The approximate de Broglie wavelength of a tennis ball is 9.4×10^(-34) m.
What is the de Broglie wavelength:
It is the wavelength that is associated with an object in relation to its momentum and mass is known as de Broglie wavelength.
A particle's de Broglie wavelength is usually inversely proportional to its force.
The formula of de Broglie wavelength:
λ = h / ( m*v )here mass of a tennis ball is given
mass, m=70 g = 0.07 kg
ball is moving with velocity
v = 10 m/s
h is Plank constant,
h=6.63×10^(-34) Js
substituting the values in formula,
λ = 6.63×10^(-34) / ( 0.070*10)
λ = 9.4 ×10^(-34) m
Hence
The approximate de Broglie wavelength of a tennis ball is 9.4×10^(-34) m
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Why should a scientist not be concerned when a fellow scientist responds to their conclusions with skepticism?
Throw two balls from the same height at the same time at an initial speed of 20 m/s. One is thrown vertically down, while the other is thrown vertically up. We want to find the time difference between their landing.
The time difference between their landing is 2.04 seconds.
Time of difference of the two balls
The ball thrown vertical upwards will take double of the time taken by the ball thrown vertically downwards.
Time difference, = 2t - t = t
t = √(2h/g)
where;
h is the height of fallg is acceleration due to gravityApply the principle of conservation of energy;
¹/₂mv² = mgh
h = v²/2g
where;
v is speed of the ballh = (20²)/(2 x 9.8)
h = 20.41 m
Time of motiont = √(2 x 20.41 / 9.8)
t = 2.04 s
Thus, the time difference between their landing is 2.04 seconds.
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The eyes of a farsighted person are Blank______ than normal; therefore light rays tend to focus Blank______ the retina.
The eyes of a farsighted person are shorter than normal; therefore light rays tend to focus behind the retina.
What is Farsightedness:
It is also known as hyperopia, is a common vision condition in which you can see distant objects clearly, but objects nearby may be blurry. The degree of your farsightedness influences your focusing ability.Farsightedness is caused by a cornea that isn't curved enough or It causes by an eyeball that's too short.
These two problems prevent light from focusing directly on the retina.
so, the light rays focus behind the retina.
Hence,
The eyes of a farsighted person are shorter than normal; therefore light rays tend to focus behind the retina.
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4. An applied force of 6.2 N acts on a 2.1-kg object, pushing it horizontally across a surface
where the coefficient of kinetic friction is 0.15.
a. Draw a free body diagram for the body
b. Determine the frictional force acting.
c. Determine the net force acting on the body
d. Determine the object's acceleration.
Step by step, thank you
[2C]
[2A]
[2A]
[2A]
The frictional force on the body is 0.25 N the net force on the body is 5.95 N while the acceleration produced is 2.83 m/s^2
What is the net force?The net force is used to describe the resultant force that acts on the object.
The frictional force that acts on the object s obtained from;
Ff = 0.15 * 2.1-kg * 9.8 m/s^2
Ff = 0.25 N
The net force that acts on the body = F - Ff where F is the applied the force
= 6.2 N - 0.25 N
Net force = 5.95 N
Now;
Net force = ma
a= Net force /m
a = 5.95 N/2.1-kg
a = 2.83 m/s^2
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If f is the magnitude of the force on the test charge due to only one of the other charges, what is the magnitude of the net force acting on the test charge due to both of these charges?.
The net force acting on test charge is resultant of two force i.e., √2 F
The magnitude of force on test charge due to one charge Q is:
F = k qQ / r²
where k = 9 × 10^9
Similarly the force due to second charge will be:
F = k qQ / r²
Now both the charges of magnitude Q is present and it is pointed at right angle to each other then the force will be resultant i.e.,
F = √ F² + F² = F√2
Hence, net force acting on test charge is √2 F.
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At which points on the roller coaster is the car not moving?
A
A and E
B
B and C
C
C and D
D
D and E
The maximum potential energy of the car or the when car is not moving at points A and E; option A.
What is potential energy?Potential energy is the energy of a body due to its state or position.
In the rollercoaster motion shown, the maximum potential energy occurs when the car is no longer moving.
At different points other than at maximum potential energy, the energy is a combination of potential and kinetic energy.
The maximum potential energy occurs at A and E, at which point the car is not moving.
In conclusion, at maximum potential energy, the car is not moving.
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A brightfield microscope equipped with a ____________ ___________ stop is able to block light from the illuminator creating bright objects on a dark background.
Answer: A Brightfield microscope equipped with an opaque light stop is able to block light from the illuminator and creating bright objects on a dark background.
Explanation: To find the correct statement about Brightfield microscope, we need to know more about the Brightfield microscope.
what is Brightfield microscope?The Brightfield microscopic technique is generally used with the compound microscopes and, these are one, among the widely used optical illumination techniques. The Brightfield microscope or Compound light microscope is an optical instrument, which uses light rays and produces dark images with a bright background.How Bright field microscope works?A brightfield microscope equipped with a opaque light stop, will produce bright objects on a dark background by blocking the light illuminated from the light source. This opaque light stop, blocks the light rays that arising from the illuminator to the objective lenses and, only allows the reflected or transmitted light of the specimen. As a result, we get bright objects with a dark background.Thus, we can conclude that, A Brightfield microscope equipped with an opaque light stop is able to block light from the illuminator and creating bright objects on a dark background.
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A vehicle starts from rest and accelerates uniformly for 12 seconds to a
velocity of 10ms1. It then runs at a constant velocity and is finally came to rest
in 66m with constant . The total distance covered by the vehicle is
580m. Find the value of acceleration and the time taken.
The value of the acceleration is 0.76 m/s² and the total time taken by the vehicle is 39 seconds.
Acceleration of the vehicleThe acceleration of the vehicle before coming to rest is calculated as follows;
v² = u² + 2as
where;
v is the final velocityu is the initial velocitya is the accelerations is the distance traveled before stoppingthe car came to rest with constant velocity attained after 12 seconds.
the initial velocity of the car before 12 seconds is zero.
v² = 0 + 2as
a = v²/2s
a = (10²)/(2 x 66)
a = 0.76 m/s²
Time of motion of the vehicled = ut + ¹/₂at²
where;
d is the total distance traveledt is the time of motiona is accelerationu is initial velocity of the vehicle580 = 0 + ¹/₂(0.76)t²
580 = 0.38t²
t² = 580/0.38
t² = 1,526.3
t = √1,526.3
t = 39 seconds
Thus, the value of the acceleration is 0.76 m/s² and the total time taken by the vehicle is 39 seconds.
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1. An object moving with velocity 15 ms accelerates at 2 m/s for 5 seconds. Calculate the final velocity.
Answer:
final velocity=u+at 15+2×5=25
You hear three beats per second when two sounds tones are generated. The frequency of one tone is known to be 610 Hz. The frequency of the other is:
The frequency of the other wave is 613 Hz or 607 Hz.
The difference between the frequencies of two waves is called the beat frequency.
Here, one wave has a frequency 610 Hz and the beat frequency is 3 beats per second.
Which has a higher frequency is not mentioned. Therefore, there are two possibilities.
Δf = | 610 - 613 | = 3
or
Δf = | 610 - 607 | = 3
Therefore, the frequency of the other wave is 613 Hz or 607 Hz.
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The table shows the percentage of carbon dioxide in the Earth’s atmosphere in the years 1800 and 2013. Calculate the difference in the mass of carbon dioxide in 500 kg of air in 2013 compared to 1800. Select the correct answer.
The difference in the mass of carbon dioxide in 500 kg of air in 2013 compared to 1800 is 0.06 Kg
Data obtained from the questionYear 1800 percent = 0.028%Year 2013 percent = 0.040%Mass of air = 500 KgDifference =?How to determine the mass of CO₂ in 500 Kg in year 1800Year 1800 percent = 0.028% Mass of air = 500 KgMass of CO₂ =?Mass = percent × mass of air
Mass of CO₂ = 0.028% × 500
Mass of CO₂ = 0.14 Kg
How to determine the mass of CO₂ in 500 Kg in year 2013Year 1800 percent = 0.040% Mass of air = 500 KgMass of CO₂ =?Mass = percent × mass of air
Mass of CO₂ = 0.040% × 500
Mass of CO₂ = 0.2 Kg
How to determine the differenceMass of CO₂ in year 1800 = 0.14 KgMass of CO₂ in year 2013 = 0.2 KgDifference =?Difference = mass in 2013 - mass in 1800
Difference = 0.2 - 0.14
Difference = 0.06 Kg
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Calculate the frequency of the 3rd harmonic of a 0.5 meter string with a linear density of 0.03 kg/m at a tension of 50 newtons.
Frequency of third harmonic is 122.48 HZ
Theory:
Frequency of third harmonic with length L , linear density μ , Tension
T is given as
fₙ= n v/2L
fₙ = n/2L × √T/√μ
plugging in the values here-
fₙ = 3/ 2×0.5 × √50 / √0.03
fₙ = 122.48 Hz
Thus, the frequency of third harmonic is 122.48 HZ
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Which statement is true of a piece of ice at 0°C that is put into a freezer at
-18°C?
The statement which is true about a piece of ice at 0°C which is put into a freezer at -18°C is it having the temperature of the freezer.
What is Temperature?This is referred to the degree of hotness or coldness of a body and the unit is Celsius or Kelvin.
The ice at 0°C will experience a change in temperature of the freezer when put in it in this scenario.
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Mass of empty density bottle - 43.2g Mass of density bottle full of water 66.4g Mass of density bottle with some sand - 67.5g Filled up with water 82.3g he above data to determine the: Mass of the water that completely filled the bott Volume of water that completely filled the bottl Volume of the density bottle: Mass of sand Mass of water that filled the space above the sa Volume of the sand: Mass of empty density bottle - 43.2g Mass of density bottle full of water 66.4g Mass of density bottle with some sand - 67.5g Filled up with water 82.3g he above data to determine the : Mass of the water that completely filled the bott Volume of water that completely filled the bottl Volume of the density bottle : Mass of sand Mass of water that filled the space above the sand
Volume of the sand :
The values of the density, volume and masses are:
mass of water that completely filled the bottle = 23.2 gvolume of water that completely filled the bottle = 23.2 cm³volume of the density bottle = 23.2 cm³mass of sand = 24.4 gmass of water that filled the space above the sand = 14.8 gvolume of sand = 8.4 cm³density of sand = 2.9 g/cm³What is density?Density is defined as the ratio of mass and volume of a substance.
Density = Mass/volumeThe given data is as follows:
Mass of empty density bottle = 43.2 g
Mass of density bottle full of water = 66.4 g
Mass of density bottle with some sand = 67.5 g
Mass of density bottle with the sand
filled up with water = 82.3 g:
(a) mass of water that completely filled the bottle = 66.4 - 43.2
mass of water that completely filled the bottle = 23.2 g
(b) volume of water that completely filled the bottle = mass/density
density of water = 1 g/cm³
volume of water that completely filled the bottle = 23.3 g/1 g/cm³
volume of water that completely filled the bottle = 23.2 cm³
(c) volume of the density bottle = 23.2 cm³
(d) mass of sand = 67.5 g - 43.2 g = 24.4 g
(e) mass of water that filled the space above the sand = 82.3 - 67.5 = 14.8 g
(f) volume of sand = volume of bottle - volume of water
volume of water = 14.8 cm³
volume of sand = 23.2 cm³ - 14.8 cm³
volume of sand = 8.4 cm³
(g) density of sand = 24.4 g/8.4 cm³ = 2.9 g/cm³
Therefore, the density of sand is obtained by taking the ratio of the mass of the sand and the volume if the sand.
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A ball is launched from the ground with a horizontal speed of 30 m/s and a vertical speed of 30 m/s. How long will it take to get to the top of its trajectory?
The ball will take 3 seconds to get to the top of its trajectory When it is launched from the ground with a horizontal speed of 30 m/s and a vertical speed of 30 m/s
What is Trajectory ?When any object is thrown from horizontal at an angle θ except 90°, then the path followed by it is called trajectory, the object is called projectile and its motion is called projectile motion.
Here, we have a type of motion called projectile motion and it is pretty similar to an upside down parabola. The top of the trajectory is the vertex of the parabola and is also when v = 0.
Given :
Horizontal speed= 30m/sVertical Speed= 30 m/sThis problem is now relatively easy because we only need to find the vertical distance so we can ignore horizontal speed and use
vy = vy0 + ayt
Plug in our givens
0 = 30 - 10t
solve for t
t = 3 seconds
Hence, The ball will take 3 seconds to get to the top of its trajectory When it is launched from the ground with a horizontal speed of 30 m/s and a vertical speed of 30 m/s
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Which equation can be used to solve for the magnitude of the velocity (v) at which Frances slides backwards after giving the curling stone a push in Test 3?
A
0 = –(45 kg • v) + (20 kg • 3 m/s)
B
0 = –[(45 kg + 20 kg) • v]
C
60 kg • m/s2 = (45 kg • v) + (20 kg • 3 m/s)
D
195 kg • m/s2 = (45 kg + 20 kg) • v
A
A 0 = –(45 kg • v) + (20 kg • 3 m/s)
Not moving, means momentum is zero. Thus, momentum after = momentum before (& that momentum is zero)
V = 4/3m/s
(-45 x 4/3) + (20x3) = 0
Solving to find v :
0. = –(45 kg • v) + (20 kg • 3 m/s)
= - (45 kg • v) + 60
-60
-60 = - 45 x v
÷ — 45
4/3 = v
(4/3 = 1.33333333333)
Hope this helps!
The velocity for the entire trip is 0.4 m/s as It takes her 500 seconds to make the round trip and 60 kg • m/s2 = (45 kg • v) + (20 kg • 3 m/s).
Path 1 = 400 m В B. A Path 2 = 200 m Path 3 = 300 m. Thus, option C is correct.
What is velocity?A particle's settling velocity known as the rate at which is travels through a still fluid. The specific gravity of the particles, their size, and their shape all have an impact on settling velocity.
A particle in still air will gravitationally settle and reach its terminal velocity fairly quickly. A particle's terminal velocity in a still fluid is referred to as the settling velocity (also known as the "sedimentation velocity").
Understanding variations in the hydraulic regime and interactions between sediment and fluid in the surf zone depends heavily on the particle settling velocity at the foreshore region. In contrast to sedimentation, which is the end product of the settling process, settling is the movement of suspended particles through the liquid.
Therefore, Thus, option C is correct.
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If the radio waves transmitted by a radio station have a frequency of 79.5 MHz, what is the wavelength of the waves, in meters
If the radio waves transmitted by a radio station have a frequency of 79.5 MHz, the wavelength of the waves, in meters is 3.77m.
What are radio waves?The electromagnetic spectrum's longest wavelengths, which are found in radio waves, are normally found at frequencies of 300 gigahertz. They may be reflected and refracted to change direction, and they do not harm the human body if they are absorbed by it. They are perfect for communicating due to their characteristics.What are Frequency and wavelength?When describing the temporal rate of change seen in oscillatory and periodic phenomena like mechanical vibrations, radio waves, and light, frequency is a crucial parameter utilized in science and engineering.The length of a waveform signal that is propagating in space or over a wire is measured by the separation between identical points (adjacent crests) in the adjacent cycles. Frequency, or the number of wave cycles per second, is inversely related to wavelength.Given: Velocity of light, c = 3.00 x 10⁸ m/s m/s
Frequency, f = 7.95 x 10⁷/s
For wavelength, the required formula is
[tex]λ = c/ f \\λ = 3 * 10 ^{8} / 7.95 * 10^{7} \\λ = 3.77[/tex]
Hence, the required wavelength is 3.77 meters.
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. A man of mass 80 kg stands in a lift, what is the reaction from the floor of the lift if the lift;A. moves upwards with steady speed. moves upwards with acceleration 0.5m/sC. moves downward with acceleration 0.4m/s(g - 9.8m/s')
The reaction of the floor is as follows:
R = 784 NR = 816 NWhat is the reaction from the floor for a man in a lift?The reaction of the floor on the man is equal to the weight of the man and the upward force.
Reaction, R = mg + maA) When the lift moves upwards with steady speed.
R = 80 * 9.8 + * 80 * 0
R = 784 N
B) When the lift moves upwards with acceleration 0.4m/s
R = 80 * 9.8 + * 80 * 0.4
R = 816 N
In conclusion, the reaction of the floor changes as the body accelerates upwards.
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While standing at the top of an 100 m high observatory, you accidentally dropped your phone through the grates (it is now falling straight down to the ground). a. What is the velocity of the phone after 4.0 s? b. How far does the phone fall during this time? C. Will your phone hit the ground? If not, how long more before it hits the ground?
Using the idea of motion under gravity, it will take 0.5 seconds more before the phone hits the ground.
What is the distance covered?The distance covered is obtained form the equations of kinematics under gravity.
Now;
v = u + gt
Recall that it was dropped from a height hence u = 0 m/s
v = gt
v = 9.8 m/s^2 * 4 s
v = 39.2 m/s
Now;
h = ut + 1/2gt^2
but u = 0 m/s
h = 1/2gt^2
h = 1/2 * 9.8 * (4)^2
h = 78.4 m
The phone will not hit the ground within this time
For the phone to hit the ground;
h = 1/2gt^2
if h = 100 m
100 = 1/2 * 9.8 * t^2
2 * 100/9.8 = t^2
t = √2 * 100/9.8
t = 4.5 seconds
It will take about 0.5 seconds more before the phone will hit the ground.
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The density of a certain type of plastic is 0.77 g/cm3. If a sheet of this plastic is 10.0 m long, 1.0 m wide, and 1 cm thick, what is its mass
Mass is 77,000 g
Given
The density of a certain type of plastic = 0.77 [tex]$\mathrm{g} / \mathrm{cm}^{3}$[/tex]
Length = 10.0 m = 1000 cm
Width = 1 m = 100 cm
Thickness = 1 cm
FormulaVolume = length x width x thickness
Mass = Density x Volume
ExplanationDensity
Density is how much "stuff" is contained in a specific quantity of space is determined by its density. For instance, a block of the harder, lighter element gold (Au) will be denser than a block of the heavier element lead (Pb) (Au). Styrofoam blocks are less dense than bricks. Mass per unit volume serves as its definition.Volume V = 1000 x 100 x 1
V = 100,000 [tex]cm^{2}[/tex]
Mass M= 0.77 [tex]$\mathrm{g} / \mathrm{cm}^{3}$[/tex] x 100,000 [tex]cm^{2}[/tex]
M = 77,000 g
So, Mass is 77,000 g
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In which of the following scenarios is the total momentum of the system conserved?
Answer:
The total momentum of a system is conserved only when the system is closed.
Explanation:
a ray of light incident on a mirror, at an angle of 45°. Another mirror is placed at an angle of 45° to the first ones as shown. Sketch the patch of the ray until it emerges from the two mirrors (2mks) 45° www.www 45°
Answer:
If the ray of light is deflected by 45 degrees by the first mirror its total deflection by mirror (I) is 90 deg. (incident = 45 and exit ray equals 45 deg)
The second mirror will cause a net deflection of 90 degrees and the total deflection will be 180 deg or in opposite direction to the incident ray.
A mass is placed on a smooth inclined plane with an angle of 37e to the horizontal. If theinclined plane is 5.0-m long, how long does it take for the mass to reach the bottom of theinclined plane after it is released from rest?
The time taken for the mass to reach the bottom of the inclined plane after it is released from rest is 0.78 s.
Height of the inclined plane
The height of the inclined plane is calculated as follows;
h = L(sin 37)
where;
h is height of the planeL is length of the planeh = 5 x sin(37)
h = 3.01 m
Time of motion of the masst = √(2h/g)
t = √(2 x 3.01 / 9.8)
t = 0.78 s
Thus, the time taken for the mass to reach the bottom of the inclined plane after it is released from rest is 0.78 s.
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What is the gravitational potential energy stored in a rock of 100 kg
mass, placed on top of a hill, which is 100 m high?
O 45 kJ
O 98 kJ
O 50 kJ
O 100 kJ
A car approaches a stationary police car at 36 m/s. The frequency of the siren (relative to the police car) is 500 Hz. What is the frequency (in Hz) heard by an observer in the moving car as he approaches the police car
The frequency (in Hz) heard by an observer in the moving car as he approaches the police car is 552 Hz
The Doppler Effect equation says
f = f0 c+v / c
Given
f0 = 500 Hz
c = 343 m/s
v = 36 m/s
by plugging the given data, we get
500 (343+36) / 343 = 552 Hz
The frequency (in Hz) heard by an observer in the moving car as he approaches the police car is 552 Hz
Doppler effect is the apparent difference between the frequency at which sound or light waves leave a source and that at which they reach an observer, caused by the relative motion of the observer and the wave source.
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An aircraft executes a horizontal loop of radius 1.00 km with a steady speed of 900 km/h.Compare it's centripetal acceleration with the acceleration due to gravity.
Answer:
900 km/hr = 9E5 m / 3600 sec = 250 m/sec
ah = v^2 / R = 250^2 m/s / 1000 m = 62.5 m/s^2 horizontal acceleration
av = 9.8 m/s^2 vertical acceleration (due to gravity)
ah / av = 62.5 / 9.8 = 6.38 centripetal acceleration is greater
A dog running at 10 m/s is 30m behind a rabbit moving at 5 m/s. when will the dog catch up with the rabbit assuming both their velocities remain constant during the chase?
The will dog catch up with the rabbit in 6 minutes assuming both their velocities remain constant during the chase.
What time will the dog catch the rabbit?The time that the dog will catch up with the rabbit is given as follows:
Let the distance covered by the rabbit be x.
Distance covered by dog = x + 30
Time taken = distance/speedThe time taken will be the same T
Time taken by dog, T = (x + 30)/10Time taken by rabbit, T = x/5Equating both times.
(x + 30)/10 = x/5
x = 30 m
Solving for T in equation (ii);
T = 30/5 = 6 minutes
In conclusion, time is obtained as a ratio of distance and speed.
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A speed skater moving across frictionless ice at 8.0 m/s hits a 5.0-m-wide patch of rough ice. She slows steadily, then continues on at 6.0 m/s. What is her acceleration on the rough patch
Her acceleration on the rough patch is 2.8m/s².
To find the answer, we need to know about the Newton's equation of motion.
What are the Newton's equation of motion?These are as follows
V= U +atS = Ut+1/2 at²V²-U²= 2aSV= Final velocity
U = initial velocity
t= time
S= distance
What will be acceleration if initial velocity, final velocity and distance are 8 m/s , 6 m/s and 5m respectively?Here U= 8 m/s, V = 6 m/s and S = 5mSo, 6²-8²= 2a× 5=10a36-64 = 10a
-28= 10 a
a= -28/10= -2.8 m/s²
Negative sign indicates the decrease of acceleration.Thus, we can conclude that the acceleration on the patch is 2.8 m/s².
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A resistor dissipates 2.00 W when the rms voltage of the emf is 10.0 V. At what voltage will the resistor dissipate 10.0 W
50 V to dissipate a power of 10 W
Given
A resistor dissipates 2.00 W when the rms voltage of the emf is 10.0 V.
Resistor and Voltage The difference in charge between two places is known as voltage. The passage of charge is measured as current. A material's propensity to oppose the flow of charge is known as resistance (current).Ohm's Law describes the relationship between current, voltage, and resistance. This asserts that, given the temperature is constant, the current flowing in a circuit is directly proportional to the applied voltage and inversely proportional to the circuit resistance.Explanation[tex]$\mathrm{V}_{r m s}=\mathrm{P}$[/tex]
10 v provides 2 W of power whereas V provides 10 W of power.
10 V = 2 W
x V=10 W
By using cross multiplication
2 x = 100
x = 50 V
So, 50 V to dissipate a power of 10 W
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A block of mass 60kg measures 6cm by 5cm by 4cm. Calculate i) The maximum pressure it can exert. The minimum pressure. ii) A block of mass 60kg measures 6cm by 5cm by 4cm . Calculate i ) The maximum pressure it can exert . The minimum pressure . ii )
Answer:
4.9 is the answer
Explanation:
pressure is force divided by area
force = mass x acceleration due to gravity
= 60kg x 9.8 m/s
area = length x width x height
=6 x 5 x 4
= 120
pressure = 588/120
= 4.9 pascal
The correct answer for the value of (i) Maximum pressure is [tex]24.5 \dfrac{N}{cm^2}[/tex](ii)Minimum pressure is [tex]19.6 \dfrac{N}{cm^2}[/tex].
Given:
Weight of the block: [tex]60[/tex] kg
Dimensions of the box: [tex]6 *4* 5[/tex]
Value of gravitational constant: [tex]9.8[/tex] m/s²
Now,
[tex]Pressure = \dfrac{Force}{Area}[/tex]
[tex]Force = mass*accelaration[/tex]
[tex]= 60 * 9.8[/tex]
[tex]= 588 N[/tex]
Minimum pressure is when area is maximum:
[tex]= \dfrac{588}{6*5}\\\\= 19.6 \dfrac{N}{cm^2}[/tex]
Maximum pressure when area is minimum:
[tex]= \dfrac{588}{6*4} \\\\= 24.5 \dfrac{N}{cm^2}[/tex]
Maximum and minimum pressure are [tex]24.5 \dfrac{N}{cm^2}[/tex] and [tex]19.6 \dfrac{N}{cm^2}[/tex] respectively.
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