The total velocity required (Av) to transfer from one circular orbit to another is 7.086 km/s
The total velocity required (Av) to transfer from one circular orbit to another is given by Av = √(μ/p)(2/r1 - 1/r2), where μ is the standard gravitational parameter of the central body and p is the semi-latus rectum of the transfer ellipse.
Given the parameters for the two circular orbits, r1 = 12,750 km and r2 = 31,890 km, and for the transfer ellipse, p = 13,475 km and e = 0.76, we can calculate the total velocity required to transfer between the two orbits as follows:
Av = √(μ/p)(2/r1 - 1/r2)
Av = √(3.986E+13/13,475)(2/12,750 - 1/31,890)
Av = √(2.946E+10)(0.0690 - 0.0314)
Av = √(2.946E+10)(0.0375)
Av = 7.086 km/s
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A boat on a river is moving with a steady speed. The engine is running.
What would happen if the engine was turned off?
If the engine of a boat on a river is turned off while the boat is moving with a steady speed, several things would happen Loss of propulsion,Drifting,Loss of steering control and Potential hazards.
Loss of propulsion: Without the engine running, the boat would lose its power source for propulsion. The boat would gradually slow down and eventually come to a stop unless other external forces, such as currents or wind, continue to move it.
Drifting: Once the boat comes to a stop, it would start to drift with the current of the river or be affected by wind forces. The direction and speed of the drift would depend on the strength and direction of the current or wind.
Loss of steering control: When the engine is turned off, the boat's steering mechanism, such as a rudder, would also lose power. Without the ability to steer, the boat would follow the course determined by the river's current or the wind direction.
Potential hazards: Depending on the surroundings and the current conditions, there could be potential hazards for a boat that is no longer under power. These hazards might include other vessels, obstacles, shallow areas, or strong currents. The boat's crew would need to take appropriate actions to ensure the safety of the boat and its occupants.
It's important to note that the specific behavior of the boat after the engine is turned off can vary depending on factors such as the size and design of the boat, the strength and direction of the current, and the presence of wind or other external forces.
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Calculate the electrical force that acts on one plate of a parallel plate capacitor. The potential difference between the plates is 10 volts, and the plates are squares 20 cm on a side with a separation of 3 cm. If the plates are insulated so the charge cannot change, how much external work could be done by letting the plates come together
Answer:
The work done in bringing the plates together is 5.9 x 10⁻¹⁰ J.
Explanation:
Given;
potential difference between the plates, V = 10 V
length of each square side of the plates, L = 20 cm = 0.2 m
area of the plates, A = 0.2 x 0.2 = 0.04 m²
separation of the plates, d = 3 cm = 0.03 m
The work done in bringing the plates together is calculated as;
W = ¹/₂qV
\(W = \frac{1}{2} (\frac{\epsilon_0 A }{d}V) \times V\\\\W = \frac{\epsilon_0 A V^2}{2d}\\\\W = \frac{8.85 \times 10^{-12} \ \times \ 0.04\ \times \ 10^2}{2(0.03)} \\\\W = 5.9 \times 10^{-10} \ J\)
Therefore, the work done in bringing the plates together is 5.9 x 10⁻¹⁰ J.
The max power of an elevator is 20,000 J/s. How much time would it take the elevator to lift 800 Kg 30 meters?
Answer:
11.76s
Explanation:
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In an experiment to measure the temperature of a Bunsen burner flame, a 250 g piece of iron is held in the flame for several minutes until it reaches the same temperature as the flame . The hot metal is then quickly transferred to 285 g of water contained in a 40.0 g copper calorimeter at 15.0 oC. The final temperature of the copper and water is 80.0 oC.
Using your answer from determine the temperature of the Bunsen flame.
Answer:
wait
Explanation:
A car sits in an entrance ramp to a freeway, waiting for a break in the traffic. The driver sees a small gap between a van and an 18-wheel truck and accelerates with constant acceleration along the ramp and onto the freeway. The car starts from rest, moves in a straight line, and has a speed of 19.0 m/s m / s when it reaches the end of the ramp, which has length 117 m m .
If the car starts from rest, moves in a straight line, and has a speed of 19.0 m/s then it would reach the ramp in 12.32 seconds.
What are the three equations of motion?There are three equations of motion given by Newton,
v = u + at
S = ut + 1/2×a×t²
v² - u² = 2×a×s
Note that these equations are only valid for a uniform acceleration.
By using the third equation of motion,
v² - u² = 2×a×s
19² - 0² = 2×a×117
a = 1.54 meters / second²
By using the second equation of the motion,
S = ut + 1/2*a*t²
117 = 0 + 0.5*1.54*t²
117 = 0.77t²
t² = 152
t = 12.32 seconds
Thus, the car would reach the end of the ramp in 12.32 seconds.
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A constant force moves an object along the line segment from to . Find the work done if the distance is measured in meters and the force in newtons.
This question is incomplete, the complete question is;
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A constant force F = 6i+8j-6k moves an object along a straight line from point (6, 0, -10) to point (-6, 7, 2).
Find the work done if the distance is measured in meters and the magnitude of the force is measured in newtons.
Answer:
the work done is -88 J
Explanation:
Given the data in the question;
we know that;
Work done = F × S
where constant force F = ( 6i + 8j - 6k )
S = ( -6i + 7j + 2k ) - ( 6i + 0j - 10k )
S = ( (-6i - 6i) + (7j - 0j) + ( 2k - ( -10k) ) )
S = ( -12I + 7j + 12k )
so
Work force = ( 6i + 8j - 6k ) × ( -12I + 7j + 12k )
Work force = ( 6 × -12 ) + ( 8 × 7 ) + ( -6 × 12 )
Work force = -72 + 56 - 72
Work force = -88 J
Therefore, the work done is -88 J
Air at 273K and 1.01x10³Nm2 pressure contains 2.70x1025 molecules per cubic meter. How many molecules per cubic meter will there be at a place where the temperature is 223K and pressure is 1.33x10 Nm-2
The molecules of O2 that are present in 3.90 L flask at a temperature of 273 K and a pressure of 1.00 atm is 1.047 x 10^23 molecules of O2
Step 1: used the ideal gas equation to calculate the moles of O2
that is Pv=n RT where;
P(pressure)= 1.00 atm
V(volume) =3.90 L
n(number of moles)=?
R(gas constant) = 0.0821 L.atm/mol.K
T(temperature) = 273 k
by making n the subject of the formula by dividing both side by RT
n= Pv/RT
n=[( 1.00 atm x 3.90 L) /(0.0821 L.atm/mol.k x273)]=0.174 moles
Step 2: use the Avogadro's law constant to calculate the number of molecules
that is according to Avogadro's law
1 mole = 6.02 x10^23 molecules
0.174 moles=? molecules
by cross multiplication
the number of molecules
= (0.174 moles x 6.02 x10^23 molecules)/ 1 mole =1.047 x 10^23 molecules of O2
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Visible matter belonging to the Milky Way Galaxy can be traced out to about 50,000 light years from the center.
a. True
b. False
Answer:
b. False
Explanation:
The visible matter that belongs to the Milky way Galaxy are traced out to be about 50 kpc distance from the center.
Kpc stands for kiloparsec. It is the unit of measurement of distance.
A parsec is\($\text{ used to measure large distances}$\) of the astronomical objects that lies \($\text{outside the solar system}$\), mainly where galaxies are involved.
1 kiloparsec is 1000 parsec and is equal to 3260 light years.
So the visible matter is about 163,078 light years away.
Hence the answer is FALSE.
what is a crystal as applied in physics
Answer:
The correct answer is - A matter that has an ordered arrangement of atoms, molecules, or ions.
Explanation:
In physics, a crystal is a type of solid matter in which a highly arranged molecule or atoms present to form a lattice that extended in all directions. It is a lightweight clear solid which is normally is colorless.
It can be cubic, hexagonal, triclinic, monoclinic, orthorhombic, tetragonal, and trigonal that are ordered arrangments. Its internal symmetry is visible to its surface.
When a constant force acts upon an object, the acceleration of the object varies inversely with its mass. When a certain constant force acts upon an
object with mass 5 kg, the acceleration of the object is 15 m/s². If the same force acts upon another object whose mass is 3 kg
what is this object's
acceleration?
25 m/s² is the object's acceleration
m1a1=m2a2
m1=5kg
m2=3 kg
a1= 15 m/s²
a2=?
m1a1=m2a2
a2=m1a1/m2
a2=5×15÷3
a2= 25 m/s²
Acceleration is a vector variable that describes the rate at which an object changes its velocity.
An object is said to be accelerating if its velocity is changing. Occasionally, a moving object can change its velocity by the same amount each second. a moving object whose speed fluctuates by 10 m/s every second. This is referred to as a constant acceleration since the velocity is changing by a fixed amount each second.
The difference between an object with a constant acceleration and one with a constant velocity must be understood. Do not be fooled! If an object's velocity changes, whether it does so by a constant amount or a variable amount, then it is accelerating. Furthermore, something that is travelling at a steady speed is not accelerating.
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Mr. Bateman creates a standing wave in the front of the classroom with the spring. S nodes form. The distance from Mr. Bateman to the cabinet is 6m. If a student times the spring moving back and forth and gets 0.2s for one cycle of the spring, how fast are the
waves moving?
Answer:
The speed of a wave is equal to the wavelength divided by the period. The wavelength is the distance between two consecutive nodes, and the period is the time it takes for one complete cycle of the wave.
In this case, the wavelength is 6 m and the period is 0.2 s. Therefore, the speed of the wave is 30 m/s.
The answer is 30 m/s.
Explanation:
An object traveling at 1.5 rad
accelerates at 0.75d for 12
S
seconds. What is the object's final
velocity?
The object's final velocity, given the data is 10.5 rad/s
What is acceleration?This is defined as the rate of change of velocity which time. It is expressed as
a = (v – u) / t
Where
a is the acceleration v is the final velocity u is the initial velocity t is the time How to determine the final velocityThe following data were obtained from the question
Initial velocity (u) = 1.5 rad/sAcceleration (a) = 0.75 rad/s²Time (t) = 12 sFinal velocity (v) = ?The final velocity can be obtained as follow:
a = (v – u) / t
0.75 = (v – 1.5) / 12
Cross multiply
v – 1.5 = 0.75 × 12
v – 1.5 = 9
Collect like terms
v = 9 + 1.5
v = 10.5 rad/s
Thus, the final velocity of the object is 10.5 rad/s
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Can someone please help me with this question thank you!
A student lifts their 75 N backpack 0.50 m onto their chair. How much work is done?
Answer:
37.5 J
Explanation:
With work done equation: W=Fs
W=75*0.50=37.5 J
or use mgh=(75)(0.5) which is the same
A toy car of mass 600g moves through 6m in 2 seconds. The average kinetic energy of the toy car is
Answer:
12
Explanation:
I'm a beginner so am not sureeeeee
In thermodynamics energy can be transeferd to and from a system by
Three different types of energy can enter or leave a system: heat, work, and mass flow.Heat and work are the only 2 types that can exist in a closed system because no mass can cross its borders.
How does thermodynamics handle energy transfer?There are three methods that thermal energy can be transferred: conduction, convection, & radiation.Conduction is the transfer of heat energy between adjacent molecules which are in touch with one another.
What are the three ways that energy can be moved?We must become familiar with the three different ways that energy is transferred: conduction, convection, & radiation.
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Based on the diagram, which statement explains how energy is conserved during this chemical reaction? Potential energy of a system 1 11 U # I IE 1 1 II E Reaction progress I] T B II I П L C
The statement that explains how energy is conserved during this chemical reaction is option D
What is the potential energy?
Chemical reactions that release energy in the form of heat into their environment are called exothermic reactions. The total energy of the reactants is greater than the total energy of the products in an exothermic process. The temperature of the surroundings rises as a result of this energy differential being released as heat.
We can see that the reaction is exothermic as such the energy that is lost by the system is gained by the surroundings.
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an object is dropped from a height of 25 meters.at what velocity will it hit the ground?
a.)7.0 m/s
b.)11 m/s
c.)22m/s
d.) 49 m/s
e.) 70 m/s
Answer:
C. 22 m/s
Explanation:
Ignoring air friction....all of the Potential energy will be converted to Kinetic Energy
mgh = 1/2 m ^v^2
gh = 1/2 v^2
sqrt (2 gh) = v = ~ 22.1 m/s
In a local bar, a customer slides an empty beer mug down the counter for a refill. The height of the counter is 1.36 m. The mug slides off the counter and strikes the floor 1.00 m from the base of the counter.(a) With what velocity did the mug leave the counter?m/s(b) What was the direction of the mug's velocity just before it hit the floor?° (below the horizontal)
(a) Velocity with which the mug leaves the counter is approximately 5.02 m/s.
(b) The direction of velocity of mug just before it hits the floor is downwards, since mug is falling under influence of gravity.
What is velocity?Velocity is a quantity that designates how fast and also in what direction a certain point is moving.
(a) As we know, PE = m g h
\(\mathrm{PE = mgh }\)
\(\mathrm{= (m)(9.81 m/s^2)(1.36 m) }\)
= 13.4mJ
\(\mathrm{KE =\frac{1}{2} mv^2}\)
PE = KE
\(\mathrm{mgh = \frac{1}{2}mv^2}\)
\(\mathrm{v = \sqrt{2gh}}\)
\(\mathrm{v = \sqrt{2 \times 9.81 \times 1.36 m}}\)
v = 5.02 m/s
Therefore, the velocity with which the mug leaves the counter is approximately 5.02 m/s.
(b) The direction of the velocity of mug just before it hits the floor is downwards, since mug is falling under the influence of gravity. Velocity vector has a vertical component that points downwards and horizontal component that is parallel to counter.
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A student holds a water balloon outside of an open window and lets go. The window is 10 meters above the ground, and the balloon is falling under the acceleration of gravity, which is 9.8 m/s2. There are two equations that can be used to describe its motion over time:
x=x0+v0t+12at2
v=v0+at
Would the balloon hit the ground before or after 1.0 s of falling? Which equation did you use to decide, and what comparison did you make to determine that it would or would not hit the ground by then?
Use 3–5 sentences to explain your reasoning.
The water balloon that is dropped by the student 10 meters above the ground, will hit the ground after 1.0s of falling (in 1.4 s).
To find the time we need to use the equation that related the time with the height and the acceleration due to gravity, that is to say, the first equation from the two given, but in the vertical direction:
\( y_{f} = y_{0} + v_{0}t + \frac{1}{2}at^{2} \) (1)
Where:
\( y_{f}\): is the final height = 0
\( y_{0}\): is the initial height = 10 m
\( v_{0}\): is the initial velociy = 0 (it is dropped)
a: is the acceleration due to gravity = -9.8 m/s² (it is negative because its direction of motion is downward)
t: is the time =?
Solving equation (1) for t, we have:
\( 0 = 10 m - \frac{1}{2}gt^{2} \)
\( t = \sqrt{\frac{2*10 m}{9.8 m/s^{2}}} = 1.4 s \)
Therefore, the ballon will hit the ground after 1.0s of falling.
We used the first given equation but in the y-direction, because the equation x=x₀+v₀t+(1/2)at², considers a motion in the horizontal direction, and the ball is falling (y-direction). We did not use the other equation (v = v₀ + at) because we do not know the final velocity of the ball before it hits the ground (v).
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Answer:
The time it takes for the balloon to hit the ground is 1.4 seconds. So the answer is the balloon would hit the ground after 1.0 s of falling.
Explanation:
The answer can be found by using the equation \(x=x_{0}+v_{0}t+\frac{1}{2}at^{2}\)
The first step is to identify what each value is and the information we are provided with:x is the position; the balloon falls to the ground, so \(x = 0\)
x0 is the initial position; the balloon is 10 meters high, so \(x_{0} = 10\)
t is the time; we are solving for time
a is the acceleration; in this case it is gravity (9.8 m/s^2), so \(a = 9.8\)
*we do not need v0, or initial velocity, for this problem
The next step is to input the information into the equation:the equation now looks like this \(0=10-(\frac{1}{2}9.8)t^{2}\)
**because the acceleration is gravity and the balloon is falling downwards, the acceleration becomes negative (-9.8 m/s^2)
Steps to solving this equation:1) multiply 9.8 by 1/2 inside the parenthesis
\(0=10-(\frac{1}{2}9.8)t^{2}\)
\(0=10 -(4.9)t^{2}\)
\(0=-4.9t^{2} + 10\)
2) add 4.9t^2 to both sides of the equation
\(4.9t^{2} = 10\)
3) divide both sides by 4.9
\(\frac{4.9t^{2}}{4.9}= \frac{10}{4.9}\)
\(t^{2} = 2.040816\)
4) find the square root
\(\sqrt{t} = \sqrt{2.040816}\)
\(t=1.4\)
This is how we can use the equation \(x=x_{0}+v_{0}t+\frac{1}{2}at^{2}\) to find the answer 1.4 seconds.
the balloon hits the ground after falling for 1.4 seconds.
4. When scientists calculate the trajectory a satellite takes on its way to
study a planet, what do you think they use?
5 points
A Speed and velocity
B. Velocity and acceleration
O C. Speed, velocity, and acceleration
O D. Speed only
When scientists calculate the trajectory a satellite takes on its way to
study a planet, they use C. Speed, Velocity, and acceleration.
A trajectory, often known as a flight path, is the route taken by an object moving under the influence of gravity. Typically, the phrase is applied when referring to projectiles or satellites. A parabola curve is usually a decent approximation of the trajectory form when an object is propelled for in a short distance.
When scientists calculate the trajectory a satellite takes on its way to
study a planet they take the speed, velocity, and acceleration into consideration.
The formula for calculating the trajectory can be expressed as:
\(\mathbf{y = h + xtan (\alpha) - \dfrac{gx^2 }{2V_o^2cos^2 (\alpha)}}\)
where;
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In a dart gun, a spring with k = 400.0 N/m is compressed 8.0 cm when the dart (mass m = 20.0 g) is loaded.
(a) What is the muzzle speed of the dart when the spring is released? Ignore friction.
(b) If the dart gun is located on a table top 2.2 m above the ground, and once the spring is released, it remains compressed by 4 cm, what is the final speed of the dart as it hits the ground?
The muzzle speed of the dart when the spring is released is 11.3 m/s
The given parameters are
k = spring constant = 400 N/m
m = mass = 20 g = 0.02 kg
Compression = x = 8 cm = 0.08 m
According to the question,
When the dart is loaded, the potential energy is converted into kinetic energy.
Potential Energy (P.E.)
\(P.E. =\frac{1}{2} kx^{2}\)
Putting the values,
\(P.E. =\frac{1}{2} *400*(0.08)^{2} = 1.28\)
Now, Kinetic Energy (K.E.)
K. E. = \(\frac{1}{2} mv^{2}\) = \(\frac{1}{2} *0.02*v^{2}= 0.01 v^{2}\)
Now, P.E. = K.E.
\(0.01 v^{2} = 1.28 \\\\v^{2} =\frac{1.28}{0.01} \\\\v^{2} = 128\\ \\v = 11.3\)
Hence, the muzzle speed of the dart when the spring is released is 11.3 m/s
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Which two statements describe how ultrasound technology produces an
image of part of the body?
A. The amount of time it takes for a wave to return can be used to
create an image.
B. Low-frequency sound waves are reflected by tissues in the body.
C. High-frequency sound waves are reflected by tissues in the body.
D. Body tissues absorb high-frequency sound waves, producing an
image
Answer:
A,C
Explanation:
The amount of time it takes for a wave to return can be used to create an image
High--frequency sound waves are reflected by tissues in the body
Ultrasound technology produces an image of part of the body on: The amount of time it takes for a wave to return can be used to create an image and high-frequency sound waves are reflected by tissues in the body. So, option (A) and (C) is correct.
What is ultrasonic technology?Sound waves are used in ultrasound imaging to create images of the inside of the body. It aids in determining the origins of discomfort, edema, and infection in the body's internal organs as well as the examination of a developing fetus in pregnant women.
The same concepts underlie sonar used by ships, bats, and fisherman as well as those used by ultrasound imaging. A sound wave echoes or bounces back when it encounters an object. These echo waves can be measured to obtain information on the object's size, shape, and consistency as well as its distance from the source. Whether the thing is solid or liquid is included in this.
The amount of time it takes for a wave to return helps to create an image and high-frequency sound waves are reflected by tissues in the body. So, option (A) and (C) is correct.
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Ilya and Anya each can run at a speed of 7.50 mph and walk at a speed of 3.50 mph . They set off together on a route of length 5.00 miles . Anya walks half of the distance and runs the other half, while Ilya walks half of the time and runs the other half.
- Find Anya's average speed. (express in miles per hour)
- How long does it take Ilya to cover the distance? (express in minutes)
- Now find Ilya's average speed. (express in miles per hour)
The correct answer is 54.6 min.
Running speed of Ilya and Anya is v_r=7.50 mi/hr
Walking speed of Ilya and Anya is v_w=3.50 mi/hr
Total length of the route is d=5.00 miles.
Total time taken by Anya to cover the distance is
\(t_A= d/2/v_r+ d/2/v_w\)
\(t_A= d/2v_r + d/2v_w\)
\(t_A=d/2 \times1/v_r +1/v_w\)
\(t_A= d/2 \times v_r+v_w / v_rv_w\)
\(t_A\)= 5.00/2 × 7.50+3.50/7.50 × 3.50 hr
\(t_A\)= 5.00/2 × 11/26.25 hr
\(t_A\)= 55/52.5 hr
\(t_A\)=1.05 hr
Therefore Anya will take 1.05 hr to cover the distance.
Average speed of Anya is
\(v_{avg} = d/t_A\)
\(v_{avg}\)= 5.00/1.05 mi/hr
\(v_{avg}\) = 4.76 mi/hr
Therefore Anya's average speed is 4.76 mi/hr.
If total time taken by Ilya to cover the distance is t , then
(\(v_r\) × t/2+(v_w × t/2=d))
t/2 × (\(v_r+v_w\))=d
t × (\(v_r+v_w\))=2d
t = 2d/(\(v_r+v_w\))
t = 2 × 5.00/7.50+3.50
t = 10/11 hr
t = 0.91 × 60 min
t = 54.6 min
Therefore total time taken by Ilya to cover the distance is 54.6 min.
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What is the minimum compression of the spring necessary for the block to make it to the top of the hill on the other side?
Answer:
sciencephysicsphysics questions and answersA 12-kg Block Is Pressed Against A Spring (spring Constant 870 N/m ), Compressing It Some Distance. ...
Question: A 12-kg Block Is Pressed Against A Spring (spring Constant 870 N/m ), Compressing It Some Distance. The Block Is Released From Rest And Slides Across A Track As Shown In (Figure 1). While Most Of The Track Is Frictionless, There Is A 55-cm Section Of Track That Has A Coefficient Of Friction With The Block Of 0.4. A Bit Further On, The Track Ascends ...
A 12-kg block is pressed against a spring (spring constant 870 N/m ), compressing it some distance. The block is released from rest and slides across a track as shown in (Figure 1). While most of the track is frictionless, there is a 55-cm section of track that has a coefficient of friction with the block of 0.4. A bit further on, the track ascends into a hill that is 40-cm tall.
What is the minimum compression of the spring necessary for the block to slide past the section with friction?
What is the minimum compression of the spring necessary for the block to make it to the top of the hill on the other side? 40cm 12kg Ww.w.wwwww 55cm
Explanation:
PLEASE HELP
the graph shows a plot of an objects velocity versus time for 15 seconds. is the acceleration of the object constant or changing? how do you know? what does this tell you about the net force on the object?
Answer:
It cannot be constant because if it does not change and each time it increases its strength and speed.
Explanation:
A steel ball of mass 0.1kg is moving with a velocity of 4m/s along a frictionless surface of 10N/m. What is the distance through which the spring is comressed?
The distance through which the spring is compressed is equal to the kinetic energy of the steel ball divided by the spring constant, i.e. 0.1 kg x 4m/s2/10N/m = 0.04 m.
Computation of Distance movedLet us begin by Calculating the kinetic energy of the steel ball:
Kinetic Energy = 0.5 x mass x velocity2
Kinetic Energy = 0.5 x 0.1 kg x (4 m/s)2
Kinetic Energy = 0.8 J
Calculate the distance through which the spring is compressed:
Distance = Kinetic Energy/Spring Constant
Distance = 0.8 J/10N/m
Distance = 0.04 m
Compression is the process of reducing the size of a file or data to save disk space and reduce transmission times. Compression is achieved by encoding data using fewer bits than the original representation.
This is done by removing unnecessary or redundant data, such as white space, or by using data compression algorithms. Compression helps make data storage and transmission more efficient.
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What country first colonised Ghana
Answer: Colonialism is a big topic, but it can only be understood by looking at human experiences. Formal colonialism first came to the region we today call Ghana in 1874, and British rule spread through the region into the early twentieth century. The British called the territory the “Gold Coast Colony”.
Explanation: hey, hope this hlps! oh, btw you picked the wrong subject for this question it should have been history insteat of phiscics!
if you drop your textbook it falls because the earth pulls it downward. at the same time, your textbook pulls on the earth with a force that is group of answer choices zero. equal to the downward pull on the textbook and in the same direction. immeasurably small. equal to the downward pull on the textbook but in the opposite direction.
The force that your textbook exerts on the Earth is immeasurably small, since the mass of the textbook is much, much smaller than the mass of the Earth.
What is force?Force is a push or pull on an object due to its interaction with another object. Force is a vector quantity that is described by its magnitude and direction. It is typically measured in Newtons and represented by the symbol F. Force can cause an object to accelerate, change its direction, or change its shape. Force is an integral part of physics and is governed by the laws of motion.
This force is equal to the downward pull on the textbook but in the opposite direction, meaning that it is trying to pull the Earth up, canceling out the downward force. However, since the force exerted by the textbook is so small, it is not enough to counteract the pull of gravity, and thus the textbook falls.
To learn more about force
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Riding a bicycle on a flat, smooth surface is a lot easier than riding it along a bumpy surface or up a hill. A bumpy surface creates more friction with your bike tires than a smooth surface does, and going up a hill means fighting gravity.
Write about a time you had to ride a bicycle on a difficult surface. What did you have to do to adjust your riding?
Answer:
one time i was one the flat ground at my aunts house then we went on a hike so i brought my bike it had just rained that day so it was kinda muddy so there was sticks everywhere.i was riding up hill and noticed that it was very hard,then i rode down hill and it was much better
Explanation: