Acceleration is constant, so we can use the following kinematic equation and Newton's second law.
[tex]{v_f}^2 - {v_i}^2 = 2a\Delta x \implies -{v_i}^2 = -2\dfrac{F}{m}\Delta x \implies {v_i}^2 = 2\dfrac{F}{m} \Delta x[/tex]
where [tex]v_i[/tex] and [tex]v_f[/tex] are initial/final velocities, [tex]a[/tex] is acceleration, [tex]\Delta x[/tex] is displacment, [tex]F[/tex] is force, and [tex]m[/tex] is mass. Since the objects are coming to rest, the acceleration opposes the direction of motion and is negative.
Solve for [tex]\Delta x[/tex].
[tex]\Delta x = -\dfrac{m {v_i}^2}{2F}[/tex]
Compute the displacements of both objects.
[tex]\Delta x_{5.0\text{-kg}} = -\dfrac{(5.0\,\mathrm{kg}) \left(2.0\frac{\rm m}{\rm s}\right)^2}{2F} = -\dfrac{10}F \dfrac{\rm m}{\rm N}[/tex]
[tex]\Delta x_{1.25\text{-kg}} = -\dfrac{(1.25\,\mathrm{kg}) \left(4.0\frac{\rm m}{\rm s}\right)^2}{2F} = -\dfrac{10}F \dfrac{\rm m}{\rm N}[/tex]
Then both objects are displaced by the same amount.
Answer:
They stop in the same distance
Explanation:
f friction is the same for both and must overcome the KE of each object to bring each to rest.
f friction * d = 1/2 mv^2 for each object :
f * d1 = 1/2 m1 v1^2 f * d2 = 1/2 m2 v2^2
= 1/2 5 (2 )^2 = 1/2 1.25 (4)^2
Object 1 KE = 10 j Object 2 KE = 10 j
f d1 / f d2 = 10 / 10 ( f is the same for both)
d1 / d2 = 1 so d1 = d2 they stop in the same distance !
What are two forces on earth that could change the motion of the object?
Answer:
Force makes things move or, more accurately, makes things change their motion. Two natural forces that we have experienced are the force of gravity and magnetic forces magnetic forces. These two forces act at a distance and do not require direct contact between the objects to function..
A mass is undergoing simple harmonic motion. When its displacement is 0, it is at its equilibrium position. At that moment, its speed is _______ and its acceleration is _______.
When displacement is 0, an object is at its equilibrium position, at that moment, its speed is maximum and its acceleration is zero.
What is speed?Speed is a scalar quantity used in physics to indicate the distance of change of an object that travels in any direction.
Conversely, displacement and acceleration are vector quantities because they express magnitudes but not a direction.
In conclusion, when an object is at its equilibrium position its speed is maximum and its acceleration is zero.
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A toroid has a 4.63 cm square cross section, an inside radius of 15.0 cm, 527 turns of wire, and a current of 0.725 A. What is the magnetic flux through the cross section
Using the equation, B = (μoIN/2πr)
The inner radius is r = 16.2 cm,
so the field there is
B = (4π 10-7).(0.813).(535)/2π(0.162)
= 5.37 × 10-4 T
The outer radius is r = 16.2 + 5.2 = 21.4 cm.
The field there is B = (4π 10-7).(0.813).(535)/2π(0.214)
= 4.06 × 10-4 T.
A toroid having a square cross-section, 5.20cm on edge and an inner radius of 16.2cm has 535 turns and carries a current of 813mA to calculate the magnetic.
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The primary coil of a transformer is connected to a 120 V wall outlet. The secondary coil is connected to a lamp that dissipates 60W. What is the rms current in the primary coil
The rms current in the primary coil of a transformer is 0.5 Ampere .
Energy/power is not gained or lost going through an ideal transformer.
Consequently, the transformer in this situation doesn't really matter. The entire device receives 60 watts of power from the wall outlet if the lamp consumes energy at a rate of 60 watts.
∴ Power = (voltage) x (current)
60 watts = (120 v) x (current)
Current = (60 watts) / (120 v)
Current = 0.5 Ampere
Therefore, the rms current in the primary coil of a transformer is 0.5 ampere.
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what is specific latent heat mean in simple word?
Answer:
Amount of energy required to change the 1kg of a substance without changing its temperature.
Please answer number 3!! High-speed stroboscopic photographs show that the head of a 0.2kg golf club is traveling
at 55 m/s just before it strikes a 0.046kg golf ball at rest on a tee. After the collision, the
golf club head travels (in the same direction) at 40 m/s. Find the speed of the golf ball
just after impact.
The velocity of the golf ball is 65.2 m/s
What is the principle of conservation of momentum?The principle of conservation of momentum is that the momentum before collision is equal to the momentum after collision.
Given that;
(0.2kg * 55 m/s) + (0.046kg * 0 m/s) = (0.2kg * 40 m/s) + (0.046kg * v)
11 = 8 + 0.046v
v = 11 - 8/0.046
v = 65.2 m/s
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I Explain why you feel cold when tap or well water in winter. state with reason.
Answer:
We feel cold when tap or well water in winter because heat flows from hot body to cold body.
Explanation:
Our body is in optimal status is a hot body and tap or well water is a cold body. Therefore we feel cold.
When we drink tap or well water in the winter, we feel chilled because heat transfers from a hot body to a cold body.
What is convection?
Convection, often known as a heat transfer mechanism, is a method of moving heat. When a fluid, such as air or water, is moving, this heat transfer takes place. Temperature differences within the fluid cause convection.
Due to the increased kinetic energy of the particles, a fluid loses density at the area nearest to the heat source as it is heated. The less dense portion of the fluid then rises, while the denser portion sinks.
The process keeps happening because the less dense fluids sink as they move away from their heat source while the denser fluids rise as they move toward the heat source. Convection currents are produced by this.
Therefore, When we drink tap or well water in the winter, we feel chilled because heat transfers from a hot body to a cold body.
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Calculate the gravitational force of the Earth and Moon. The Earth has a mass of 5.972x 1024 kg and the Moon has a mass of 7.348x 1022 kg. They are an average of 384 million meters apart.
Answer:
The force of gravitation is 1.98×10^(20) m.
Explanation:
Note: The value of m1, m2, r, and G is taken to be 5.97×10^(24) kg, 7.348×10^(22) kg, 3.84×10^(8) m, and 6.67×10^(-11) N m^2/ kg^2 respectively.
The force of gravitation F acting between two masses is calculated using the formula,
F=Gm1m2/ r^2
where G is the gravitational constant.
Hence force F is
F=6.67×10^(-11)×5.97×10^(24)×7.348×10^(22)/ (3.84×10^(8))^2
F=1.98×10^(20) N
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lamp 1 and lamp 2 are in a series circuit
lamp 1 stops working
explain why lamp 2 also stops working
Lamps 1 and 2 are connected in series. lamp 1 stops working because the circuit is not complete.
What is a series circuit?In a series circuit, the amount of current flowing through any component is the same and the sum of the individual resistances equals the overall resistance of any series circuit.
The voltage in a series circuit is equal to the total of the individual voltage drops.
For the bulb to glow the circuit should be complete. If the circuit is not completed the bulb will not glow.
Lamps 1 and 2 are connected in series. lamp 1 stops working because the circuit is not complete.
Hence, lamp 1 stops working.
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A solid disk is rolling without slipping on a level surface at a constant speed of 2.50 m/s. (a) if the disk rolls up a 30.0o ramp, how far along the ramp will it move before it stops? (b) explain why your answer in part (a) does not depend on either the mass or the radius of the disk.
(a) By using the law of conservation of energy, the distance moved by the solid disk along the ramp is 0.96 m.
(b) The answer does not depend on the mass and radius of the disc as these values are canceled when the law of conservation of energy is applied.
What is the law of conservation of energy?
The law of conservation of energy states that the total energy of an isolated system is conserved.
The total initial energy Ei of the disc rolling down an inclined plane is given by the formula,
Ei=1/2*mv^2 + 1/2*Iω^2
where m is the mass of the disc, v is the velocity of the disc, I is the moment of inertia and ω is the angular velocity of the disc.
For a solid disc, I=1/2mr^2, and since it is rolling without slipping, the rolling velocity of the disc will be equal to its translational velocity, that is,
v=ωr or ω=v/r
So using ω=v/r and I=1/2mr^2, it can be written,
Ei=1/2*m(v)^2 + 1/2*(1/2mr^2)(v/r)^2
Ei=1/2*m(v)^2*(1+1/2*)
Ei=3/4*mv^2
If the height covered by the disc is h before stopping, then its final total energy Ef will be equal to the potential energy, that is,
Ef=m*g*h
From the law of conservation of energy, it can be written,
Ei=Ef
3/4*mv^2=m*g*h
h=3v^2/(4g)
The length of the ramp is then given by the formula,
l=hcosecθ
where θ is the inclination angle. So
l= 3v^2/(4g)*cosecθ
Here g=9.8 m/s^2, v=2.5 m/s and θ=30.0 degree. Using these values,
l= 3*(2.5 )^2/(4*9.8)*cosec( 30)
l= 0.96 m
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A 5.0 kg box slides down a 4.0 m long ramp that makes a 25 angle with the ground. If the coefficient of kinetic friction is 0.65, how much thermal energy was produced?
The thermal energy was produced is 116J
What is the thermal energy produced?Now we know that the frictional force produces the energy that is lost as heat as the body slides down the incline. The magnitude of the frictional force is obtained from;
Ff= μmgcosθ
Ff = 0.65 * 5.0 kg * 9.8 m/s^2 * cos 25
Ff = 29 N
Hence, the thermal energy is;
29 N * 4.0 m = 116J
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The magnetic field at the center of two concentric current loops is zero. The smaller loop has a radius of 0.01 m and a current of 10.0 A. The larger current loop carries a current of 29.0 A. What is the radius of the larger loop
Answer:
B = μ I / (2 * R) magnetic field at center of current carrying loop
I1 / R1 = I2 / R2 and we want to know R2
R2 = I2 / I1 * R1 = 29 / 10 * 01 = .029 m
A local FM radio station broadcasts at an energy of kJ/photon. () Calculate the frequency at which it is broadcasting. Frequency
The frequency is 94.2 Hz
We are given the energy of a photon = 6.24x10-29 kJ
We are asked to find the frequency (ν),
We use the equation E = hν where E is energy, h is Planck's constant and ν is the frequency
The quantity of events per unit of time that occur when an event repeats itself at regularly spaced intervals is known as its frequency.
6.24x10-29 kJ = 6.626x10-34 J-sec * ν
ν = 6.24x10-29 kJ x 1000 J/kJ / 6.626x10-34 Jsec
ν = 6.24x10-26 J / 6.626x10-34 Jsec
ν = 0.942x108 s-1 = 9.42x107 s-1 = 9.42x107 Hz
9.42x107 Hz x 1 MHz/106 Hz = 94.2 MHz
Hence the frequency is 94.2MHz
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An unstoppable object is heading right toward an unmovable object. What's going to happen?
Answer:
In my opinion the unstoppable object will hit the unmovable object and stop but the wheels will still be rolling and trying to move but can't.
Hope this helps.Good luck ✅.Which of the following graphs represents the equation y-2 = 3(x-1)?
Answer:
Graph b
Explanation:
I used Desmond calculator which is a calculator where you can graph an equation! :)
Hello!
The slope-intercept form is :
y - 2 = 3(x - 1)
y - 2 = 3x - 3
y = 3x - 1
Two points that are found to be on the line :
(0, -1) and (1, 2)
∴ The graph which contains both points is Graph B.
Two boys are playing with two different balls of masses m and 2 meter respectively. if first boy through vertically up and the second boy throws at an angle teta from vertical and balls remains same time in air what will be the ratio of height attend by the two balls
The ratio of the height attended by the object thrown upward to that thrown at some angle θ with the vertical is 1.
Note: It is assumed that the mass of the first object is m and the mass of the second object is 2m. Also, the first object is thrown upward and the second object is thrown at some angle θ with the vertical.
Projectile motion: Any object that is thrown in the air is called a projectile and the motion described by it under gravity is called the projectile motion.
If the air resistance is neglected, then the acceleration due to gravity g is the same for all objects irrespective of their masses. It is given that both objects remain in the air for the same time period. So first calculate the time period for the objects when they are in the air.
Time period of the first object: The first object is thrown in the air upwards so from the second kinematics equation,
h1=u1*t1-(1/2)*gt1^2
where h1 is the height, t1 is the time, and u1 is the initial velocity for the first object.
When the object is not in air h1=0, so
0=u1*t1-(1/2)*gt1^2
After solving the above quadratic equation, the values of t1 obtained are t1=0 which represents the initial time, and t1=2u1/g which represents the time period. So the time period of the first object is,
t1=2u1/g
Time period of the second object: The second object is thrown at some angle θ with the vertical as shown in the diagram. From the diagram, the initial velocity along the vertical direction is,
u2=uocos(θ)
where uo is the initial velocity and u2 is the initial velocity along the vertical direction.
From the second kinematic equation,
h2=u2*t2-(1/2)*gt2^2
where h2 is the height, t2 is the time, and u2 is the initial velocity for the second object along the vertical direction.
When the object is not in air h2=0, so using u2=uocos(θ),
0=uocos(θ)*t2-(1/2)*gt2^2
After solving the above quadratic equation, the values of t2 obtained are t2=0 which represents initial time, and t2=2uocos(θ)/g which represents the time period. So the time period of the second object is,
t2=2uocos(θ)/g
Given that the time period is the same for both cases,
t1=t2
2u1/g=2uocos(θ)/g
u1=uo cos(θ)
Calculation of the ratio of the height of the object:
The maximum height is attained when the time of the object in the air is half of the total time period. At maximum height, velocity is zero.
From the third equation of motion.
v^2=u^2-2gh
h=u^2/2g
where h is the height and u is the initial velocity of an object.
Using it and u1=uo cos(θ) and u2=uo cos(θ), the ratio of h1 and h2 is,
h1/h2= u1^2/2g÷u2^2/2g
h1/h2=(uo cos(θ))^2/(uo cos(θ))^2
h1/h2=1
The ratio of their height will be 1.
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What is typically responsible for forming volcanic island chains in the oceans?
A. Shield volcanoes
B. Fissures
Composite volcanoes
Hot spot volcanism
Hot spot volcanism is responsible for the formation of volcanic island chains in the oceans and is denoted as option D.
What is a Volcano?This is a rupture in the earth crust thereby giving rise to the release of hot lava, gases etc from below the surface.
The pushing of the lava through the cracks partially melts the plate at a subduction zone thereby forming the volcanic island chains.
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A typical muscle fiber is 2.0 cm long and has a cross-section area of 3.1 * 10-9 m2. When the muscle fiber is stimulated, it pulls with a force of 1.2 mN. What is the work done by the muscle fiber as it contracts to a length of 1.6 cm
= F.d
F = 1.2 x 10-3N
d= (2-1.6)10-2=0.004m
W=1.210-3(0.004)=4.8 x 10 -6 J
A typical muscle fiber is 2.0 cm long and has a cross-section area of 3.1 * 10-9 m2. When the muscle fiber is stimulated, it pulls with a force of 1.2 mN.
Please check the attached file for a brief answer.
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The work done by the muscle fiber is 4.80x 10 J
The work done by the muscle fiber as it contracts to a length is
W = FAL
= (1.20x103 N) (0.4x10-2 m)
= 4.80x 10 J
Muscle tissues contain something known as muscle fibers. Muscle fibers consist of a single muscle cellular.
They assist to control the physical forces in the body. When grouped collectively, they are able to facilitate organized movement of your limbs and tissues.
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Use the graph to complete the sentences about two runners.
Runner A is
than runner B.
Runner
has a head start.
Runner A is faster than runner B and Runner B has a head start.
What is a velocity time graph?The velocity time graph shows us how fast that an object can be. The slope of the graph is the acceleration.
If we look at the graph as shown in the image attached to this answer, we can see that Runner A is faster than runner B and Runner B has a head start.
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Answer:
Runner A is
✔ faster
than runner B.
Runner
✔ B
has a head start.
Explanation:
On the way to school, the bus speeds up from 20 m/s to 36 m/s in 4 seconds. What distance
does the bus cover in this time frame?
A. 64 m
B. 112 m
C. 80 m
D. 144 m
Suppose a car traveling at 8m/s is brought to rest in a distance of 20m.what is it's deceleration and time taken.
The deceleration is -8 m/s² and the time taken by the car is 2.5 s.
It is given that,
Speed, v = 8 m/s
Distance, d = 20m
We need to find the time taken by the car and deceleration.
As we know speed is defined as
v = d/t
Plugging in the values,
t = 20/8
t = 2.5s
Now from the deceleration formula,
a = - v/ t
a is the deceleration, v is the speed and t is the time taken.
a = - 20/ 2.5
a = - 8 m/s²
Thus, the time taken by the car and deceleration is 2.5 s and -8 m/s² respectively.
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A fluorescent light "hums" at
120 Hz. A bored grad student
hums at 118.4 Hz. What beat
frequency does he hear in his
head?
(Unit = Hz)
(This is on autobiographical problem buthe
The beat frequency he hears in his head will be 1.6 Hz
What is the frequency of the sound?A sound pressure wave's frequency, also known as pitch, is the number of times it repeats itself every second.
The frequency of the sound is the inverse of the period. If the wavelength of a wave is short. The wave will indeed have a lower frequency. A longer wavelength denotes a lower frequency.
Given data;
The frequency at which the fluorescent light "hums",f=120 Hz
The frequency at which the bored grad student "hums",f'=118.4 Hz
The beat frequency he hears in his head is,fₐ=?
The beat frequency he hears in his head is found as;
fₐ = f-f'
fₐ = 120 - 118.4 Hz
fₐ = 1.6 Hz
Hence the beat frequency he hears in his head will be 1.6 Hz.
At 120 Hz, a fluorescent light "hums." At 118.4 Hz, a Ph.D. student who is bored hums. He hears 1.6 Hz as the rhythm frequency in his brain.
Hence the beat frequency he hears in his head will be 1.6 Hz
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A 18 kg box slides from rest down a ramp inclined at 25° to the horizontal onto a spring with a spring constant 740 N/m
as shown in the diagram. The spring is compressed 0.32 m before the box stops. Determine how far along the ramp the
box will slide before it stops. Assume friction is negligible.
The distance traveled by the box along the ramp is 0.51 m.
Distance traveled by the box
Apply the principle of conservation of energy;
Potential energy of the box at top of incline = Elastic potential energy at bottom
mgh = ¹/₂kx²
where;
h is the height of the incline m is mass of the boxx is the compression of the springk is spring constanth = L sinθ
where;
L is length of the incline = distance traveled by the boxmg(L sinθ) = ¹/₂kx²
(18)(9.8)(L sin25) = ¹/₂(740)(0.32)²
74.55 L = 37.89
L = 37.89 / 74.55
L = 0.51 m
Thus, the distance traveled by the box along the ramp is 0.51 m.
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A box having a mass of 7.3 kg is lifted 0.60 m from ground level to the top of a desk. What is the gravitational potential energy in J of the box on the desktop?
Considering the definition of potential energy, the potential energy of the box is 42.9678 J.
Potential energyPotential energy is the energy that measures the ability of a system to do work based on its position. In other words, this is the energy that a body located at a certain height above the ground has.
Gravitational potential energy is the energy associated with the gravitational force. This will depend on the relative height of an object to some reference point, the mass, and the strength of gravity.
So for an object with mass m, at height h, the expression applied to the gravitational energy of the object is:
Ep= m×g×h
Where:
Ep is the potential energy in joules (J).m is the mass in kilograms (kg).h is the height in meters (m).g is the acceleration of fall in m/s² (approximately 9.81 m/s²).Potential energy in this caseIn this case, you know:
m= 7.3 kgg= 9.81 m/s²h= 0.60 mReplacing in the definition of potential energy:
Ep= 7.3 kg × 9.81 m/s² × 0.60 m
Solving:
Ep= 42.9678 J
Finally, the potential energy of the box is 42.9678 J.
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Suppose that the engine of a 1,500 kg automobile has a maximum power output of 30 hp. What is the maximum grade (in percent) that the automobile can climb at 41 km/h if the drag force on it is 410 N
The maximum grade (in percent) is 14%.
Suppose angle be theta.
Power = force * velocity
35*746 = [1300*9.8* sin theta + 410]* (43*1000/3600)
sin theta = (35*746/(43*1000/3600) - 410)/(1300*9.8) = 0.139
grade = tan ( arcsin 0.139 )
= 0.140
= 14%
Drag is a mechanical pressure. it is generated via the interplay and speaks to a stable body with a fluid (liquid or fuel). It is not generated via a force area, within the feel of a gravitational field, or an electromagnetic field, wherein one item can have an effect on some other item without being in physical touch.
For instance drag on a ship moving in water or drag on an aircraft shifting in the air. therefore drag pressure is the resistance pressure resulting from the motion of a body thru a fluid like water or air. This drag force acts opposite to the path of the oncoming float velocity.
Drag is generated via the distinction in speed between the strong item and the fluid. There needs to be motion among the object and the fluid. If there's no movement, there's no drag. It makes no difference whether or not the item moves via a static fluid or whether or not the fluid movements beyond a static stable item.
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The maximum grade (in percent) is 14%.
Suppose angle be theta.
Power = force * velocity
35*746 = [1300*9.8* sin theta + 410]* (43*1000/3600)
sin theta = (35*746/(43*1000/3600) - 410)/(1300*9.8) = 0.139
grade = tan ( arcsin 0.139 )
= 0.140
= 14%
Drag is a mechanical pressure. it is generated via the interplay and speaks to a stable body with a fluid (liquid or fuel). It is not generated via a force area, within the feel of a gravitational field, or an electromagnetic field, wherein one item can have an effect on some other item without being in physical touch.
For instance drag on a ship moving in water or drag on an aircraft shifting in the air. Therefore drag pressure is the resistance pressure resulting from the motion of a body thru a fluid like water or air. This drag force acts opposite to the path of the oncoming float velocity.
Drag is generated via the distinction in speed between the strong item and the fluid. There needs to be a move between the object and the fluid. If there's no movement, there's no drag. It makes no difference whether or not the item moves via a static fluid or whether or not the fluid moves beyond a static stable item.
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A projectile is launched with a velocity of 13.2 m/s at an angle of 37.0° above the horizontal.
How high above the point from which it was fired will the projectile be 1.0 s after firing?
a.
3.0 m
b.
12.8 m
c.
5.6 m
d.
13.2 m
The height reached by the projectile from the point of projection is determined as 12.84 m.
Height reached by the projectileH = vt + ¹/₂gt²
where;
v is vertical velocityt is time of motiong is acceleration due to gravityH = (13.2 x sin37)(1) + (0.5)(9.8)(1)²
H = 12.84 m
Thus, the height reached by the projectile from the point of projection is determined as 12.84 m.
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An uncharged metal rod brought close to but not touching the cap of a charged electroscope caused decrease in the divergence of the leaf. Explain this observation.
-ItzKenzo
-CuteSiKenzo(Oo cute)
Alam kung wala kang pake tángina mo!!
A top fuel dragster with a mass of 500.0 kg starts from rest and completes a quarter mile (402 m) race in a time of 5.0 s. The dragster's final speed is 130 m/s. Neglecting friction, what average power was needed to produce this final speed
The average power needed to produce this final speed is 1069.1 hp.
Mass of the dragster, m = 500.0 kg,
Displacement travelled by the dragster, s = 402 m,
Time taken in this travel, t = 5.0 s,
Final velocity of the dragster, v = 130 m/s.
Let the initial velocity of the dragster be u and acceleration be a.
Using kinematical equation, s = ut + (1/2)at^2.
402 = u*5 + (1/2)*a*5^2
10*u + 25*a = 804. ...........(1)
Using kinematical equation, v = u +at.
130 = u + 5*a
5*u + 25*a = 650. .............(2)
Solving (1)and (2), we get,
u = 30.8 m/s.
According to work-energy theorem,
Work done = change in kinetic energy
W = (1/2)*m*(v^2 - u^2)
W = (1/2)*500*(130^2 - 30.8^2)
W = 3987840. J
Therefore power rating of the dragster is given by,
P ⇒ W/t. = 3987840/5 = 797568 watt.
P ⇒ 797568/746 = 1069.1 hp.
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The answer is 1069.1 hp.
Mass of the dragster, m = 500.0 kg,
Displacement travelled by the dragster, s = 402 m,
Time taken in this travel, t = 5.0 s,
Final velocity of the dragster, v = 130 m/s.
Let the initial velocity of the dragster be u and acceleration be a.
Using kinematical equation, s = ut + (1/2)at^2.
402 = u*5 + (1/2)*a*5^2
10*u + 25*a = 804. ...........(1)
Using kinematical equation, v = u +at^.
130 = u + 5*a^
5*u + 25*a = 650. .............(2)
Solving (1)and (2), we get,
u = 30.8 m/s.
According to work -energy theorem,
Work done = change in kinetic energy
W = (1/2)*m*(v^2 - u^2)
, W = (1/2)*500*(130^2 - 30.8^2)
, W = 3987840. J
therefore Power rating of the dragster is given by,
P = W/t. = 3987840/5 = 797568 watt.
, P = 797568/746 = 1069.1 hp.
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A 2.00 kg block hangs from a spring. A 300 g body hung below the block stretches the spring 2.00 cm farther. (a) What is the spring constant
The spring constant is 147 N/m
Given the mass of the block is 2.00 kg , the mass of the body is 300 g and the length of the spring is 2.00 cm
We need to find the spring constant
A spring is an object that can be deformed by a force and then return to its original shape after the force is removed.
The force required to stretch an elastic object such as a metal spring is directly proportional to the extension of the spring
We know that F = kx
300(9.8)= k (0.02)
k = 147.15 N/m
Rounding off to the nearest is 147N/m
The spring constant is 147N/m
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A spring is hung from the ceiling. A 2.0-kg mass suspended hung from the spring extends it by 6.0 cm. A downward external force applied to the mass extends the spring an additional 10 cm. What is the work done by the force
The work done by force on a spring hung from the ceiling will be 1.67 J
Any two things with mass are drawn together by the gravitational pull. We refer to the gravitational force as attractive because it consistently seeks to draw masses together rather than pushing them apart.
Given that a spring is hung from the ceiling with a 2.0-kg mass suspended hung from the spring extends it by 6.0 cm and a downward external force applied to the mass extends the spring an additional 10 cm.
We need to find the work done by the force
Given mass is of 2 kg
So let,
F = 2 kg
x = 0.1 m
Stiffness of spring = k = F/x
k = 20/0.006 = 333 n/m
Now the formula to find the work done by force will be as follow:
Workdone = W = 0.5kx²
W = 0.5 x 333 x 0.1²
W = 1.67 J
Hence the work done by force on a spring hung from the ceiling will be 1.67 J
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