A point charge of -3. 00 μC is located in the center of a spherical cavity of radius 6. 80 cm inside an insulating spherical charged solid. The charge density in the solid is 7. 35 x 10^-4 C/m^3. The magnitude of the electric field inside the solid at a distance of 9.20 cm from the center of the cavity is 394.6857x10⁻⁶ N/C.
Charge inside the cavity = -3μC = -3x10⁻⁶ C
Charge density of the sphere, ρ = 7.35x 10⁻⁴ C/m³
Radius of he cavity = 6.90 cm = 0.0690 m
We have to find the electric field at a radius of 9.20 cm, therefore, the effective radius,
Effective radius = 0.092-0.069 = 0.023 m
Effective Volume = [tex]\frac{4}{3}[/tex] π[tex]r^{3}[/tex]= 0.5093 x10⁻⁴ m³
The charge, Q = volume x charge density = 374.33 x 10⁻⁶ C
The net charge enclosed,
[tex]Q_{in} =(374.33-3)x10^{-6} =371.33 x10^{-6}[/tex]C
We know according to the Gauss law,
⇒E=[tex]\frac{Q_{in} }{e_{o} }[/tex]
E=394.6857 X[tex]10^{-6}[/tex]N/C
Hence, the magnitude of the electric field inside the solid at a distance of 9.20 cm from the center of the cavity is 394.6857x10⁻⁶ N/C.
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How many valence electrons does the ground-state electron configuration of nitrogen have?.
Answer:5 valence electrons
Explanation:
Which substance is insoluble and Which substance is insoluble and floats in water?
A wood chips
B an ice cube
C a sugar cube
D a handful of salt
The substante that is insoluble and floats is the wood chip and the one that is insoluble is a sugar cube.
What is solubility?Solubility is a property that a substance has to dissolve in another that is called a solvent, this can also depend on the conditions of the temperature to which the solution can be exposed and also the pressure. The solubility will depend on the properties of each of the components, it is not always the same in all types of solutions.
When no more of the solute can dissolve in a solution, the solution is said to be saturated.
Therefore, we can confirm that the substante that is insoluble and floats is the wood chip and the one that is insoluble is a sugar cube.
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Determine the thermal voltage for a diode at a temperature of 20 degrees celsius
Based off the image below, identify where the kinetic or potential energy is the highest?
That is the answer
So in the position of 1 and 5 height is maximum from the ground and here velocity is also minimum so potential energy is maximum and in the position of no 3 velocity maximum so kinetic energy will be maximum.
Briefing:We know total energy that is kinetic energy + potential energy is always constant so where kinetic energy is maximum there potential energy will be minimum and where potential energy is maximum there kinetic energy will be minimum.
we know potential energy=m g h
where h=height
So in the position of 1and 5 height is maximum from the ground and here velocity is also minimum so potential energy is maximum and in the position of no 3 velocity maximum so kinetic energy will be maximum.
What are some examples of kinetic energy?Kinetic energy, which may be seen in the motion of an object, particle, or group of particles, is the electricity of motion. Any moving item uses kinetic energy, such as a person who walks, a baseball being thrown, a piece of food falling from a table, or a charged particles in an electric field.
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A star is moving directly towards earth at a large fraction of the speed of light. How does the light we observe from this star compare to the light received from the same star if it were at rest relative to earth?.
A star is moving directly toward earth at a large fraction of the speed of light. The light we observe from this star compared to the light received from the same star if it were at rest relative to earth will be blueshifted.
The amount of light emitted by a star coming towards earth would have a shorter wavelength as compared to other sources of light. This refers to the doppler shift principle. According to the doppler shift principle, if an observer is moving in relation to a source, then the change in frequency in the light waves would be relative to that observer.
In this case, we will observe that the star approaching earth will shift its spectral lines toward the blue end of the light series.
Since blue light has a shorter wavelength, therefore, the large fraction of the speed of light coming out from a star would correspond to the blue end of the light spectrum hence called blueshift.
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A_____is a short-circuited winding, which consists of a single turn of copper wire and acts on only a portion of the stator windings.
A shaded pole is a short-circuited winding, which consists of a single turn of copper wire and acts on only a portion of the stator windings.
Most single-phase motors just require the two start leads to be switched in order to turn the rotor in the opposite direction.
Is a copper wire winding that has been short-circuited and has only one turn acceptable?An AC motor that starts using a shaded stator pole. A partial short-circuited winding that affects only some of the stator windings, made up of a single turn of copper wire.
What does the stator do?The motor is momentarily changed into a two-phase motor during the starting time to get around this problem and make it self-start. A single-phase motor’s stator is equipped with an additional winding, referred to as the starting winding, for this function (or auxiliary).
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newton’s law of cooling says that the temperature of a body changes at a rate proportional to the difference between its temperature and that of the surrounding medium (the ambient temperature) where t
The temperature from t=0 to 10min with a step size of 2 min is as follows:
t (in min) 0 2 4 6 8 10
T ( in °C ) 70 68.1 66.27 64.514 62.82 61.2
We know that,
The solution of first order differential equation is
dT / dt = - k ( T - [tex]T_{a}[/tex] )
where,
T = Initial temperature
[tex]T_{a}[/tex] = Ambient temperature
Given that,
k = 0.019 / min
[tex]T_{a}[/tex] = 20 °C
Using Euler method
[tex]T_{i} ^{'}[/tex] = - 0.019 ( [tex]T_{i}[/tex] - 20 )
[tex]T_{i+1}[/tex] = [tex]T_{i}[/tex] + ( [tex]T_{i+1}[/tex] * h )
With t = 0 to 10, h = 2, the iterative solution is as follows
At i=1, [tex]t_{i}[/tex] = 0, [tex]T_{i}[/tex] = 70
[tex]T_{i} ^{'}[/tex] = - 0.019 ( 70 - 20 ) = - 0.95
[tex]T_{i+1}[/tex] = 70 - ( 0.95 * 2 ) = 68.1
At i=2, [tex]t_{i}[/tex] = 2, [tex]T_{i}[/tex] = 68.1
[tex]T_{i} ^{'}[/tex] = - 0.019 ( 68.1 - 20 ) = - 0.914
[tex]T_{i+1}[/tex] = 68.1 - ( 0.914 * 2 ) = 66.27
At i=3, [tex]t_{i}[/tex] = 4, [tex]T_{i}[/tex] = 66.27
[tex]T_{i} ^{'}[/tex] = - 0.019 ( 66.27 - 20 ) = - 0.88
[tex]T_{i+1}[/tex] = 66.27 - ( 0.88 * 2 ) = 64.514
At i=4, [tex]t_{i}[/tex] = 6, [tex]T_{i}[/tex] = 64.514
[tex]T_{i} ^{'}[/tex] = - 0.019 ( 64.514 - 20 ) = - 0.846
[tex]T_{i+1}[/tex] = 64.514 - ( 0.846 * 2 ) = 62.82
At i=5, [tex]t_{i}[/tex] = 8, [tex]T_{i}[/tex] = 62.82
[tex]T_{i} ^{'}[/tex] = - 0.019 ( 62.82 - 20 ) = - 0.8136
[tex]T_{i+1}[/tex] = 62.82 - ( 0.8136 * 2 ) = 61.2
At i=6, [tex]t_{i}[/tex] = 10, [tex]T_{i}[/tex] = 61.2
Euler's method is used for solving ordinary differential equation with a given initial value. It is a first order numerical procedure.
The given question is incomplete. The complete question is:
Newton's law of cooling says that the temperature of a body changes at a rate proportional to the difference between its temperature and that of the surrounding medium (the ambient temperature), dT/dt = -k(T-Ta) where T = the temperature of the body (°C), t = time (min), k = the proportionality constant (per minute), and Ta = the ambient temperature (°C). Suppose that a cup of coffee originally has a temperature of 70 °C. Use Euler's method to compute the temperature from t = 0 to 10 min using a step size of 2 min if Ta = 20 °C and k = 0.019/min.
Therefore, the temperature from t =0 to 10 min with a step size of 2 min is as follows:
t (in min) 0 2 4 6 8 10
T ( in °C ) 70 68.1 66.27 64.514 62.82 61.2
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two parallel conducting plates, each of cross sectional area 460 cm2, are 4.0 cm apart and uncharged. if 4.9 ✕ 1012 electrons are transferred from one plate to the other, what are the following? (a) the charge density on each plate (enter the magnitude in c/m2.) c/m2 (b) the electric field between the plates (enter the magnitude in n/c.) n/c
Surface Charge density on each plate sigma =+- Q/a = +4.9×10²⁴/4.6×10⁻²C/m² and Electric field is found to be 0.12×10³⁸N/C.
Given, here the cross sectional area of the conducting plate (A)
A = 460 cm² = 460 x 10-4 m² = 4.6 x10-²m² and distance between plate, d = 4.0 cm = 4x10² m²
Charge on the plate, Q = ne ,where n= 1×10¹²
Q = ± 10¹² x4.9x 10¹²c = ± 4.9x 10²⁴ c
Now, we have to find charge density on each plate sigma =+- Q/a = +4.9×10²⁴/4.6×10⁻² C/m²
σ= 1.09× 10²⁶ C/m²
Now we have to find electric field between the plates.
E= σ /E°= 1.09× 10²⁶/ 8.85×10⁻¹²
E= 0.12×1038N/C
Charge density is a measure of the amount of electric charge present in a space per unit volume in one, two, or three dimensions. The quantity of electric charge per surface area or volume, respectively, is known as the linear surface or volume charge density.
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A simple pendulum, consisting of a mass m and a string of length l, swings upward, making an angle θ with the vertical. The work done by the tension force is?.
The work done by tension force in pendulum is zero.
We need to know about pendulum experiment to solve this problem. The pendulum is a simple experiment to calculate gravitational acceleration.
The work done is proportional to force and displacement. It can be determined as
W = F . s
where W is work done, F is force and s is displacement.
From the question above, the given parameters are
mass is m
string of length is l
angle is θ
F = T
s = 0
In pendulum motion, there is no displacement so the work done by the tension force is zero.
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A soccer ball rolls to a stop from a speed of 25 m/s with an acceleration of * 5 points
-5 m/s/s. How far does it roll until it stops?
O 62.5 m
8.33 m
2.5 m
104.2 m
937.5 m
The distance the soccer ball cover, until it stops, is 62.5m
The distance the ball rolls up to before coming to rest may be calculated using Newton's third equation of motion:
v² = u² + 2as .......(i)
Where v ⇒ Final velocity
u ⇒ Initial velocity
a ⇒ Acceleration
s ⇒ Displacement
Now as per the question,
v = 0 m/s
u = 25 m/s
a = -5m/s²
Putting these values in equation (i)
0 = (25)² - 2× 5 × s
0 = 625 - 10s
10s = 625
s = 625/10
s = 62.5m
Therefore, the distance the ball cover, until it stops, is 62.5m.
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What makes a question able to be answered by science
Question 1 (4 points)
A ball was kicked 50 m/s and is rolling to a stop. It is slowing
down at a rate of -3 m/s². How long was the ball in motion?
(YOU MUST FILL IN ALL THE BLANKS)
Givens:
Initial velocity=
Final Velocity=
Acceleration=
m/s
m
m/s²
Answer: (Use one decimal place) Time=
S
The ball was in motion for 16.7 seconds before coming to rest.
Let us assume that the ball was in motion for a time equal to 't' seconds before coming to rest.
Given in the question
Initial velocity = u = 50 m/s
Final Velocity = v = 0 m/s ( As the ball stops moving after some time, its final velocity hence becomes zero. )
Acceleration = a = -3 m/s²
To find
Time = t seconds
Using the first equation of motion,
v = u + at
Put in the value in the equation, we get
0 = 50 + (-3)(t)
3t = 50
t = 50/3
t = 16.66 seconds
t = 16.7 seconds ( rounding off to one decimal place )
Therefore, the ball was in motion for 16.7 seconds before coming to rest.
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HEY GUYS!! I NEED YOUR HELP URGENTLY ON THIS PHYSICS QUESTION. I WILL GIVE BRAINLIEST.
Answer:
Q=Mcø
528=m×4200×50
m=528÷(4200×50)
m=0.002514kg
2.514g
gasoline vapor is injected into the cylinder of an automobile engine when the piston is in its expanded position. the temperature, pressure, and volume of the gas-air mixture are t1
Answer:
bad for the environment
Explanation:
what is causing air pollution and killing animals and killing trees which is bad for the environment
Calculate the weight force of the Mars vehicle when it is on Earth and when it is on Mars.
The mass of the Mars vehicle is 2500 kg.
Weight of vehicle on Earth, where g = 10 N kg-¹.
Weight of the vehicle on Earth is 6736.72 kg.
What is the weight of an object?
The weight of an object is the force exerted on it owing to gravity in science and engineering. Weight is defined as a vector quantity in certain standard textbooks, as the gravitational force exerted on the item. Others describe weight as a scalar quantity, the magnitude of gravity.
Weight on Mars = (Weight on earth/10) x 3.711
Given, weight on mars = 2500 kg
Therefore, weight on Earth = 6736.72 kg.
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Compare the energy of a 1-megaton hydrogen bomb to the energy released by a major earthquake.
Compare the energy of a 1-megaton hydrogen bomb to the energy released by a major earthquake. Earthquake/Ebomb = 5.
How many megatons is a hydrogen bomb?Although hydrogen bombs with an explosive force of more than 50 megatons have exploded, the weapons mounted on strategic missiles typically have a range of 100 kilotons to 1.5 megatons.The "Tsar Bomba," commonly known, was around 3,300 times more powerful than the 15-kiloton nuclear weapon dropped on Hiroshima and produced an explosion of 50 megatons.The area that is destroyed varies as the square of the distance, but the volume that the weapon's energy spreads into varies as the cube of the distance. Therefore, 80 square miles would be destroyed by a single bomb having a yield of 1 megaton. In contrast, eight bombs with a combined yield of 125 kilotons would decimate 160 square miles.
Compare the energy of a 1-megaton hydrogen bomb to the energy released by a major earthquake. Earthquake/Ebomb = 5.
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Which object has the least thermal energy?
OA. 5 kg of oxygen gas at 30°C
OB. 5 kg of liquid oxygen at -225°C
OC. 2 kg of oxygen gas at 20°C
OD. 2 kg of liquid oxygen at -225°C
OD. 2kg of liquid oxygen at -225 *C has the least thermal energy.
The thermal energy for a phase of matter is given by the formula
Q = mCΔT where
m = mass
C = Specific heat constant
T = Temperature
For Oxygen, the Specific heat constant = 0.92 J/gK
Let's calculate the thermal energy for each case, assuming that the initial temperature is 0 Kelvin
OA.
m= 5000 g
ΔT = 303 K
So Q =5000×0.92×303 = 1393800 J = 1393.8 KJ
OB.
m= 5000 g
ΔT = 48 K
So Q =5000×0.92×48 = 220800 J = 220.8 KJ
OC.
m= 2000 g
ΔT = 293 K
So Q =2000×0.92×293 = 539120 J = 539.12 KJ
OD.
m= 2000 g
ΔT = 48 K
So Q =2000×0.92×48 = 88320 J = 88.32 KJ
Therefore, OD 2kg of liquid oxygen at -225 *C has the least thermal energy which is 88.32 KJ.
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Answer:
2 kg of liquid oxygen at -225°C
Explanation:
B has the same temperature, yet it's more mass. So, to use this as an example--B has the most thermal energy. If the mass decreased, then the thermal energy would decrease. More mass=more thermal energy. Hence, why the mass decreasing to 2 kg would equal the least amount of thermal energy.
Ammonia is a weak bas. Potassium hydroxide is a strong base. Which statement is correct?
A solution of potassium hydroxide will always have greater productivity than a solution of ammonia
A solution of potassium hydroxide will always have a lesser concentration of salute in a solution of ammonia
A solution of potassium hydroxide will always have a higher pH of a solution of ammonia
A solution of potassium hydroxide will always have more OH negative ions in a solution of ammonia
Ammonia is a weak base and Potassium hydroxide is a strong base, then then the solution of potassium hydroxide will always have a higher pH than a solution of ammonia, therefore the correct answer is option C.
What is a Chemical compound?A chemical compound is a combination of two or more either similar or dissimilar chemical elements.
As given in the problem, ammonia is a weak base. Potassium hydroxide is a strong base then we have to find out which statement is correct from the given list of statements,
Since ammonia is a weak base and potassium hydroxide is a strong base, their solutions will always have a higher pH than ammonia solutions. Option C is the right response.
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THE FIRST PERSON TO GIVE ANSWERS GETS BRAINLIEST
PLEASE HELP URGENT
Answer:
number 3
Explanation:
the effort is almost equal to the load
how do you prevent mother to child transmission
The prevention factors required to prevent mother-to-child transmission are:
Get antenatal administrations and HIV testing amid pregnancy. Have got to antiretroviral treatment (ART) Practice secure childbirth hones and fitting newborn child feeding. Make utilization of newborn child HIV testing and other post-natal healthcare administrations.Mother-to-child transmission also referred to as Perinatal is described as when HIV is transmitted from the mother to the child during pregnancy, childbirth, or breastfeeding. For ladies who are pregnant, HIV counseling and testing is the primary and most basic step in lessening mother-to-child transmission. Prenatal Treatment , when taken all through pregnancy, conveyance, and breastfeeding, can decrease the chance of HIV transmission to a newborn child to less than 5%. Ladies who test HIV-positive amid their pre-birth visits at EGPAF-supported locales are quickly connected to long-lasting Craftsmanship and counseled on the benefits of enrollment and maintenance on treatment. Safe childbirth is an imperative thought for all hopeful moms and especially for HIV-positive ladies. Post-natal care is eesential for both the mother and the baby especially since it includes child wellness services such as nutrition, immunization, and growth monitoring.
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a human expedition lands on an exoplanet. one of the explorers is able to jump a maximum distance of 18.5 m with an initial speed of 2.30 m/s. find the gravitational acceleration on the surface of the exoplanet. assume the planet has a negligible atmosphere. (enter the magnitude in m/s2.)
The exoplanet where the human expedition landed has a surface gravity acceleration of -0.20 meters per square second.
What is the equation of motion?The relationship between a moving body's distance, velocity, acceleration, and time is known as the equation of motion.
Here, is the beginning body, is the final body, is the acceleration of the body, and is the distance travelled. This is the third equation of motion for distance.
The greatest distance leapt by one of the explorers is 19.5 metres, as stated in the problem.
The starting velocity is 2.80 m/s.
It is necessary to determine the gravitational acceleration on the exoplanet's surface.
The explorers' velocity is currently zero. As a result, according to the third equation of motion, the explorers are moving downhill as shown by the negative sign.
Thus, the gravitational acceleration on the surface of the exoplanet is -0.20 meter per squared second.
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A block of mass 5.00 kg rests on a horizontal surface where the coefficient of kinetic friction between the two is 0.200. a string attached to the block is pulled in direction 25 degrees with horizontal, resulting in a 2.50-m/s2 acceleration by the block. find the tension in the string. (g = 9.80 m/s2)
The tension in the string is 19.8 N.
What is mass?Mass is a physical body's total amount of matter. It also serves as a gauge for the body's inertia or resistance to acceleration (change in velocity) in the presence of a net force. The strength of an object's gravitational pull to other bodies is also influenced by its mass.The kilogram is the SI unit of mass (kg). In science and technology, a body's weight in a given reference frame is the force that causes it to accelerate at a rate equal to the local acceleration of free fall in that frame.For instance, a kilogram mass weighs around 2.2 pounds at the surface of the planet. However, the same kilogram mass would weigh just about 0.8 pounds on Mars and about 5.5 pounds on Jupiter.An object's mass is a crucial indicator of how much stuff it contains. Weight is a measurement of an object's gravitational pull. It is influenced by the object's location in addition to its mass. As a result, weight is a measurement of force.Find the tension in the string:
The net force on the block will be:
T - Fr = Fnet = ma
Where T is the tension and Fr is the frictional force
T - ukR = ma
Where R is the upwards normal force equal to the block's weight. R = mg
therefore,
T - uk mg = ma
T - 0.2(5 x 9.8) = 5(2)
=> T = (0.2 x 5 x 9.8) + 5(2) = 19.8 N
This is the tension in the string.
For the free body diagram, draw a block and show a rope connected horizontally to it. The tension in the string will be along the rope outwards from the block. Now, the frictional force will be between the ground and the block and it will be in the opposite direction to the direction of tension of the rope. The force due to gravity will act vertically downwards while the normal force R will act vertically upwards.
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A typical laboratory centrifuge rotates at 4300 rpm. test tubes have to be placed into a centrifuge very carefully because of the very large accelerations. what is the acceleration at the end of a test tube that is 10 cm from the axis of rotation? for comparison, what is the magnitude of the acceleration a test tube would experience if dropped from a height of 1.0 m and stopped in a 1.0-ms-long encounter with a hard floor?
The first acceleration is 20276.6 m/s² and the second one is 4427.2 m/s².
First of all, you need to calculate the angular velocity of the centrifuge as it is needed in the formula for centripetal acceleration that is given by [tex]a = r\omega^2[/tex].The frequency is given in rotations per minute and the calculation would look like the following,[tex]\begin{aligned}\\\\\omega &= 2\pi.f\\\\&= 2\pi.\big(\frac{4300\,\text{rpm}}{60\,s})\\\\&= 450.29\, rad/s\end{aligned}[/tex]
Then simply the acceleration that the test tube experiences is calculated as follows,[tex]\begin{aligned}\\\\a &= r\omega^2 \\\\&= 0.1\, m\,\times\,(450.29\,rad/s)^2\\\\&= 20276.1\, m/s^2 \end{aligned}[/tex]
For the second part, you need to calculate the speed the test tube would acquire just before hitting the surface as a result of free fall under gravity. And that is simple as [tex]v = \sqrt{2gh}[/tex].Once it is calculated you need to divide the speed difference by the time it took until the tube came to a rest.[tex]\begin{aligned}\\\\v &= \sqrt{2.\times 9.8\,m/s^2\,\times 1.0\,m }\\\\&= 4.427\,m/s\\\\a &=\frac{\Delta v}{\Delta t}\\\\&= \frac{0-4.427\,m/s}{0.001\,s}\\\\&= -4427\, m/s^2\cdots(\text{- indicates a deceleration})\end{aligned}[/tex]
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he offense can gain the 10 yards by running, or passing the ball, or by a ________________ assessed against the opposition
Answer:
penalty
Explanation:
Penalties can gain your team yards if assessed against the opposition...their are multiple penalties which are assessed 10 yards
you are driving a car at 30\text{ m/s}30 m/s when a bad driver traveling at a constant speed of 20\text{ m/s}20 m/s swerves into your lane 2020 meters ahead. as soon as you see the car in front of you, you begin to brake. what is the smallest deceleration needed, in \text{m/s}^2,m/s 2 , to avoid rear-ending the car? you may assume your acceleration is constant
the smallest deceleration needed, in m/s^2, to avoid rear-ending the car
will we of a = -12.5m/s^2.
What is deceleration?A reduction in velocity, or momentum, is referred to as a deceleration. The slowing of a rise is referred to as a deceleration. For instance, the wind has a certain amount of momentum. We may say that the wind is decelerating when its momentum slows down, i.e., when it blows less forcefully.
To avoid rear-ending the car
Reduce final velocity of the car to less or equals to the speed of bad driver's car before car travels 20 m
Because if final velocity of car is less than or equal to the speed of bad driver's car then car can't able to cross bad driver's car.
Final velocity of the car (Vf) = 20 m/s
Using equation of motion
[tex]v^{2} _{f} = v^{2} _{i} + 2as\\ a = v^{2} _{f} - v^{2} _{i} / 2s\\= 20 ^{2} - 30 ^{2}/ 2 *20\\= -12.5m/s^{2}[/tex]
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Two vectors have length v1 = 3. 5 km and v2 = 4. 0 km. What are the maximum and minimum magnitudes of their vector sum? discuss how you decided on those magnitudes
Answer:
77ruririfusjd u 6r7r u y9r8
the van of hans and frans is stuck on slippery ice, so they must get out and move it by hand. hans pushes rightwards with force magnitude f hf h f, start subscript, h, end subscript and frans pulls rightwards. the van of mass mmm accelerates with magnitude aaa rightwards.
The expression for horizontal magnitude acceleration of van is a = ( [tex]F_{H}[/tex] + [tex]F_{F}[/tex] ) / m
Given that,
The magnitude of the force applied by Hans on van = [tex]F_{H}[/tex]
The magnitude of the force applied by Frans on van = [tex]F_{F}[/tex]
Let us consider the horizontal magnitude of acceleration as a and the mass of van as m.
The horizontal forces acting on the van are [tex]F_{H}[/tex] and [tex]F_{F}[/tex].
∑[tex]F_{Hor}[/tex] = [tex]F_{H}[/tex] + [tex]F_{F}[/tex]
m a = [tex]F_{H}[/tex] + [tex]F_{F}[/tex]
a = ( [tex]F_{H}[/tex] + [tex]F_{F}[/tex] ) / m
The vertical forces here are the force due to weight and the normal force due to its weight on ground.
Therefore, the expression for horizontal magnitude acceleration of van is a = ( [tex]F_{H}[/tex] + [tex]F_{F}[/tex] ) / m
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A 500 kg pumpkin is dropped 15 m on
top of a school bus.
How much gravitational potential
energy was gained when it was
winched up 15 m?
Assuming all of this GPE is turned
into kinetic energy as the pumpkin
drops, work out its speed as it hits
the school bus.
gravitational potential energy was gained when it was winched up 15 m is 73500 J.
GPE is turned
into kinetic energy as the pumpkin drops
Then,
pumpkin speed is 17.14 m/s.
What is gravitational potential energy?The potential energy that a huge item has in relation to another massive object due to gravity is known as gravitational energy or gravitational potential energy. When two objects descend toward one another, the potential energy associated with the gravitational field is released (transformed into kinetic energy). Bringing two things farther apart increases the gravitational potential energy.m, g, and h stand for mass, the gravitational constant (with a value of around 9.8 m/s), and height, respectively.To learn more about : gravitational potential energy
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a jet fighter pilot wishes to accelerate from rest at 5.00 g to reach mach-3 (three times the speed of sound) as quickly as possible. experimental tests reveal that he will black out if this acceleration lasts for more than 5.00 s . use 331 m/s for the speed of sound.
With an acceleration 5 g pilot can reach the greatest speed of 240 m/s before blacking out.
The acceleration due to gravity is g.
g = 9.8 m/s²
The jet's acceleration is,
[tex]Acceleration = Mass × Gravity[/tex]
[tex]a = m \times g[/tex]
[tex]a = 5 \times 9.8[/tex]
[tex]a = 49 \: m/s ^{2}[/tex]
The jet's acceleration is 49 m/s².
The speed of sound be v.
v = 331 m/s
Let the initial velocity of the jet be
[tex]v _i[/tex]
[tex]v _i = 0[/tex]
The time at which pilot wil black out = 4.9 seconds
Let the time taken by the pilot to reach Mach 3 be t.
The final velocity will be,
[tex]v _f = 3 \times v[/tex]
[tex]v _f = 3 \times 331[/tex]
[tex]v _f = v _i + at[/tex]
[tex]993 = 0 + (49) \times t[/tex]
[tex]t = 20.26 \: seconds[/tex]
So, he time taken to reach Mach 3 is 20.26 seconds. Thus, the pilot will black out.
With an acceleration 5 g pilot can reach the greatest speed of,
[tex]v _f = 0 +( 49 \times 4.9)[/tex]
[tex]v _f = 240.1 \: m/s[/tex]
Therefore, the with an acceleration 5 g pilot can reach the greatest speed of 240 m/s before blacking out.
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The complete question is-
a jet fighter pilot wishes to accelerate from rest at 5.00 g to reach mach-3 (three times the speed of sound) as quickly as possible. experimental tests reveal that he will black out if this acceleration lasts for more than 5.00 s . use 331 m/s for the speed of sound.
A swimmer swims at 5 m/s. How long would it take to swim 5 laps of a 50m pool?