It will take the model rocket 0.2 seconds to accelerate from rest to a velocity of 208 ft/sec.
In order to answer the question of how long it will take a model rocket to accelerate from rest to a velocity of 208 ft/sec, we need to understand the acceleration of the rocket.
According to the given parameters, the acceleration of the model rocket is proportional to the difference between 260 ft/sec and the rocket's velocity.
Therefore, when the rocket is initially at rest and its initial acceleration is 260 ft/sec, the time it takes to accelerate to 208 ft/s can be calculated as follows:
Acceleration = (260 ft/sec - 208 ft/sec) / time
Rearranging the equation, we get:
Time = (260 ft/sec - 208 ft/sec) / acceleration
Plugging in the values given, we get:
Time = (260 ft/sec - 208 ft/sec) / 260 ft/sec
Time = 0.2 seconds
Therefore, it will take the model rocket 0.2 seconds to accelerate from rest to a velocity of 208 ft/sec.
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a constant force fa is applied to an object of mass m, initially at rest. the object moves in the horizontal x-direction, and the force is applied in the same direction. after the force has been applied, the object has a speed of vf. which mathematical routines can be used to determine the time in which the force is applied to the object of mass m? select two answers.
t=Δpx/FA
Δt=Mvf/FA
F = ma, Newton's second rule of motion, may be used to calculate when a force is applied to an object of mass m.
The time may also be calculated using the equation of motion, vf = vi + at, where an is the acceleration and t is the time.
There are two mathematical procedures that may be used to figure out when a force is applied to an object with mass m: Newton's Second Law of Motion states that F = ma, where ma is the object's mass and an is its acceleration, and F is the applied force.
The acceleration may be found by rearranging the equation, which can then be used to calculate how long it will take the item to reach its final speed vf.
Kinematic Equation for Constant Acceleration: where vi is the starting velocity, t is the time, and an is the acceleration, vf = vi + at The time t may be calculated from the final velocity vf and the acceleration a by replacing the initial velocity with 0 (because the object is initially at rest).
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The complete question is:
A constant force FA is applied to an object of mass M, initially at rest. The object moves in the horizontal x-direction, and the force is applied in the same direction. After the force has been applied, the object has a speed of vf. Which mathematical routines can be used to determine the time in which the force is applied to the object of mass M? Select two answers.
two cars in an inelastic collision . A small car with a mass of 1600kg is moving east at 55 km/hr a van with a mass of 3300 kg is moving west 65 km/hr . calculate the speed and direction of the combined vehicles just after impact
The speed and direction of the combined vehicles just after impact is 7.2 m/s west.
What is final speed of the cars after the collision?
The final speed of the cars after the collision is obtained by applying the principle of conservation of linear momentum.
m₁u₁ + m₂u₂ = v ( m₁ + m₂ )
where;
m₁ is the mass of the first carm₂ is the mass of the second caru₁ is the initial velocity of the first car = 55 km/h = 15.28 m/su₂ is the initial velocity of the second car = 65 km/h = 18.1 m/sv is the final velocity of the two cars(1600 x 15.28) - ( 3300 x 18.1) = v ( 1600 + 3300)
-35,282 = 4,900v
v = -35,282 / 4,900
v = -7.2 m/s
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during a flight where the distance was 1200 km, an aircraft was slowed down due to bad weather. its average speed for the trip was reduced by 400 km/hr. and the time of flight increased by 60 minutes. the duration of the flight is:
The duration of the flight is 3 hours if its speed is reduced by 400 km/h, and the time of flight increased by 60 minutes.
The length of the trip, d = 1200 km
Reduced average speed of the aircraft, v = 400 km/h
Increased time of flight, = 60 minutes = 1 hour
Let the initial time of the flight = t hours
Total time during reduced speed, = (t+1) hours
So, total distance, 1200 km = 400 × (t+1)
t+1 = 1200/400 = 3 hours
So total time of flight is 3 hours.
Initial time of flight, t = 3 - 1 = 2 hours
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if we assume that all of the u 235 and u 238 undergo fission and roughly 200 mev of energy is emitted per fission, for how many days can this fuel power a 3 gw power plant, assuming the plant is able to convert fuel energy to power with roughly 30% efficiency?
The plant can convert fuel energy into power with an efficiency of about 30% for 241 days .
As per the data given in the above question are as bellow,
We need to know how much fuel is in the plant in order
To calculate how many days a 3 GW power plant can run on fuel containing U-235 and U-238.
Assume the facility contains 1 metric tonne of U-235 and U-238-containing fuel.
The total energy released from the fission if all of the U-235 and U-238 split would be
200 MeV/fission * 2.51021 fissions/metric tonne = 51023 MeV.
The total energy transferred to electricity would be
51023 MeV * 0.3 = 1.51023 MeV = 1.51016 J
if we assume a 30% efficiency.
It would require 1.51016 J / 3109 J/s or 5106 s or 5,800 hours or 241 days to generate 3 GW at the power plant.
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PLEASE ANSWER MY OTHER QUESTION
An airplane travels down a
runway for 4.0 seconds with an
acceleration of 9.0 m/s2. What
its change in velocity during
this
time?
Someone helpppp!!
Answer:
We can use the equation of motion of an object under constant acceleration to calculate the change in velocity during this time. The equation is:
Δv = at
Where:
Δv = change in velocity (m/s)
a = acceleration (m/s^2)
t = time (s)
Given that the acceleration of the airplane is 9.0 m/s^2 and the time it travels down the runway is 4.0 seconds we can substitute these values into the equation:
Δv = 9.0 m/s^2 * 4.0 s = 36 m/s
Therefore, the change in velocity of the airplane during this time is 36 m/s.
a satellite travels at constant speed in a circular orbit at a height just 100 miles above sea level. without doing any calculations, what is the acceleration of the satellite?
The acceleration of the satellite is just below the acceleration of gravity
i.e 9.8 m/s^2.
Given that,
the speed of the satellite in a circular orbit at a height just 100 miles above sea level.
The acceleration of the satellite is due to the force of gravity. The force of gravity is equal to the mass of the satellite times the acceleration of gravity, which is 9.8 m/s^2. The mass of the satellite is very small, so the force of gravity is also very small. However, the satellite is also moving very fast, so the centripetal force is also very small. As a result, the acceleration of the satellite is just a hair less than 9.8 m/s. The satellite is accelerating because it experiences a net force acting on it, and also because its velocity is changing.
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A proton is at the origin. One electron is at the point (2m,4m) and the other is at the point (-2m,-4m). What is the net force on the proton. F=
The net force on the proton is 2.88 x 10⁻³⁰ N.
What is the net electric force between the particles?The net electric force between the particles is calculated by applying Coulomb's law as shown below.
F = ( kq₁q₂ ) / r²
where;
k is the Coulomb's constantq₁ is the charge of the electronq₂ is the charge of the protonr is the distance between the chargesThe distance between the charges is calculated as;
r = √ [ (-2 - 2)² + ( -4 - 4 )² ]
r = 8.94 m
F = ( 9 x 10⁹ x 1.6 x 10⁻¹⁹ x 1.6 x 10⁻¹⁹ ) / ( 8.94² )
F = 2.88 x 10⁻³⁰ N
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a ringing alarm clock is put under a glass jar. the air is slowly removed from the space around it. what will happen as the air is removed? (1 point)
As the air is removed from space around a ringing alarm clock that is placed under glass jar, sound of the alarm will become progressively quieter, and the clock's mechanical parts will move more smoothly.
What will happen as the air is removed?As the air is slowly removed from space around a ringing alarm clock that is placed under a glass jar, the sound of alarm will become progressively quieter. This is because sound is a mechanical wave that travels through a medium.
The air particles in jar vibrate as the sound waves pass through them, which allows us to hear the sound. As the air is removed, number of air particles available to vibrate and carry the sound waves decreases, which reduces the intensity of sound. As the air is almost completely removed, there will be very few air particles left to vibrate, then the sound will be almost inaudible.
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A 1. 0 kg toy car is released at the top of a frictionless track on the left and rolls off of the track from its right
side ramp. The car starts at a height of 0. 80 m, goes through a 0. 50 m diameter loop, and exits the ramp at a
height of 0. 25 m.
The change in gravitational potential energy is 5.39 J.
Describe gravitational potential energy?Gravitational potential energy is a type of potential energy that an object possesses due to its position in a gravitational field. It is the energy that an object has due to its height and the force of gravity acting upon it. The formula for gravitational potential energy is given by mgh, where m is the mass of the object, g is the acceleration due to gravity, and h is the height of the object above a reference point. When an object falls from a height, it converts its gravitational potential energy into kinetic energy.
To find the change in gravitational potential energy, we can use the formula: ΔU = mgh, where m is the mass of the object (1.0 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the change in height.
The change in height is 0.80 m - 0.25 m = 0.55 m. Plugging in the values, we get:
ΔU = 1.0 kg * 9.8 m/s^2 * 0.55 m = 5.39 J
So the change in gravitational potential energy is 5.39 J.
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The complete question is:
A 1.0kg toy car is released at the top of a frictionless track on the left and rolls off of the track from its right side ramp. The car starts at height of 0.80 m, goes through a 0.50 m diameter loop, and exits the ramp at a height of 0.25 m 0.80 m 0.50 mi 0.25 m What is the change in the car's gravitational potential energy from A to B? Round answer to two significant digits.
Four mass–spring systems oscillate in simple harmonic motion. Rank the periods of oscillation for the mass–spring systems from largest to smallest.m = 2 kg , k = 2 N/mm = 2 kg , k = 4 N/mm = 4 kg , k = 2 N/mm = 1 kg , k = 4 N/m
The order of the periods of oscillation from largest to smallest is:
T1 > T3 > T2 > T4
How to find the smallest?The period of oscillation for a mass-spring system is given by the formula:
T = 2π √(m/k)
Where T is the period, m is the mass of the oscillating object and k is the spring constant.
Using this formula, we can find the periods of oscillation for the four mass-spring systems:
m = 2 kg, k = 2 N/m: T = 2π √(2 kg / 2 N/m) = 2π √(1) = 2π secondsm = 2 kg, k = 4 N/m: T = 2π √(2 kg / 4 N/m) = 2π √(0.5) = π secondsm = 4 kg, k = 2 N/m: T = 2π √(4 kg / 2 N/m) = 2π √(2) = 2π √(2) = 2π √(2) secondsm = 1 kg, k = 4 N/m: T = 2π √(1 kg / 4 N/m) = 2π √(0.25) = π/ √(2) secondsSo the order of the periods of oscillation from largest to smallest is:
T1 > T3 > T2 > T4
Where T1 is the period of the mass-spring system with m = 2 kg and k = 2 N/m, T3 is the period of the mass-spring system with m = 4 kg and k = 2 N/m, T2 is the period of the mass-spring system with m = 2 kg and k = 4 N/m, and T4 is the period of the mass-spring system with m = 1 kg and k = 4 N/m.
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what is the line charge density on a long wire if the electric field 45 cm from the wire has magnitude 240 kn/c and points toward the wire?
The line charge density on a long wire if the electric field 45 cm from the wire has magnitude 240 kn/c and points toward the wire is 2.08 x 10-7 C/m.
Calculating the line charge density of a long wire is a simple process. To begin, we need to know the magnitude of the electric field at a given distance from the wire. In this case, the electric field 45 cm from the wire has a magnitude of 240 kn/c and points toward the wire.
Using this information, we can calculate the line charge density of the wire using the equation:
Line charge density (λ) = Electric field (E) * Permittivity of free space (ε0) / 2π * distance (r)
Plugging in the values, we get:
λ = 240 kn/c * 8.85 x 10-12 C2/Nm2 / 2π * 0.45m
Solving this equation, we get a line charge density of 2.08 x 10-7 C/m.
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an 8 3 4 -inch circular power saw rotates at 4,500 revolutions per minute. (round your answers to two decimal places.) (a) find the angular speed of the saw blade in radians per minute. incorrect: your answer is incorrect. rad/min (b) find the linear speed (in feet per minute) of one of the 24 cutting teeth as they contact the wood being cut. incorrect: your answer is incorrect. ft/min
(a) Angular speed of the saw blade in radians per minute is:
(b) The linear speed (in feet per minute) of one of the 24 cutting teeth as they contact the wood being cut is 286.68 ft/min.
What is angular speed?Angular speed formula is used to calculate the distance that the body covers in terms of rotations or revolutions to the time taken.
The angular speed of the saw blade in radians per minute is calculated as 785.39 rad/min. (b) linear speed of one of the cutting teeth in feet per minute is 286.68 ft/min
(a) Given, 4,500 revolutions per minute
rad/min = RPM * 2 * pi / 60
So, angular speed of the saw blade in radians per minute is:
rad/min = 4,500 * 2 * pi / 60 = 785.39 rad/min (rounded to two decimal places)
(b) Given, 24 cutting teeth. Circumference of a circle is given by the formula: C = 2 * pi * r
r is radius of the circle.
linear speed (v) is given by the formula: v = r * w
w is angular speed in radians per second.
So, you can find linear speed of one of the cutting teeth in feet per minute as: r = (8 3/4 inches) / 2 = 4.375 inches
C = 2 * pi * 4.375 inches = 27.54 inches
w = 785.39 rad/min
v = r * w = 4.375 inches * 785.39 rad/min = 3440.21 inches/min
3440.21 inches/min = 286.68 ft/min
So, linear speed of one of the cutting teeth in feet per minute is 286.68 ft/min.
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cody's car acce;rates from 0m/s to 45 m/s northward in 15 seconds. what is the acceleration of the car
The acceleration of the Cody's car when moving from 0 m/s to 45 m/s is calculated to be 3 m/s².
The acceleration is described to be the rate of change of velocity. It is a vector quantity. Its S I unit is m/s². If an object in a straight line accelerates up or slows down, it is said to be accelerated. Mathematically, a = Δv/t
where, a = acceleration
Δv = change of velocity = 45 - 0 = 45 m/s
t = time = 15 sec
After plugging the values into the equation above, we have,
a = Δv/t = 45/15 = 3 m/s²
Thus, the required acceleration of the car is calculated to be 3 m/s².
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A hollow, conducting sphere with an outer radius of 0. 240 m and an inner radius of 0. 200 m has a uniform surface charge density of +6. 47 × 10−6 C/m2. A charge of -0. 900 μC is now introduced into the cavity inside the sphere.
a. What is the new charge density σ on the outside of the sphere? Express your answer with the appropriate units.
b. Calculate the strength of the electric field E just outside the sphere. Express your answer with the appropriate units.
c. What is the electric flux through a spherical surface just inside the inner surface of the sphere? Express your answer with the appropriate units
Consider the following explanation for defining surface charge density:
= Q/surface sphere's area.Answers: (a) 5.73 * 10-6 C/m2, (b) 648 * 10 N/C, and (c) 56.5 * 10 3 Nm2/C.
What is the exterior of the sphere's new charge density?Consider the following explanation for defining surface charge density:
= Q/surface sphere's area
Q thus equals * * 4* * R2.
The new Q is as follows: Q initial -0.5 microC =5 micro C at the outer radius.
The new outside charge on the electric is equal to:
E=k*Q/R2outer=9 109*5 microC/(0.25m)2=648*103 N/C.
Last but not least, the flux equivalent to the internal charge is
The total flux is equal to net change divided by 0 according to Gauss's law.
0.5 microC/0 = 56.5 * 10 3 Nm2/C, therefore.
Answers: (a) 5.73 * 10-6 C/m2, (b) 648 * 10 N/C, and (c) 56.5 * 10 3 Nm2/C.
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How does genetics affect our chance of developing certain diseases and disorders?
Answer:
A genetic predisposition results from specific genetic variations that are often inherited from a parent. These genetic changes contribute to the development of a disease but do not directly cause it. Some people with a predisposing genetic variation will never get the disease while others will, even within the same family.
Explanation:
Hope this helps!
HELPP ME ASPP PLEASE
Point A represents the b.crest.
Transverse wavePoint A in a transverse wave represents the wavelength. The wavelength is the distance from a point on a wave to the next identical point on the wave. It is the measure of the length of the wave, and is measured in meters (m). A wave's wavelength is the distance between two identical points on the wave, such as peak-to-peak, or crest-to-crest. The crest of the wave is the highest point of the wave. It is the point where the wave reaches its maximum displacement from its undisturbed position. The amplitude of the wave is the maximum distance the wave deviates from its undisturbed position. It is the measure of the maximum displacement of the wave and is measured in meters (m). The trough of the wave is the lowest point of the wave. It is the point where the wave reaches its minimum displacement from its undisturbed position. The particles move at right angles to the direction of the wave. This means that they move up and down as the wave moves forward.To learn more about Transverse wave refer to:
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If you were to go into outer space, your ______ would stay the same, but your ______ would change question 7 options: a. age, spaceship b. weight, mass c. mass, weight d. inertia, magnitude
B is the answer ur welcome
Copper wire with a diameter of 0. 5 cm is covered with a 0. 65-cm layer of insulating material having a thermal conductivity of 0. 242W/m⋅K. The air adjacent to the insulation is at 290 K. If the wire carries a current of 400 A, determinea. The convective heat-transfer coefficient between the insulation surface and the surrounding airb. The temperatures at the insulation-copper interface and at the outside surface of the insulation
The convective heat-transfer coefficient between the insulation surface and the surrounding air is approximately 19.59 W/m²K; b.
The required details for insulation surface in given paragraph
The temperatures at the insulation-copper interface is approximately 324.96 K, and at the outside surface of the insulation is approximately 305.57 K.
What is the temperature of the air surrounding the insulated copper wire?
The temperature of the air surrounding the insulated copper wire is 290 K (or 17°C or 62.6°F). This is specified in the problem statement as being the temperature of the air adjacent to the insulation. This information is important for determining the heat transfer between the insulation and the surrounding air, which will affect the temperature at the insulation-copper interface and at the outside surface of the insulation.
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a metal conductor is connected to a battery. which statement accurately describes the current in the metal conductor?
The statement accurately describes the current in the metal conductor is it is a flow of electrons from negative to positive terminal. The correct option is A.
Conductors allow the current to pass through them. The given statements are about movement of subatomic particles.
When linked to a battery, the electric field (created by the battery) accelerates free electrons, and because they are negatively charged, they travel from the negative terminal to the positive terminal due to the electric field, gaining energy and speed. The ion is ultimately the winner (of energy) in this collision because the transit is not smooth and the electrons bump into the lattice ion.
The given question is incomplete. The complete question contains options. They are 'A.It is a flow of electrons from negative to positive terminal B.It is a flow of electrons from positive to negative terminal C.It is a flow of protons from negative to positive terminal D.It is a flow of protons from positive to negative terminal
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Need this ASAP please because it's for a homework!
Make a time line where you organize the events since the Big Bang in sequence.
After the universe's 13.7 billion year old Big Bang, it has gone through numerous phases or epochs.
What is Big bang theory?It may have experienced more activity and change in the first second than in all the billions of years since because of the harsh environment and violence of its very early beginnings.
We can construct a rough timeline as follows using our current understanding of how the Big Bang might have developed, taking into account notions about inflation, the Grand Unification, etc.
The Planck Epoch (or Planck Era), which spans from 0 to around 10-43 seconds (1 Planck Time), is the closest that modern physics can come to the beginning of time itself.
Therefore, After the universe's 13.7 billion year old Big Bang, it has gone through numerous phases or epochs.
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a point charge is located on the -axis at and a second point charge is on the -axis at what is the total electric flux due to these two point charges through a spherical surface centered at the origin and with radius (a) 0.500 m, (b) 1.50 m, (c) 2.50 m?
The total electric flux due to these two point charges through a spherical surface centred at the origin and with radius 0.500 m, 1.50 m and 2.50 m are respectively Ф₁ = 0, Ф₂ = -519.8 N m²/ C, Ф₃ = -67.8 N m²/ C.
Charge q₁ = 4 nC = 4 × 10⁻⁹ C
Charge q₂ = -4.6 nC = -4.6 × 10⁻⁹ C
The radius of various spheres is R₁ = 0.5 m, R₂ = 1.5 m, R₃ = 2.5 m
∈₀ is the electric permittivity of free space
It is observed that the sphere of radius R₁ does not contain any charges. Therefore, the flux through that sphere is zero.
Ф₁ = 0
The sphere of radius R₂ contains only charge q₂. So, the flux through that sphere is given as, Ф₂ = q₂/∈₀ = (-4.6 × 10⁻⁹)/(8.85 × 10⁻¹²) = -519.8 N m²/ C
The sphere of radius R₃ contains both charges q₁, q₂. So, the flux through that sphere is
Ф₃ = (q₁ + q₂)/∈₀ = (-4.6 × 10⁻⁹ + 4 × 10⁻⁹)/(8.85 × 10⁻¹²) = (-0.6 × 10⁻⁹)/(8.85 × 10⁻¹²) = -67.8 N m²/ C
The given question is incomplete. The complete question is 'A point charge q₁ = 4 nC is located on the x-axis at x = 2 m and a second point charge q₂ = -4.6 nC is on the y-axis at y = 1. What is the total electric flux due to these two point charges through a spherical surface centred at the origin and with radius (a) 0.500 m, (b) 1.50 m, (c) 2.50 m?'
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if the contacts on a manual starter cannot be closed immediately after a motor overload has tripped them open, what is the probable reason?
If the contacts on a manual starter cannot be closed immediately after a motor overload has tripped them open, the probable reason is that the thermal overload device (a type of overload relay) has not had enough time to cool down.
The thermal overload device is designed to protect the motor from damage by sensing the temperature of the motor windings and tripping the starter if the temperature becomes too high.
The contacts of the starter will remain open until the thermal overload device has cooled down enough to reset, which typically takes several minutes.
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A car drives 40 miles at an angle of 35 degrees north of east then drives for 50 miles due north and finally 10 miles at an angle of 20 degrees north of west what is the cars resultant magnitude and direction
Answer:
Below
Explanation:
FIND TOTAL NORTH DISPLACEMENT
40 sin 35 + 50 + 10 sin 20 = 76.363 mi
FIND TOTAL E-W DISPLACEMENT
40 cos 35 - 10 cos 20 = 23.369 mi
Total resultant displacement ( using pythag theorem)
sqrt ( 76.363^2 + 23.369^2 ) = 79.86 mi
angle = arctan ( 79.23 / 23.37) = 73 ° North of East
a hockey puck slides off the edge of a table with an initial velocity of 28.0 m/s. and experiences no air resistance. the height of the tabletop above the ground is 2.00 m. what is the angle below the horizontal of the velocity of the puck just before it hits the ground?
The angle which is calculated below the horizontal of the velocity of the puck just before it hits the ground is 12.60°.
Given the following data:
Initial velocity = 28.0 m/s
Acceleration due to gravity = 9.81
Displacement (height) = 2.00 meters.
Let us calculate and determine the angle below the horizontal of the velocity of the puck just before it hits the ground:
First of all, we need to determine the horizontal and vertical components of the hockey puck.
For horizontal component of this:
[tex]V^{2}= U^{2}+ 2 as\\V^{2}= 0^{2}+ 2 (9.8)(2)\\V^{2}=39.24\\ V=6.26 m/s[/tex]
For vertical component:
[tex]V_{x}=U_{x} \\V_{x}= 28.0 m/s[/tex]
Now, we can find the angle by using the following formula:
θ= [tex]tan^{-1} =\frac{Vx}{Vy}\\[/tex]
Substituting the values, we get.
Angle = 12.60 degrees.
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A billiard ball traveling at 4 m/s has an elastic head-on collision with a billiard ball of equal mass that is initially at rest. The first ball is at rest after the collision. What is the speed of the second ball after the collision?
__m/s
Following the collision, the initial ball is now at rest. 4 m/s is the speed of the second ball after the collision.
This is an elastic collision, which means that both kinetic energy and momentum are conserved.
We can use the conservation of momentum to find the velocity of the second ball after the collision:
momentum initial = momentum final
m1v1 + m2v2 = m10 + m2v'2
where m1 and m2 are the masses of the balls, v1 is the initial velocity of the first ball, and v'2 is the final velocity of the second ball.
Since the mass of the two balls are equal and the first ball is initially moving, and the second ball is initially at rest, we can simplify the equation as:
m1v1 = m2v'2
v'2 = v1*(m1/m2)
Therefore, the final velocity of the second ball is 4 m/s * (m1/m2) = 4 m/s
We can conclude that the final velocity of the second ball is 4 m/s , and the speed of the first ball is 0 m/s after the collision, which confirms that it is an elastic collision and the total kinetic energy is conserved.
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Two children pull a 320 kg sled on an ice rink - Child A pulls with a force of 450 N [S 15° E]
while child B pulls with 380 N [W].
If the sled starts from rest, how far will it go in 2.0s, and in what direction?
Please add a diagram!
Acceleration (a) is the change in velocity (Δv) over the change in time (Δt), represented by the equation a = Δv/Δt. This allows you to measure how fast velocity changes in meters per second squared (m/s^2).
How do you calculate frictional force pulling?The resistive force of friction (Fr) divided by the normal or perpendicular force (N) pushing the objects together yields the coefficient of friction (fr), which is a numerical value. The formula fr = Fr/N serves as a representation of it.
The force of gravity acting on an object is known as its weight, and it can be determined using the formula w = mg, which equals the mass times the acceleration of gravity. The newton is the SI unit for weight since it is a force.
The formula F = m a, where F the is the force acting on the object, m is the object's mass, and an is the acceleration, can be used to express Newton's second equation of motion. This formula's subject, m, must be changed in order to get the object's mass.
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part 1 of 2
Objects with masses of 111 kg and 290 kg are separated by 0.396 m. A 23.9 kg mass is placed midway between them.
Find the magnitude of the net gravitational force exerted by the two larger masses on the 23.9 kg mass. The value of the universal gravitational constant is 6.672 × 10^-11 N • m^2 /kg^2.
Answer in units of N.
part 2 of 2
Leaving the distance between the 111 kg and the 290 kg masses fixed, at what distance from the 290 kg mass (other than infinitely remote ones) does the 23.9 kg mass experience et force of zero?
Answer in units of m.
(a) The gravitational force between the two large masses is 1.37 x 10⁻⁵ N.
(b) The distance from 290 kg mass in which 23.9 kg mass experiences a zero net force is 0.24 m.
What is the gravitational force between the two masses?The gravitational force between the two large masses is calculated as follows;
F = (Gm₁m₂ ) / R²
where;
G is universal gravitation constantm is the mass of the objectR is the distance between the massesF = ( 6.672 x 10⁻¹¹ x 111 x 290 ) / ( 0.396² )
F = 1.37 x 10⁻⁵ N
The distance from 290 kg mass in which 23.9 kg mass experiences a zero net force is calculated as;
Let the distance between 290 kg and 23.9 kg = d
F1 = ( 6.672 x 10⁻¹¹ x 23.9 x 290 ) / ( d² )
F1 = ( 4.62 x 10⁻⁷ ) / d²
F2 = ( 6.672 x 10⁻¹¹ x 23.9 x 111 ) / ( 0.396 - d )²
F2 = ( 1.77 x 10⁻⁷ ) / ( 0.396 - d )²
( 4.62 x 10⁻⁷ ) / d² = ( 1.77 x 10⁻⁷ ) / ( 0.396 - d )²
( 4.62 x 10⁻⁷ )(0.396 - d )² = ( 1.77 x 10⁻⁷ )d²
2.6(0.396 - d )² = d²
(0.396 - d )² = d²/2.6
0.396 - d = √ (d²/2.6)
0.396 - d = 0.62d
0.396 = 1.62d
d = 0.396 / 1.62
d = 0.24 m
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a ping-pong ball covered with a conducting graphite coating has a mass of 5.0 x 10-3 kg and a charge of 4 uc. what electric field will balance exactly the weight of the ball? give magnitude and direction of the e.
The ball's weight will be precisely balanced by the electric field is E = 12.25 x 10³ N/C = 12.25 KN/C
The electrostatic force caused by the electric field must be equal to the weight of the body or charge in order to balance the object's weight. Therefore,
Weight = Electrostatic Force
E q = mg
where,
Electric field = (E) =?
5 x 103 kg is the mass of the charge, or m.
g = 9.8 m/s2 is the acceleration caused by gravity.
Q equals the charge's magnitude, which is 4 C (4 x 106 C).
Therefore,
E(4 x 10⁻⁶ C) = (5 x 10⁻³ kg)(9.8 m/s²)
E = 0.049 N/4 x 10⁻⁶ C
E = 12.25 x 10³ N/C = 12.25 KN/C
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A car that weighs 10,500N drives at a constant speed of v = 49.0m/s for a distance of d =
5,110m. What is the net work done on the car during this drive?
a. Wnet = 0
b. Wnet = 1.28 · 106J
c. Wnet = 2.05J
d. Wnet = 5.37 · 107J
The work that has been done in driving the car is 5.37 * 107J.
What is the net work done?We know that the force that moves the car in the forward direction is the force that is provided by the engine of the car and as such we have to look at the work that is done as the product of the force and the distance that is covered here.
We know that;
Work done = force * distance
Work done = 10,500N * 5,110m
Work done = 5.37 * 107J
A total work that has a magnitude of 5.37 * 107J has been done by the car.
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