The acceleration and velocity of the plane is 78.57 m/s² and 157.14 m/s respectively
To calculate the acceleration of the plane, we use the formula below.
Formula:a = F/m..................... Equation 1Where:
a = Acceleration of the planeF = Force applied to the planem = mass of the plane.From the question,
Given:
F = 550000 Nm = 7000 kgSubstitute these values into equation 1
a = 550000/7000a = 78.57 m/s²To calculate the velocity, we use the formula below.
v = u+at............. Equation 2Where:
v = Final velocityu = initial velocitya = accelerationt = time.From the question,
Given:
u = 0 m/sa = 78.57 m/st = 2.0 secondsSubstitute these values into equation 2
v = 0+2(78.57)v = 157.14 m/sHence, The acceleration and velocity of the plane is 78.57 m/s² and 157.14 m/s respectively.
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What is the minimum diameter necessary for a radio telescope working at f=1×10^10 Hz to be able to separate two objects 1 deg apart?
[a] 0.03 m
[b] 0.037 m
[c] 0.47 m
[d] 2.1 m
[e] 6.3 m
Re arrange the formula to make v the subject
Answer:
[tex]v = \sqrt{ \frac{2 e}{m} } [/tex]
Explanation:
[tex]e = \frac{1}{2} m {v}^{2} [/tex]
[tex]e = \frac{m {v}^{2} }{2} [/tex]
[tex]2e = m {v}^{2} [/tex]
[tex] \frac{2e}{m} = {v}^{2} [/tex]
[tex] \sqrt{ {v}^{2} } = \sqrt{} \frac{2e}{m} [/tex]
[tex]v = \sqrt{ \frac{2e}{m} } [/tex]
1. Explain what happens during a solar eclipse
Answer:
when the moon moves in front of the Sun as seen from a location on Earth.
Explanation:
A solar eclipse happens when the moon moves in front of the Sun as seen from a location on Earth. During a solar eclipse, it gets dimer and dimmer outside as more and more of the Sun is covered by the Moon. During a total eclipse, the entire Sun is covered for a few minutes and it becomes very dark outside.
What is nature of light Definition
What is the sources of light
What is the behaviours of light
Reflection and refraction
Colours and light
Concave and convex mirrors
Please help fast
(25 points)
Answer:
1.Light is a transverse, electromagnetic wave that can be seen by the typical human.
2.The Sun is the major source of light for the earth.
3.An inclusive list of the basic categories of behavior for light would most likely include transmission, absorption, reflection, refraction, diffraction, scattering, polarization and interference.
4.Spherical mirrors contain curved surfaces which are painted on either one of the sides. When an inward surface of a spherical mirror is painted, then this type of mirror is considered a convex mirror. In the meantime, when the outward surface of the spherical mirror is painted, this mirror is called a concave mirror.
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A net force of 250 N is applied to a 20 kg object, what will it's acceleration be?
Answer:
1N is the same as 1kgm/s2. F= ma. For m and a, enter the numbers in the appropriate units.For m, solve F = ma: m=F/a. Plug in some numbers for F and a, with proper units, starting with m = F/a. Substitute what I said the N is equivalent to for the N. Cancel as much as you can until kg = kg. It's effective!*Good luck!
A car weighing 14,700 N is speeding down a highway with a velocity of 99 km/h. What is the
momentum of this car?
Answer: 148348.6239 kg•m/s
Explanation: Firstly, we need to convert the 14700 N into kilograms, and to do so, use the formula net force is equal to mass times acceleration and rearrange the formula to find mass like shown below...
F = ma
F/a = m
14700/9.81 = 1498.470948 kg, this is your mass
Now that we convert it into kilograms, plug all the numbers into the variable of the momentum formula.
Momentum formula is P = mass x velocity
Like this:
P = 1498.470948 x 99
p = 148348.6239 kg•m/s.
I believe that is your answer, hope that helps you even a bit out.
Thanks.
What happen when the rest of the container is filled with water?
Answer:
the water will float over mercury and the steel ball will be in it's respected place
Two charges (Q1 = 0.00368 ; Q2 = 0.00059 ) are separated by a distance of 0.4 meters. What is the Force between them?
Coulomb force
[tex]\tt F=k\dfrac{q_1.q_2}{r^2}[/tex]
F = coulomb force (N)
Q1,Q2 = the charge (C)
r = distance between two charges (m)
k = constant = 9.10⁹ Nm² / C²
Input the value:
[tex]\tt F=9.10^9\dfrac{0.00368\times 0.00059}{0.4^2}=122130~N[/tex]
Please help me Thanks!
l will help and if you get it correctly I will say you are welcome
Please urgent
7. The average force applied by the engine of a 1200 kg car is 900 N. If the car is
initially traveling at 15 m/s, how fast will it be traveling 300 m farther down the road?
final speed will be 25.98 m/s Explanation Force applied by the engine on the car is We have to find final speed of the car after travelling 300 m , so displacement of car is given as so wor done by the engines on the car
Answer:
[tex]F=ma \\ 900 = 1200a \\ a = \frac{3}{4} m. {s}^{ - 2} \\ {v}^{2} = {u}^{2} +2 as \\ {v}^{2} = {(15)}^{2} + 2( \frac{3}{4} )(300) \\ = 675 \\ v = \sqrt{675} \: m. {s}^{ - 1} [/tex]
Explain how the amount of acceleration will change if the net force or the mass of the object increases?
Answer:
The acceleration of an item is proportional to the net force exerted on it and inversely proportional to its mass. When the force pushing on an item rises, so does the object's acceleration. The acceleration of an item decreases as the mass of the thing increases.
If the force on an object remains constant, increasing mass will decrease acceleration. In other words, force and acceleration are directly proportional, while mass and acceleration are inversely proportional. ( How Newton's Laws of Motion Work" 29 July 2008.)
If the mass of an object is held constant, increasing force will increase acceleration. If the force on an object remains constant, increasing mass will decrease acceleration.
This net force is the resultant force of all forces acting on the body at any given instant. The larger the magnitude of this net force, the larger the acceleration of the body. One of the definitions of mass is the resistance of a body to motion in general and acceleration in particular due to the application of force.
Explanation:
What is the net force on this object A: 0 newtons b: 200 newtons c:400 newtons D:600 newtons
Answer:
100kg
Explanation:
Total force =400N+600N=1000N
As we have,
F=m×g
1000N=m×10m/s²
1000N÷10m/s²=m
m=100kg
USATESTPREP
Examine the image here showing the different characteristics of the components of the electromagnetic spectrum. Ultraviolet light is often used to sterilize water in a water treatment facility. Which characteristics of ultraviolet light allow it to sterilize water? Select ALL that apply.
A) It is not visible, so pathogens cannot detect its presence.
B) It has low enough energy to be safe to humans but harmful to pathogens.
C) Its energy level is high enough to cause ionizing damage to DNA and molecules.
D) It has wavelengths too large to cause molecular changes to the molecules of water.
E) Its wavelengths are small enough to interact with bacteria and viruses, which can be pathogenic in the water.
Answer:
C, D, and E
Explanation:
Answer:it’s C, D, E
Explanation:
What is an independent variable?
Answer:
a variable (often denoted by x) whose variation does not depend on that of another.
Explanation:
If I do 140j work in 7 seconds, how much did I produce
you produce about 20 W
Explanation:
Determine the time in which the work will be done. Here, we can take t = 60 s . Power is equal to work divided by time. In this example, P = 9000 J / 60 s = 150 W(example)
Answer:
Explanation:
To determine the power (or rate of work) produced, we can use the formula:
Power = Work / Time
Where:
Work = 140 J (joules)
Time = 7 seconds
So, we can plug in the values into the formula:
Power = 140 J / 7 s = 20 W (watts)
So, you produced 20 watts of power in 7 seconds.
It's important to note that work and power are different physical quantities. Work represents the amount of energy transferred, while power represents the rate at which that energy is transferred.
the question is on he attachment
Answer:
A. put water in a container under the pot water will disappear from the container .
Block A has mass 1.00 kg and block B has mass 3.00 kg. The blocks collide and stick together on a level, frictionless surface. After the collision, the kinetic energy (KE) of block A is
Answer:
1/2mv²=0
1/2(4kg)(v²)=0
2=-v²
square root -2=v
v=1.414
Block A of mass 1 kg and block B has a mass of 3 kg, then the blocks collide and stick together so the kinetic energy of block A is one-third of the kinetic energy of the block.
What is Kinetic energy?Kinetic energy is a type of power that an item or particle possesses as a result of motion. When an item undergoes work—the transfer of energy—by having subject to a net force, it accelerates and consequently obtains kinetic energy. An object in motion or particle's kinetic energy, which depends on both mass and speed, is one of its features. Any combination of motions, including translation, rotation about an axis, and vibration, may be used as the type of motion.
A body's translational kinetic energy is equal to 1/2mv², or one-half of the sum of its mass, m, and the square of its velocity, v.
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Brainliest if correct Question 6 of 10
According to the law of conservation of energy, which statement must be
true?
A. There is only one form of energy.
B. Energy that is transformed is neither destroyed nor created.
C. The total energy in a system can only increase over time.
D. Energy can change only from nuclear to chemical
Answer:
B. Energy that is transformed is neither destroyed nor created.
Explanation:
Hope this helped! :)
who discovered the law of gravity and motion?
Answer:
Physically, Sir Isaac Newton was not a large man. However, he had a large intellect, as shown by his discoveries on gravity, light, motion, mathematics, and more. Legend has it that Isaac Newton came up with gravitational theory in 1665, or 1666, after watching an apple fall.
Explanation:
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Define the term Distance.
Maths :
The length between 2 points.
Physics =
Distance = speed × time
Distance is defined as the space between two points in space.
What is distance?Distance is defined as the space between two points in space.
The unit of distance is metres or kilometres.
Distance can be measured directly usind instruments such as metre rule and tape rule.
The formula for calculating distance is given below:
Distnce = velocity × time.
Therefore, distance is defined as the space between two points in space.
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When the reflection of an object is seen in a concave mirror the image will: Select one: a.will always be enlarged. O b.may be either real or virtual. c.always be real. d.always be virtual.
A baseball leaves the bat of Henry Aaron with a speed of 34 m/s at an angle of 370 above the horizontal. The ball is 1.2 m off the ground when it leaves the bat. To be a home run, the ball must clear a fence that is 3.0 m high and 106 m from the batter. (a.) At what times after being hit will it reach the height of the fence? (b.) How far from the batter will the ball be at these times? (c.) Will Henry have a home run? Explain
(a) The time taken for the ball to reach the height of the fence is 4.1 s.
(b) The position of the ball from the batter at the time is 111.33 m.
(c) Henry will have a home run.
Time to reach the height of the fence
The time taken for the ball to reach the height of the fence is calculated as follows;
[tex]h = v_0_y t - \frac{1}{2} gt^2\\\\(3 - 1.2) = (34 \times sin37)t - \frac{1}{2} (9.8)t^2\\\\1.8 = 20.46t- 4.9t^2\\\\4.9t^2 - 20.46t+ 1.8 = 0\\\\a = 4.9, \ b = -20.46, \ c = 1.8\\\\t = \frac{-b\pm \sqrt{b^2 - 4ac} }{2a} \\\\t = 4.1 \ s[/tex]
Position of the ball at the timeThe position of the ball at the time is calculated as follows;
[tex]X = v_x t\\\\X = (34 \times cos37) \times 4.1\\\\X = 111.33 \ m[/tex]
Since, 111.33 m is greater than 106 m, Henry will have a home run.
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When you push on a stationary object on a horizontal surface and it doesn't move, what kind of force is most likely opposing your efforts
Find the charge on capacitor, C2 , in the diagram below if V_ab=24.0 volts,〖 C〗_1=6.00 μF,〖 C〗_2= 3.00 μF,and C_3=10.0 μF.
The charge on the capacitor 2 (C₂) is 48μF.
Explanation of the circuit diagram
In the circuit diagram, C₁ and C₂ are in series connection, while C₃ is parallel to C₁ and C₂.
Same charge flows in a series arrangement (charge on C₁ = charge on C₂).Same voltage flows in a parallel arrangement (voltage on C₁, C₂ = voltage on C₃).Charge on capacitor C₁ and C₂[tex]Q_1 = Q_2 = Q\\\\V = \frac{Q}{C_1} + \frac{Q}{C_2} \\\\V = Q(\frac{1}{C_1} + \frac{1}{C_2} )\\\\Q = V (\frac{C_1C_2}{C_1 + C_2} )\\\\Q = 24 \times (\frac{6 \times 10^{-6} \times 3 \times 10^{-6}}{6\times 10^{-6} \ + \ 3 \times 10^{-6}} )\\\\Q = \frac{4.32 \times 10^{-10}}{9\times 10^{-6}} \\\\Q = 48 \ \mu C[/tex]
Thus, the charge on the capacitor 2 (C₂) is 48μF.
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When you apply a horizontal force of 51 N to a block, the block moves across the floor at a constant speed. (a) When you apply a force of 40 N to the block initially at rest, what is the magnitude of the horizontal component of the force that the floor exerts on the block? (b) When you apply a force of 59 N, what is the magnitude of the horizontal component of the force that the floor exerts on the block?
(a) When you apply a force of 40 N, the magnitude of the horizontal component of the force that the floor exerts on the block is 51 N.
(b) When you apply a force of 59 N, the magnitude of the horizontal component of the force that the floor exerts on the block is 51 N.
Net force on the blockThe horizontal component of the force that the floor exerts on the block is determined by applying Newton's second law of motion;
[tex]F = ma[/tex]
When the block moves at a constant speed[tex]\Sigma F = 0\\\\F - F_f = 0\\\\F = F_f[/tex]
[tex]F_f = F\\\\F_f = 51 \ N[/tex]
where;
[tex]F_f[/tex] is the frictional force exerted on the block by the floorSince the surface (floor) is constant, the frictional force (horizontal component of the force that the floor exerts on the block ) will be the same on the cases.
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Based on the definition of limiting frictional force;
the magnitude of the horizontal component of the force that the floor exerts on the block is equal to 40 N when a force of 40 N is applied to the block, the magnitude of the horizontal component of the force that the floor exerts on the block is equal to 51 N, the limiting frictional force, when a force of 59 N is applied to the block.What is force?Force is any push or pull effect which tends to cause a change in the motion or shape of an object.
Based on Newton's third law, every force produces an equal and opposite reaction.
Therefore, the force that the floor exerts on the block is oppositely direction to the force exerted by the horizontal force acting on the block.
Since a horizontal force of 51 N applied to the block is required to keep the block moving across the floor at a constant speed, this is the limiting frictional force.
a) when a force of 40 N is applied to the block, the magnitude of the horizontal component of the force that the floor exerts on the block is equal to 40 N
b) when a force of 59 N is applied to the block, the magnitude of the horizontal component of the force that the floor exerts on the block is equal to 51 N, the limiting frictional force.
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A fighter jet is being tested before it is delivered. The jet took off and had its initial velocity of 280 m/s. As the pilot ask permission from the flight control to test its capacity the jet reached its final velocity of 1250 m/s in just 20 seconds. What is the acceleration of the fighter jet?
Answer:
48.5 m/s².
Explanation:
We can use the following kinematic equation:
[tex]\displaystyle v_f = v_i + at[/tex]
Substitute in the appropriate values and solve for acceleration a:
[tex]\displaystyle \begin{aligned} (1250\text{ m/s}) & = (280\text{ m/s}) + a(20\text{ s}) \\ \\ a & = 48.5\text{ m/s$^2$} \end{aligned}[/tex]
Therefore, the acceleration of the fighter jet was 48.5 m/s².
How far will 60J of work push a hockey puck if the applied force is 5N? Question 15 options: 0.1 meters 300 meters 12 meters 65 meters
The distance travelled by the hockey puck, given that 60 J of work was done by an applied force of 5 N is 12 meters
How do I determine the distance travelled by the hockey puck?We know already that work is the product of force and distance moved in the direction of the force:
Work done (Wd) = force (F) × distance (d)
Using the above formula, we can determine the distance travelled by the hockey puck. Details below:
Workdone (Wd) = 60 JForce applied (F) = 5 NDistance travelled (d) =?Work done (Wd) = force (F) × distance (d)
60 = 5 × distance travelled
Divide both sides by 5
Distance travelled = 60 / 5
Distance travelled = 12 meters
Thus, from the above calculation, we can conclude that the distance travelled is 12 meters
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The speed of a rocket just after being launched is 12 m / s.
The mass of the rocket is 0.05 kg.
Calculate the kinetic energy of the rocket just after being launched. Use the equation: Kinetic energy = 0.5 x mass x velocity squared
As the rocket moves upwards, it gains gravitational potential energy.
State the maximum gravitational potential energy gained by the rocket. Ignore the effect of air resistance.
Maximum gravitational potential energy =
Calculate the maximum height the rocket will reach.
Ignore the effect of air resistance.
Gravitational field strength = 10 N/kg.
This question involves the concepts of the equations of motion, kinetic energy, and potential energy.
a. The kinetic energy of the rocket at launch is "3.6 J".
b. maximum gravitational potential energy of the rocket is "3.6 J".
a. KINETIC ENERGY AT LAUNCHThe kinetic energy of the rocket at launch is given by the following formula:
[tex]K.E=\frac{1}{2} m v_i^2[/tex]
where,
K.E = initial kinetic energy = ?m = mass of rocket = 0.05 kg[tex]v_i[/tex] = initial speed = 12 m/sTherefore,
[tex]K.E=\frac{1}{2}(0.05\ kg)(12\ m/s)^2[/tex]
K.E = 3.6 J
b. MAXIMUM GRAVITATIONAL POTENTIAL ENERGY
First, we will use the third equation of motion to find the maximum height reached by rocket:
[tex]2gh=v_f^2-v_i^2[/tex]
where,
g = -9.81 m/s²h = maximum height = ?vf = final speed = 0 m/sTherefore,
2(-9.81 m/s²)h = (0 m/s)² - (12 m/s)²
h = 7.34 m
Hence, the maximum gravitational potential energy will be:
P.E = mgh
P.E = (0.05 kg)(9.81 m/s²)(7.34 m)
P.E = 3.6 J
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A man pulls a box with an applied force of 500N. He pulls the box 100.0 m. About how much work does he do? Question options: 0.5 J 50,000 J 5,000 J 2 J
Workdone:-
[tex]\\ \tt\longrightarrow Force\times Displacement [/tex]
[tex]\\ \tt\longrightarrow 500(100)[/tex]
[tex]\\ \tt\longrightarrow 50000J[/tex]
A horizontal force of 15N is needed to keep a box sliding on a horizontal floor at a constant speed of 0.5m/s. The speed of the box is increased to 1.5m/s by some means (which is not the concern of this question). What horizontal force would be needed to keep the same box sliding on the same horizontal floor at a constant speed of 1.5m/s?
The horizontal force needed to keep the same box sliding on the same horizontal floor at a given constant speed is 45 N.
Horizontal forceThe horizontal force needed to keep the same box sliding on the same horizontal floor at a given constant speed is calculated as follows;
[tex]F= \frac{mv}{t} \\\\mv = Ft\\\\m = \frac{Ft}{v} \\\\\frac{F_1}{v_1} = \frac{F_2}{v_2}[/tex]
where;
v1 is the initial velocityv2 is the final velocity[tex]F_2 = \frac{F_1v_2}{v_1} \\\\F_2 = \frac{15 \times 1.5}{0.5} \\\\F_2 = 45 \ N[/tex]
Thus, the horizontal force needed to keep the same box sliding on the same horizontal floor at a given constant speed is 45 N.
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