be at as it moves along the track. Identify at which point in the roller coaster's journey does it have the least potential energy and explain why. w
Answer:
I got this question on Ap3x. The answer is Car C...I got it correct
Explanation:
This is because Car C is at the lowest point with the lowest amount of potential energy. Potential energy is stored at it's highest when it has the "potential" to fall, move, or etc. Car C seems to have gone farther down from the high point of the slope, meaning that most of the potential energy transformed into kinetic energy. All in all, Car C has the least potential energy. (Please give Brainliest, or not...your choice but this is the first question that I answered)
if a block weight 60N and is lying on a side with area 2m by 3m. what is the pressure exerted on the surface?
Answer:
10 Pa
Explanation:
Pressure is defined as force per unit area.
Mathematically, P= F/A where P is pressure, F is force and A is area.
Force, F = 60 N
Area, A = 2*3 = 6m²
P = 60/6 = 10 Pa
What is the gravitational force between two identical 5000 kg asteroids whose centers of mass are separated by 100 m?
Answer: 1.67 x 10^-7N
Explanation:
After concluding his research, which statements would Virchow agree with? Check all that apply.
O Living things come from nonliving things.
Cells can come from nonliving materials.
Frogs can come from mud.
O Living things can only
comeyom living things.
Cells come from pre-existing cells.
Answer:
A Living things can only come from living things.
B Cells come from pre-existing cells.
Explanation:
sorry if wrong e d g e 2022
Answer: D,E
Explanation:
I JUST TOOK IT
A projectile is fired vertically from Earth's surface with an initial speed of 3.1 km/s. Neglecting air drag, how far above the surface of Earth will it go
Answer:
The maximum height reached by the projectile is 490.31 km.
Explanation:
Given;
initial velocity of the projectile, u = 3.1 km/s = 3100 m/s
acceleration due to gravity, g = 9.8 m/s²
The maximum height reached by the projectile is calculated as;
v² = u² - 2gh
where;
v is the final velocity of the projectile at maximum height = 0
h is the maximum height reached by the projectile
0 = u² - 2gh
2gh = u²
h = u² / 2g
h = (3100²) / (2 x 9.8)
h = 490306.122 m
h = 490.31 km
Therefore, the maximum height reached by the projectile is 490.31 km.
3. If an x-ray imaging system is operated at 800 mA, 2000 ms, the total
mAs will be?
Point charges are located at 3, 8, and 11 cm along the x-axis(+q, -2q, +q). What is the x-component of the force on the charge located at x=8 cm given that q=1.15nc?
Answer:
Approximately [tex]1.69 \times 10^{-23}\; {\rm N}[/tex].
Explanation:
Look up the value of Coulomb's Constant: [tex]k \approx 8.988 \times 10^{-9}\; {\rm N \cdot m^{2} \cdot C^{-2}}[/tex].
Consider point charges of magnitude [tex]q_{1}[/tex] and [tex]q_{2}[/tex]. If the distance between these charges is [tex]r[/tex], the magnitude of the electrostatic force between them would be [tex](k\, q_{1}\, q_{2}) / (r^{2})[/tex].
In this question, the two [tex](+q)[/tex] charges are [tex]5\; {\rm cm}[/tex] and [tex]3\; {\rm cm}[/tex] away from the center [tex](-2\, q)[/tex] charge, respectively. Convert units to standard unit of distance (meters, [tex]{\rm m}[/tex]) and charge (coulombs, [tex]{\rm C}[/tex]):
[tex]q = 1.15 \; {\rm nC} = 1.15 \times 10^{-9}\; {\rm C}[/tex].
[tex]\begin{aligned} 5\; {\rm cm} = 5\; {\rm cm} \times \frac{1\; {\rm m}}{100\; {\rm cm}} = 0.05\; {\rm m} \end{aligned}[/tex].
[tex]\begin{aligned} 3\; {\rm cm} = 3\; {\rm cm} \times \frac{1\; {\rm m}}{100\; {\rm cm}} = 0.03\; {\rm m} \end{aligned}[/tex].
The magnitude of the electrostatic forces on the [tex](-2\, q)[/tex] charge would be:
[tex]\begin{aligned}\frac{k\, q_{1}\, q_{2}}{r^{2}} &\approx \frac{1}{(0.05\; {\rm m})^{2}} \\ &\quad \times (8.988 \times 10^{-9}\; {\rm N \cdot m^{2} \cdot C^{-2}})\\ &\quad \times ((-2) \, (1.15\times 10^{-9}\; {\rm C}))\, (1.15\times 10^{-9}\; {\rm C})) \\ &\approx 9.509\times 10^{-24}\; {\rm N}\end{aligned}[/tex].
[tex]\begin{aligned}\frac{k\, q_{1}\, q_{2}}{r^{2}} &\approx \frac{1}{(0.03\; {\rm m})^{2}} \\ &\quad \times (8.988 \times 10^{-9}\; {\rm N \cdot m^{2} \cdot C^{-2}})\\ &\quad \times ((-2) \, (1.15\times 10^{-9}\; {\rm C}))\, (1.15\times 10^{-9}\; {\rm C})) \\ &\approx 2.641\times 10^{-23}\; {\rm N}\end{aligned}[/tex].
Since the charges are of opposite sign, the [tex](-2\, q)[/tex] charge would attract both of the [tex](+q)[/tex] charges. In particular, the (approximately) [tex]9.509\times 10^{-24}\; {\rm N}[/tex] force would point to the left. The (approximately) [tex]2.641 \times 10^{-23}\; {\rm N}[/tex] force would point to the right.
As a result, the net force on the [tex](-2\, q)[/tex] charge would point to the right. The magnitude of the net force on this charge would be approximately [tex]2.641 \times 10^{-23}\; {\rm N} - 9.509\times 10^{-24}\; {\rm N} \approx 1.69 \times 10^{-23}\; {\rm N}[/tex].
how much power an appliance uses
Answer:
it depends on the appliance
Explanation:
bigger appliances will use more power, smaller appliances will use a lesser amount
as an object falls a greater distance, does the speed at which it is falling increase or decrease
Answer:
Increase
Explanation:
If an object is falling for a greater distance, it will gain momentum and speed up.
I need help with magnitudes of net force
How do objects at rest and in motion respond in the presence of an external, unbalanced force?
Explanation:
Objects at rest and in motion respond to the presence of an external unbalanced force by simple changing their magnitude of motion or position.
We have this knowledge from Newtons first law of motion "a body will remain in a state of rest or continue with uniform motion unless if it is acted upon by an external force".
When an external force acts on a body at rest, it will change the position of the body or set it motion. For a body in motion, an external force can make they come to rest or change the motion of the bodyDuring a group project, two students constructed a simple machine to add to their Rube Goldberg project. They were told to create one that demonstrates the concept of force-distance tradeoff. One student created model A and the other students created model B.
image
Using the CER method, which model best demonstrates the force-distance tradeoff and why?
Answer:
Image B
Explanation:
although I'm not exactly sure, i've recently gotten this question as well. but model B demonstrates the force- distance trade off because you can see how in that image them distance is increased in the force is decreased with the object being shorter. hopefully this helps in some way
Which one of the following statements contain a simile?
Answer:
I don't know if its just me, but I don't see the statements so I'm not sure how to help answer your question.
0.0884 moles of a diatomic gas
are in a piston. When the piston
is compressed, the temperature
drops by 18.8 K, and 83.7 J of
heat flow out. Find W.
(Be careful with + and - signs.
+W = expansion, +Q = added,
+AU = temp goes up)
(Unit = J)
Answer:
W = - 118.24 J (negative sign shows that work is done on piston)
Explanation:
First, we find the change in internal energy of the diatomic gas by using the following formula:
[tex]\Delta\ U = nC_{v}\Delta\ T[/tex]
where,
ΔU = Change in internal energy of gas = ?
n = no. of moles of gas = 0.0884 mole
Cv = Molar Specific Heat at constant volume = 5R/2 (for diatomic gases)
Cv = 5(8.314 J/mol.K)/2 = 20.785 J/mol.K
ΔT = Rise in Temperature = 18.8 K
Therefore,
[tex]\Delta\ U = (0.0884\ moles)(20.785\ J/mol.K)(18.8\ K)\\\Delta\ U = 34.54\ J[/tex]
Now, we can apply First Law of Thermodynamics as follows:
[tex]\Delta\ Q = \Delta\ U + W[/tex]
where,
ΔQ = Heat flow = - 83.7 J (negative sign due to outflow)
W = Work done = ?
Therefore,
[tex]-83.7\ J = 34.54\ J + W\\W = -83.7\ J - 34.54\ J\\[/tex]
W = - 118.24 J (negative sign shows that work is done on piston)
Answer:
-49.2
Explanation:
Trust me bro
A supersonic aircraft flies at 3 km altitude at a speed of 1000 m/s on a standard day. How long after passing directly above a ground observer is the sound of the aircraft heard by the ground observer
Answer:
After 3secondsExplanation:
A supersonic aircraft flies at 3 km altitude at a speed of 1000 m/s on a standard day. How long after passing directly above a ground observer is the sound of the aircraft heard by the ground observer
Using the formula for calculating speed expressed as;
Speed = Distance/Time
Given;
Distance = 3km = 3000m
Speed = 1000m/s
Required
How long after passing directly above a ground observer is the sound of the aircraft heard by the ground observer (Time)
From the formula;
Time = Distance/speed
Time = 3000/1000
Time = 3seconds
Hence the sound of the aircraft is heard after 3 seconds
What are the precautionary measures against earthquake?
Answer:
Seek shelter under stable tables or under door frames.
If outside, stay away from buildings, bridges and electricity pylons and move to open areas.
Avoid areas at risk from secondary processes, such as landslides, rockfall and soil liquefaction.
From
Measures against earthquakeswww.planat.ch › earthquake › measures-eb
Mistakes were made" is a classic passive voice confession. President Ronald Reagan famously said "mistakes were made" by his administration during the Iran Contra scandal. But some people said he wasn't directly taking the blame. This is why some writers choose to use the passive voice. It can sidestep the question of who did something.
Based on this passage, a writer may use the passive voice to
A
give the reader all known information about a criminal and his or her crimes.
B
clarify the person who was responsible for a crime.
C
describe a crime without blaming anyone for committing it.
D
take responsibility for committing a crime.
Indentify the following types of reflection.
Answer:
first is regular reflection and 2nd is irregular
A cannonball is fired at a 45.0° angle and an initial velocity of 670 m/s. Assume no air resistance. What is the vertical component of the cannonball’s velocity? What is the horizontal component of the cannonball’s velocity?
473.8 m/s; 473.8 m/s
-525.2 m/s; 435.5 m/s
0 m/s; 670 m/s
-378 m/s; 378 m/s
Answer:
473.8 m/s; 473.8 m/sExplanation:
Given the initial velocity U = 670m/s
Horizontal velocity Ux = Ucos theta
Vertical component of the cannon velocity Uy = Usin theta
Given
U = 670m/s
theta = 45°
horizontal component of the cannonball’s velocity = 670 cos 45
horizontal component of the cannonball’s velocity = 670(0.7071)
horizontal component of the cannonball’s velocity = 473.757m/s
Vertical component of the cannonball’s velocity = 670 sin 45
Vertical component of the cannonball’s velocity = 670 (0.7071)
Vertical component of the cannonball’s velocity = 473.757m/s
Hence pair of answer is 473.8 m/s; 473.8 m/s
A circular parallel plate capacitor is constructed with a radius of 0.52 mm, a plate separation of 0.013 mm, and filled with an insulating material with dielectric constant of 40. If a potential is applied to this device of 2.0 mV, how much charge will accumulate on this capacitor, in terms of the number of charge carriers?
Answer:
289282
Explanation:
r = Radius of plate = 0.52 mm
d = Plate separation = 0.013 mm
A = Area = [tex]\pi r^2[/tex]
V = Potential applied = 2 mV
k = Dielectric constant = 40
[tex]\epsilon_0[/tex] = Electric constant = [tex]8.854\times 10^{-12}\ \text{F/m}[/tex]
Capacitance is given by
[tex]C=\dfrac{k\epsilon_0A}{d}[/tex]
Charge is given by
[tex]Q=CV\\\Rightarrow Q=\dfrac{k\epsilon_0AV}{d}\\\Rightarrow Q=\dfrac{40\times 8.854\times 10^{-12}\times\pi \times (0.52\times 10^{-3})^2\times 2\times 10^{-3}}{0.013\times 10^{-3}}\\\Rightarrow Q=4.6285\times 10^{-14}\ \text{C}[/tex]
Number of electron is given by
[tex]n=\dfrac{Q}{e}\\\Rightarrow n=\dfrac{4.6285\times 10^{-14}}{1.6\times10^{-19}}\\\Rightarrow n=289281.25\ \text{electrons}[/tex]
The number of charge carriers that will accumulate on this capacitor is approximately 289282.
what kind of law of motion A car still moves for a short period even after the brakes
have been applied.
Answer:
inertia of motionExplanation:
.... ...