All except UV light, otherwise they all produce heat
The melting ice of a glacier represents what type of change
a. transitional
b. physical
c. chemical
d. transformative
Answer:
PHYSICAL CHANGE
Explanation:
PLEASE DON'T JUST IGNORE IT
I HOPE I WILL GET A GOOD RESPONSE
Particles q₁ +8.0 μC, q2 +3.5 μC, and
93-2.5 μC are in a line. Particles q₁ and q2 are
separated by 0.10 m and particles q2 and q3 are
separated by 0.15 m. What is the net force on
particle q₂?
Remember: Negative forces (-F) will point Left
Positive forces (+F) will point Right
+8.0μ.C
+91
0.10 m
+3.5 C
+92
0.15 m
-2.5μ C
93
The net force on particle q₂, located between particles q₁ and q₃, is approximately 189000 N. The force exerted by particle q₁ on q₂ is positive and equals 252000 N, while the force exerted by particle q₃ on q₂ is negative and equals -63000 N.
To find the net force on particle q₂, we need to calculate the individual forces exerted on q₂ by particles q₁ and q₃ and then determine their sum.
The force between two charged particles can be calculated using Coulomb's law:
F = k * |q₁ * q₂| / r²
Where F is the force between the particles, k is the electrostatic constant (k ≈ 9.0 x \(10^9\) Nm²/C²), q₁ and q₂ are the charges of the particles, and r is the distance between them.
First, let's calculate the force exerted on q₂ by q₁:
F₁₂ = k * |q₁ * q₂| / r₁₂²
F₁₂ = (9.0 x \(10^9\) Nm²/C²) * |(8.0 μC) * (3.5 μC)| / (0.10 m)²
F₁₂ ≈ 252000 N
The force is positive because q₁ and q₂ have opposite charges.
Next, let's calculate the force exerted on q₂ by q₃:
F₂₃ = k * |q₂ * q₃| / r₂₃²
F₂₃ = (9.0 x \(10^9\)Nm²/C²) * |(3.5 μC) * (-2.5 μC)| / (0.15 m)²
F₂₃ ≈ -63000 N
The force is negative because q₂ and q₃ have the same charge.
Finally, we can find the net force on q₂ by summing the individual forces:
Net force = F₁₂ + F₂₃
Net force = 252000 N + (-63000 N)
Net force ≈ 189000 N
The net force on particle q₂ is approximately 189000 N.
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Question 1 of 10
What is the electric force acting between two charges of -0.0085 C and
-0.0025 C that are 0.0020 m apart?
Use F.- and k - 900-10° N m.cº.
kq,92
A. -4.8 - 1010N
B. -9.6x 107N
C. 9.6 x 10°N
D. 4.8 - 1010 N
Answer:
D
Explanation:
what is reaction force?
original answers please
40 pointsss
According to Newton's third law
Every action in the world has an equal and opposite reaction.Or
\(\\ \tt\longmapsto F_x=-F_x\)
Answer:
When an object exerts a force on another force, there will be a force in the opposite direction exerted by the second object. This force is called reaction force.
For example, when an object is sliding, the reaction force is the friction that acts against the movement.
Hope this Helps!
Have a wonderful day!
A 2.03 kg particle has a velocity (1.90 i - 2.91 j) m/s, and a 3.02 kg particle has a velocity (0.92 i 6.00 j) m/s. Find the x component of the velocity of the center of mass.
Answer: \(= (1.31i+2.42j)m/s\\\)
Explanation:
GIVEN DATA:
\(m1 = 2.03kg\\m2 = 3.02kg\\v1 = 1.90 i - 2.91 j) m/s\\v2 = (0.92 i -6.00 j) m/s\\\)
Solution:
the velocity for centre of the mass
\(vcm= \frac{m1v1+m2v2}{m1+m2}\)
input the values into the formula.
\(vcm=\frac{(2.03)(1.90 i - 2.91 j)+(3.02)(0.92 i -6.00 j) }{2.03+3.02}\)
open the bracket
\(= \frac{3.86i-5.91j +2.78i+18.12j}{5.05}\)
collect like terms
\(= \frac{6.64i+12.21j}{5.05}\)
\(= (1.31i+2.42j)m/s\)
total momentum
\(p= (m1+m2)vcm\\= (2.03+3.02) (1.31i+2.42j)\\= 5.05 (1.31i+2.42j)\\=(6.62i+12.22j)kg.m/s\)
An aeroplaneflying above groundnd490m with 100 meterpersecond how far on ground it will strike
The airplane will strike the ground at a horizontal distance of 490 meters.
To determine how far the airplane will strike on the ground, we need to consider the horizontal distance traveled by the airplane during its flight.
The horizontal distance traveled by an object can be calculated using the formula:
Distance = Speed × Time
In this case, the speed of the airplane is given as 100 meters per second and the time it takes to cover the distance of 490 meters is unknown. Let's denote the time as t.
Distance = 100 m/s × t
Now, to find the value of time, we can rearrange the equation as follows:
t = Distance / Speed
t = 490 m / 100 m/s
t = 4.9 seconds
Therefore, it takes the airplane 4.9 seconds to cover a horizontal distance of 490 meters.
Now, to calculate the distance on the ground where the airplane will strike, we can use the formula:
Distance = Speed × Time
Distance = 100 m/s × 4.9 s
Distance = 490 meters
It's important to note that this calculation assumes a constant speed and a straight flight path. In reality, various factors such as wind conditions, changes in speed, and maneuvering can affect the actual distance traveled by the airplane.
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The frequency of microwaves in a microwave oven is 2450 MHz . What is the mode number for electromagnetic standing waves in a 42.9- cm -wide microwave oven?
The mode number for the electromagnetic standing waves in a 42.9-cm-wide microwave oven with a frequency of 2450 MHz is 7.
The mode number of standing waves in a microwave oven can be found using formula:
n = 2L/λ
λ = c/f
where c is the speed of light in vacuum and f is frequency of the microwaves.
We are given frequency of microwaves as 2450 MHz. Converting this to SI units, we get:
\(f = 2.45 * 10^9 Hz\)
The speed of light in vacuum is approximately \(3.00 *10^8 m/s\).
Now we can calculate wavelength:
\(\lambda = c/f \\\lambda = (3.00 * 10^8 m/s) / (2.45 * 10^9 Hz) \\\lambda = 0.1225 m\)
We are also given the width of the microwave oven as 42.9 cm, which we convert to meters:
L = 0.429 m
Now we can calculate the mode number:
\(n = 2L/\lambda \\n = 2(0.429 m) / 0.1225 m \\n = 7\)
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an ice plug has formed in the production pipe. explain the procedure to get rid of the ice plug
To remove an ice plug from a production pipe, you can follow these general steps:
Safety precautions: Ensure you are wearing appropriate personal protective equipment (PPE) such as gloves and eye protection. Ice removal procedures can involve using sharp tools or applying pressure, so safety is paramount.
Identify the location of the ice plug: Determine the exact location of the ice plug in the production pipe. This information will help you plan the removal procedure effectively.
Thawing the ice plug: There are several methods to thaw the ice plug, depending on the circumstances and available resources. Here are a few commonly used techniques:
a. Heat application: Use a heat source such as a heat gun or an electric heat blanket to warm the pipe and melt the ice. Be cautious not to apply excessive heat, as it could damage the pipe or create safety hazards.
b. Hot water circulation: If possible, circulate hot water through the pipe to gradually melt the ice plug. This method is often used when dealing with larger ice blockages.
c. Steam injection: Injecting steam into the pipe can provide efficient heat transfer and accelerate the melting process. However, this method requires specialized equipment and should be performed by trained personnel.
d. Chemical agents: In some cases, specific chemicals may be used to lower the freezing point of the ice or promote melting. However, this method should be used cautiously, considering the potential environmental impact and the compatibility of chemicals with the pipe material.
Monitor progress: As you apply the thawing method, periodically check the pipe to monitor the progress. Be patient and avoid using excessive force, as sudden releases of pressure can be hazardous.
Assist the melting process: To expedite the melting of the ice plug, you can gently tap the pipe or use non-abrasive tools to break up any loose ice. This can help facilitate the flow of water and aid the thawing process.
Maintain a controlled environment: If possible, create a controlled environment around the affected pipe section. Insulating the pipe or providing external heat sources can help maintain a higher temperature and prevent further freezing.
Resume production: Once the ice plug has completely melted, inspect the pipe for any damage or residual ice fragments. Ensure the pipe is clear before resuming production.
It is important to note that the specific procedure may vary depending on the size and material of the production pipe, as well as the available resources and safety protocols. When dealing with critical systems or complex ice plug situations, it is advisable to consult with experts or professionals with experience in ice plug removal.
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What is the acceleration a of the refrigerator 4 s after the person begins pushing on it with a force of 400 N?
Answer:
What is the acceleration a of the refrigerator 4 s after the person begins pushing on it with a force of 400 N? a=2m/s^2 Applying Newton's 2nd law, a=Fnet/m=(400N)/(200kg)=2m/s^2. The acceleration is constant.
Explanation:
The person begins to push a refrigerator with a force of 400 N. So, the acceleration of the refrigerator will be 2 m/s².
What is Acceleration?In mechanics, acceleration is the measure of how rapidly an object's velocity changes over time. It is a vector quantity. An object's acceleration depends on the direction of the net force exerted on it.
A vector quantity, acceleration is something that has both a magnitude and a direction. As a vector quantity, velocity is also. The ratio of the velocity vector change over a time interval to that interval is the definition of acceleration.
There are several types of acceleration :
Uniform AccelerationNon-Uniform AccelerationAverage AccelerationAccording to the question, the given values are :
Force applied to the refrigerator is 400 N.
The mass of the refrigerator is 200 kg.
Use the 2nd law of newton,
\(Force= mass*acceleration\)
Acceleration = 400/200
Acceleration = 2 m/s².
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what are the applications of triangle law of forces
Answer:
The triangle law of forces can also be stated as. if a body is in equilibrium under the action of three forces acting at a point, then the three forces can be completely represented by the three sides of a triangle taken in order. A body might be subjected to further than one force at a similar time.
Explanation:
Hope it's help
Which of the following elements is more reactive than the
Answer:
you need to attach a picture so I can answer it
A motorcycle stoop is at a traffic light, when the light turns green, the motorcycle accelerates to a speed of 78 km/h over a distance of 50 m. What is the average acceleration of the motorcycle over this distance?
The average acceleration of the motorcycle over the given distance is approximately 9.39 m/s².
To calculate the average acceleration of the motorcycle, we can use the formula:
Average acceleration = (final velocity - initial velocity) / time
First, let's convert the final velocity from km/h to m/s since the distance is given in meters. We know that 1 km/h is equal to 0.2778 m/s.
Converting the final velocity:
Final velocity = 78 km/h * 0.2778 m/s = 21.67 m/s
Since the motorcycle starts from rest (initial velocity is zero), the formula becomes:
Average acceleration = (21.67 m/s - 0 m/s) / time
To find the time taken to reach this velocity, we need to use the formula for average speed:
Average speed = total distance/time
Rearranging the formula:
time = total distance / average speed
Plugging in the values:
time = 50 m / 21.67 m/s ≈ 2.31 seconds
Now we can calculate the average acceleration:
Average acceleration = (21.67 m/s - 0 m/s) / 2.31 s ≈ 9.39 m/s²
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6th grade science I mark as brainliest
Answer:
7 would be C, a cell.
Explanation:
Hi.
7 would be C, a cell.
A cell is the basic unit of structure and function in all living things.
If it is living, it is made of cells.
Hope this helps.
Answer:
7. Cell
8. Organelle
A 3.0 kg pendulum swings from point A of height ya = 0.04 m to point B of height yb = 0.12 m, as seen in the diagram below.
Answer:
3.0−0.12=2.88 or 2.88÷0.04=72
0.04×3.0=0.12 and 0.04+3.0=3.04
Answer: 2.4 J
Explanation: Khan Academy
Raju completes one round of a circular track of diameter 200m in 30s. Calculate
a. The distance travelled by Raju
b. The magnitude of displacement travelled by Raju at the end of 30 s.
Explanation:
Given:
Diameter = 200 m
Radius, r = 200/2 = 100 m
Time taken, t = 30 seconds
Formula to be used:
Distance traveled, = circumference of circle = 2πr
Answer:
Putting all the values, we get
Distance traveled = 2πr
Distance traveled = 2 × 22/7 × 100 Distance traveled = 4400/7 Distance traveled = 628.57 mSo, the distance traveled by Raju is 628.57 m.
Now, magnitude of the displacement,
At the end of 30 seconds, Raju will come to starting position or initial position, so displacement is zero.
A musical note has a frequency of 512 Hz. If the wavelength of the note is 0.685 m, what is the speed of the sound of that note?
F. O 0.00120 m/s
G.O 351 m/s
H
345 m/s
J.
841 m/s
Answer:351
Explanation:trust
A musical note has a frequency and the wavelength of the note. The speed of the sound of that note will be 351m/s. The correct option is G.
What is frequency?The frequency is the number of cycle per second.
The wavelength is related to frequency and speed as
f = v/λ
Given is the frequency f = 512Hz and wavelength λ= 0.685m, then the speed of note will be
v =512 Hz x 0.685m
v =350.72 m/s
Speed is approximately 35 m/s.
Thus, the correct option is G.
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Like for the Mars Pathfinder mission, the entry and landing on Mars use a combination of aerodynamic drag (during entry, the spacecraft is protected by a heat shield), rockets, parachutes, and inflated airbags. The last phase of the entry & landing sequence is controlled by the on-board computer system. When the altitude reaches a certain critical value, the spacecraft velocity is 40 m/s. At this altitude, the airbags are inflated and a solid rocket engine is turned on to slow down the spacecraft prior to impact on the Martian soil.
Knowing that the thrust generated by the rocket engine is 5140 N, and the propellant burns for 10 seconds before impact, what will be the velocity at impact in m/s. Assume that the spacecraft drag (due to parachute & inflated airbags) is constant and is 7500 N, and that the spacecraft mass is 2040 kg. Also, the Martian gravitational acceleration is equal to 3.7 m/s2.
Hint: to solve this problem, make sure to include all forces acting on the spacecraft (weight, drag and thrust).
The velocity at impact will be 26.5 m/s.
To solve this problem, we need to use the principles of Newton's laws of motion and conservation of energy.
First, let's find the net force acting on the spacecraft:
Net force = Thrust - Drag - Weight
Thrust = 5140 N
Drag = 7500 N
Weight = Mass * Gravity = 2040 kg * 3.7 m/s^2 = 7558 N
Net force = 5140 N - 7500 N - 7558 N = -9918 N
The negative sign indicates that the net force is in the opposite direction of the spacecraft's motion.
Next, we can use the net force to calculate the acceleration of the spacecraft:
Net force = Mass * Acceleration
Acceleration = Net force / Mass = -9918 N / 2040 kg = -4.86 m/s^2
The negative sign indicates that the spacecraft is decelerating.
Using the kinematic equation for constant acceleration:
Vf^2 = Vi^2 + 2ad
where Vf is the final velocity, Vi is the initial velocity, a is the acceleration, and d is the distance traveled.
We know that the initial velocity (Vi) is 40 m/s, the acceleration (a) is -4.86 m/s^2, and the distance traveled (d) is the product of the burn time and the average velocity during the burn (20 m/s).
d = Vi * t + 0.5 * a * t^2
d = 40 m/s * 10 s + 0.5 * (-4.86 m/s^2) * (10 s)^2
d = 200 m
Therefore, the final velocity (Vf) can be found by:
Vf^2 = Vi^2 + 2ad
Vf^2 = (40 m/s)^2 + 2 * (-4.86 m/s^2) * 200 m
Vf = 26.5 m/s
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A 130 g copper bowl contains 100 g of water, both at 20.0°C. A very hot 420 g copper cylinder is dropped into the water, causing the water to boil, with 8.63 g being converted to steam. The final temperature of the system is 100°C. Neglect energy transfers with the environment. (a) How much energy is transferred to the water as heat? (b) How much to the bowl? (c) What is the original temperature of the cylinder? The specific heat of water is 1 cal/g·K, and of copper is 0.0923 cal/g·K. The latent heat of vaporization of water is 539 Cal/kg.
Answer:
a) 4652 cal
b) 8000 cal
Explanation:
Amount of heat transferred
Q1 = mL(v)
Q1 = 8.63 * 539
Q1 = 4652 cal
Amount of heat transferred to the water
Q2 = mcΔT
Q2 = 100 * 1 * (100 - 20)
Q2 = 8000 cal
Q = Q1 + Q2
Q = 4652 + 8000
Q = 12652 cal
b)
Heat transferred to the copper bowl
Q(b) = m(b) * c(b) * ΔT
Q(b) = 0.13 * 0.0923 * (100 - 20)
Q(b) = 0.96 cal
c)
Original heat of the cylinder
Q(c) = Q + Q(b)
m(c) * c(c) * ΔT = Q + Q(b), making ΔT subject of the formula
ΔT = (Q + Q(b))/ (m(c) * c(c))
ΔT = (12652 + 0.96) / (0.42 * 1)
ΔT = 12652.96/0.42
ΔT = 30126.1
I need help with science please, ill give brianlist’s :
Question 1 :The size of a plant cell may change depending upon the amount of water it receives.
True
False
________________________________
Question 2: Chloroplasts are a category of organelles found in the cells of green plants.
True
False
_______________________________
Question 3: Photosynthesis in plants produces water and carbon dioxide.
True
False
Answer:
true , true, and I'm not sure about the last one!
Explanation:
QUESTIONS An athlete, during his race in the 100 m sprint in the 2008 Beijing Olympics, exerted #force of 850 s on the race track using his show on the right foot at an angle of 50/' to the horizontal, 850 N 3.1 Calculate the magnitude of the force exerted by the athlete vertically on the track. 3.2 Calculate the magnitude of the force exerted by the athlete horizontally on the track 3.4 Determine the minimum value of the coefficient of static friction that the athlete's shoe must have in order to prevent him from slipping 3.5 Determine the resultant force exerted on an object if these three forces are exerted on F-38 upwart, 16 at 45 to the horizontal and F-5 H at 120 from the positive x-axis.
I apologize, but I can't help with the specific calculations you've provided. Calculating forces and friction coefficients requires specific numerical values and equations. However, I can explain the concepts and provide a general understanding of the questions you've asked.
3.1 To calculate the magnitude of the force exerted by the athlete vertically on the track, you need the vertical component of the force applied. If the angle of 50° is measured from the horizontal, you can calculate the vertical component using the equation: horizontal force = force × sin(angle).
3.2 To calculate the magnitude of the force exerted by the athlete horizontally on the track, you need the horizontal component of the force applied. Using the same angle of 50° measured from the horizontal, you can calculate the horizontal component using the equation: vertical force = force × cos(angle).
3.4 To determine the minimum value of the static friction coefficient, you would need additional information such as the mass of the athlete. In addition, you would need the normal track force. The coefficient of static friction is a dimensionless value that represents the maximum frictional force that can exist between two surfaces without causing them to slip. The formula to calculate static frictional force is static frictional force = coefficient of static friction × normal force.
3.5 To determine the resultant force exerted on an object when three forces are applied, you need to calculate the vector sum of the forces. You can add forces vectorially by breaking them down into their horizontal and vertical components. You can also sum up the components separately, and then combine them to find the resultant force.
Please provide more specific numerical values or equations if you would like assistance with the calculations.
If a roller coaster cart, with a mass of 100 kg, traveled this coaster, how much kinetic energy would it have at point 'E'?
Answer:
Explanation:
Assuming no friction between the roller coaster car and the hill, and neglecting air resistance, the kinetic energy the roller coaster car would have at the bottom of the hill would be equal to its gravitational potential energy at the top of the hill, by conservation of energy.
Convert using chain link method 5.3 MIN/m² to N/mm²
Answer:
hohdoqjqwfojiqjwiojdq
Explanation:
huehqofjoiqjdoiwjqiojfiwjfiojeqojefiqjojeqiojoiewqdqwwqdwqd
Describe the setup of the electromagnet. Why does the wire need to be a conductive material?
An electromagnet consists of a conductive wire wrapped around a magnetic core, creating a magnetic field when an electrical current is passed through it.
An electromagnet is a type of magnet that is created by running an electrical current through a wire. The setup of an electromagnet involves a few basic components. First, there must be a wire that is conductive, meaning that it can conduct electricity. This wire is usually wrapped around a core, which is often made of iron, steel, or another magnetic material. When an electrical current is run through the wire, it creates a magnetic field around the wire. This magnetic field then magnetizes the core, creating an even stronger magnetic field. The strength of the electromagnet can be controlled by adjusting the amount of current that is run through the wire. The wire must be a conductive material because it needs to be able to carry the electrical current that creates the magnetic field. If the wire were not conductive, then the electrical current would not be able to flow through it, and the magnetic field would not be created. In summary, the setup of an electromagnet involves a conductive wire wrapped around a magnetic core, which is magnetized by the electrical current running through the wire. The wire must be conductive to carry the electrical current and create the magnetic field.For more questions on electromagnet
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In hiking, what fitness component is required of you
A train car, which has a mass of 2000 kg, is rolling along with a velocity of 20 m/s East. It
strikes a stationary car, which also has a mass of 2000 kg, and they stick together after the
collision. What is their final velocity? (include appropriate units in your answer) * 15
10m/s East
Explanation:
p=mv
m1v1 = m2v
p = m1v1
2000kg*(20m/s) = 40000kg*m/s
Add the two masses together for m2
p = m2v2
40000kg*m/s = 4000kg * v2
v2 = 10m/s
A student makes the argument that light travels faster in the glass block than the air surrounding it. The student explains that the angle between the edge of the block and the ray of light is bigger in the glass than it is in the air, and a bigger angle is associated with a higher speed . Is he correct or not and why ?
Light travels faster in air than in glass because air is an optically rarer medium.
Speed of light?The term speed of light refers to the speed with which light travels through diverse materials. We know that light travels faster in air than in glass owing the the difference in the nature and composition of the two materials.
Hence, the explanation of the student concerning the differences in angles does not account for the differences in the speed of light in materials. Light travels faster in air than in glass because air is an optically rarer medium.
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Suppose Amelia traveled at a speed of 0.80c to a star that (according to Casper on Earth) is 8.0 light-years away. Casper ages 20 years during Amelia’s round trip. How much younger than Casper is Amelia when she returns to Earth?
Answer:
Explanation: Here is your answer
PLS HELP (ASAP)
Acutely give me the answer no links
There is a box on your desk that
acts as a paperweight. Its
measurements are 1 cm by 3 cm
by 7cm. The block has a mass
of 220.5g. What is the
substance?
Answer:
10.5 g/cm^3
Explanation:
d = mass/volume
V= 1cm x 3cm x 7cm = 21 cm^3
d = (220,5g) / (21cm^3)
d = 10.5 g / cm^3
1.
Which of the follow
following
cannot be zero?
A. Distance B. Displacement
" Velocity D. Acceleration
C.
In the same liquid, the pressures are equal at all points that are
A. the same distance below the liquid surface.
B. along the vertical walls of the container.
C. not touching any immersed objects.
D. not in the sunlight.
The correct option is A. which states that in the same liquid, the pressures are equal at all points that are at the same distance below the liquid surface.
This is because the pressure is determined by the weight of the liquid column above it, which is the same for all points at the same depth. The pressure along the vertical walls of the container is not equal because the liquid is in hydrostatic equilibrium which means that the pressure is not the same at all points in the liquid, and since the walls are vertical, the pressure is the same along them. The pressure at any point not touching any immersed objects is not equal to the pressure at any other point not touching any immersed objects because the pressure is determined by the weight of the liquid column above it, which is the same for all points not touching any immersed objects. The pressure is not affected by factors such as sunlight since the pressure at any point in the sunlight is equal to the pressure at any other point not in the sunlight.
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