The angle (θcornea) when light passes from air to cornea is 16.86°
What is Snell's law?It states that the ratio of sine of angle of incidence and angle of refraction is equal to the refractive index of second medium to the first medium.
sini/sinr =n₂ / n₁
Most of the bending of light in the eye is done at the air-cornea interface. The beam of light deviated as it passes from air to the cornea if the incident angle is θi = 23.6°.
Given the angle of incidence i = 23.6°, refractive index of air n₁ =1, refractive index of cornea n₂ = 1.38, then the angle of refraction at cornea is
sinr = sini x (n₁/n₂)
Plug the values, we get
sinr = sin23.6 x (1/1.38)
sinr = 16.86°
The angle of refraction is less than angle of incidence due to refraction.
Thus, the angle (θcornea) when light passes from air to cornea is 16.86°
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Calculate the magnitude of the linear momentum for the following cases.
(a) a proton with mass 1.67 x 10-27 kg, moving with a speed of 5.05 x 106 m/s
(d) the Earth (mass - 5.98 x 1024 kg) moving with an orbital speed equal to 2.98 x 10 m/s.
Answer:
8.4335 x 10^-²¹, 1.78204 x 10²⁶
Explanation:
p = mv
p = (5.05 x 10⁶)(1.67 x 10^-²⁷)
p = 8.4335 x 10^-²¹
p = mv
p = (5.98 x 10²⁴)(2.98 x 10)
p = 1.78204 x 10²⁶
a car is traveling along a straight road at a velocity of 30m/s when its engine cuts off. For the next ten seconds, the car slows down, and its average acceleration is a1. For the next five seconds, the car slows down further at a velocity of 24m/s, and its average acceleration is a2. The ratio of the average acceleration values is a1/a2=1.5. Find the velocity of the car at the end of the initial ten-second interval.
Answer:
See below....the question is unclear....pick the answer you think they want
Explanation:
Car goes from 30 m/s to 24 m/s in 10 seconds
acceleration = change in velocity/change in time
= -6 m/s / 10 s = - 3/5 m/s^2
The wording of the question is not very good.....perhaps they meant to say that after the 5 seconds the velocity is now 24 m/s ?
a1 *10 + a2 *5 = -6 m/s
( 6 m/s is the change in velocity from 30 to 24 m/s)
a1/a2 = 1.5 so a2 = a1/1.5 (substitute this in)
10 a1 + 5 ( a1/1.5) =-6
15 a1 + 5 a1 = -9
a1 = - .45 m/s^2
The velocity of the car at the end of the initial ten-second interval, given that the car further slows down for the next five seconds, is 25.5 m/s
How to determine the velocity of the car at the end of the 10 s interval?From the first statement, we have:
Initial velocity (u₁) = 30 m/sTime (t₁) = 10 secondsVelocity at the end of 10 s (v₁) = ?v₁ = u₁ + a₁t₁
v₁ = 30 + (a₁ × 10)
v₁ = 30 + 10a₁ ......(1)
From the second statement, we have:
Time (t₂) = 5 sFinal velocity (v₂) = 24 m/sRation of acceleration (a₁/a₂) = 1.5Initial velocity (u₂) = Velocity at the end of 10 sv₂ = u₂ + a₂t₂
24 = u₂ + (a₂ × 5)
24 = u₂ + 5a₂ .......(2)
But,
u₂ = v₁ = 30 + 10a₁
Thus, we have:
24 = 30 + 10a₁ + 5a₂
But,
a₁/a₂ = 1.5
a₁ = 1.5a₂ ..... (3)
Thus, we have:
24 = 30 + 10a₁ + 5a₂
24 = 30 + 10(1.5a₂) + 5a₂
24 = 30 + 15a₂ + 5a₂
24 = 30 + 20a₂
Collect like terms
24 - 30 = 20a₂
-6 = 20a₂
Divide both sides by 20
a₂ = -6 / 20
= -0.3 m/s²
Substituting the value of a₂ into equation 3, we have
a₁ = 1.5a₂
= 1.5 × -0.3
= -0.45 m/s²
Substitute the value of a₁ into equation 1 to obtain the velocity, v₁ at the end of the initial to 10 s
v₁ = 30 + 10a₁
= 30 + (10 × -0.45)
= 30 - 4.5
= 25.5 m/s
Thus, velocity of the car at the end of the initial ten-second interval is 25.5 m/s
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n Step 5, you will calculate H+/OH– ratios for more extreme pH solutions. Find the concentration of H+ ions to OH– ions listed in Table B of your Student Guide for a solution at a pH = 2. Then divide the H+ concentration by the OH– concentration. Record these concentrations and ratio in Table C.
What is the concentration of H+ ions at a pH = 2?
mol/L
What is the concentration of OH– ions at a pH = 2?
mol/L
What is the ratio of H+ ions to OH– ions at a pH = 2?
: 1
At pH = 2 the ratio of the concentration of the hydrogen to the hydroxide ion is 10¹⁰. The hydrogen ion concentration at pH, 2 is 10⁻² and the hydroxide ion is 10⁻¹².
What is pH?The parameter of measuring the concentration of the hydroxide and the hydrogen ion in the solution in order to determine the basic and the acidic nature is known as pH.
The concentration of H⁺ ions is calculated as:
pH = -log [H⁺]
2 = -log [H⁺]
[H⁺] = 10⁻²
The concentration of OH⁻ ions is calculated as:
pH + pOH = 14
pOH = 14 - 2
pOH = 12
Solving further:
pOH = -log [OH⁻]
12 =-log [OH⁻]
[OH⁻] = 10⁻¹²
At a pH = 2, the ratio of hydrogen ions to hydroxide ions:
10⁻² ÷ 10⁻¹² = 10¹⁰
Therefore, at pH = 2 the ratio is 10¹⁰.
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Water's specific heat capacity is 4.184 J/(g · °C). How does this compare to the heat capacities of most common substances?
Answer:
Explanation:
Comment
It's high. It takes quite a bit of heat to get water to change its heat content. Most substances don't reach much more the 0.5 J/(gram*oC)
A ball of mass 500g is dropped from a height 1.5m . It rebounds the floor and reached the height 1.2m. Calculate the impulse given to the bàll by the floor
The impulse given to the ball by the floor is 0.2865 kg.m/s.
What is impulse?The change in momentum is equal to the product of impact force applied while colliding and time for that impact.
Impulse F. t = m (Vf -Vi)
where, Vf is the final velocity and Vi is the initial velocity.
A ball of mass 500g is dropped from a height 1.5m . It rebounds the floor and reached the height 1.2m.
The initial velocity u = √2x 9.81 x 1.5 = 5.425 m/s
The final velocity v = √2x 9.81 x 1.2 = 4.852 m/s
Substitute the values into the expression, we get
Impulse = m(v- u)
Impulse=0.5 x (4.852- 5.425 )
Impulse = - 0.2865 kg.m/s
Thus, the magnitude of impulse given to the ball by the floor is 0.2865 kg.m/s.
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5. A uniform solid ring of mass M = 10 kg and radius R = 2 m is initially rotating in mid-air with angular speed 20 rad/s [the Before state in figure (I) below]. It is then placed in contact with a surface (assume that the surface immediately feels the full weight of the ring). (1) M Before Wi (II) wf V Ө ə Ө After Ө wf H ? (a) What kind of frictional force acts on the ring upon contact with the surface? (b) If the ring begins to roll smoothly to the right with speed v = 10 m/s after 5 seconds of contact (the After state), find the coefficient of friction corresponding to the frictional force you mentioned in (a). (c) What is the increase in the thermal energy of the ring/surface system from right before the ring contacts the surface, to when the ring starts to roll smoothly? The ring then rolls smoothly up a ramp of 0 = π/6 rad and H = 5 m [see figure (II)] (d) What is the horizontal distance, from the end of the ramp, at which the ring lands?
(a) The kind of frictional force acts on the ring upon contact with the surface is kinetic friction before rolling and afterwards it is static friction.
(b) If the ring begins to roll smoothly to the right with speed v = 10 m/s after 5 seconds of contact (the After state), the coefficient of friction corresponding to the frictional force is 0.612
(c) The increase in the thermal energy of the ring/surface system from right before the ring contacts the surface, to when the ring starts to roll smoothly is 7500 J.
(d) The horizontal distance, from the end of the ramp, at which the ring lands is 7.78 m
What is frictional force?When an object is moving on a rough surface, it experiences opposition. This opposing force is called the friction force.
Given is a uniform solid ring of mass M = 10 kg and radius R = 2 m is initially rotating in mid-air with angular speed 20 rad/s [the Before state in figure (I) below]. It is then placed in contact with a surface (assume that the surface immediately feels the full weight of the ring).
(a) The kind of frictional force acts on the ring upon contact with the surface is kinetic friction before rolling and afterwards it is static friction.
(b) angular frequency = velocity / radius
ωf = 10/2 = 5 rad/s and ωi = 20 rad/s
Angular acceleration, α = ωf - ωi /t
Put the values, we get
α = -15/5 = -3 rad/s²
Coefficient of friction, μ = a/g = rα/g
Plug the values, we get
μ = 0.612
Thus, the coefficient of friction corresponding to the frictional force is 0.612.
(c) The energy lost = heat generated
energy lost = 1/2 Iω² + 1/2 Mv²
energy lost = 1/2 MR²ω² + 1/2 Mv²
Plug the values, we get
energy lost = 7500 J
Thus, the increase in the thermal energy is 7500 J.
(d) The horizontal distance, from the end of the ramp, at which the ring lands
s= (v- 2gH) sin2θ /g
s = (10 - 2x (-9.81)x5 ) sin (2x π/6) / 9.81
s = 7.78m
Thus, the horizontal distance is 7.78 m
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What must the charge (sign and magnitude) of a particle of mass 1.45 g be for it to remain stationary when placed in a downward-directed electric field of magnitude 600 N/C ? Use 9.80 m/s2 for the magnitude of the free-fall acceleration.
[tex]q = -21 * 10^{-6} C[/tex]
What is Free-fall acceleration?The acceleration is constant and equal to the gravitational acceleration g which is 9.8 m/s at sea level on the Earth.
As we know that charge and its sign that remains in equilibrium is under gravity must be such that it will balance the gravitational force by electric force,
mg =qE
[tex]1.45 * 10^{-3}(9.80)=q(600)\\\\q = 21 *10^{6} C\\\\[/tex]
and its sign must be negative so that it will have upward electric force
so it is
[tex]q = -21 * 10^{-6} C[/tex]
The charge of a particle of mass is [tex]-21 * 10^{-6} C[/tex]
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A bike accelerates uniformly from rest to a speed of 10 m/s over a distance of 50 m. (a) Determine the acceleration of the bike. (b) how long will take to do that?
Answer:
a) 0.2m/s
b) 5m/s
Explanation:
a) acceleration=∆v/∆t
v=10
t=50
10/50=1/5
=0.2m/s
b) time= d/s
d=50m
s=10
50/10
=5m/s
Answer:
See below
Explanation:
Average speed = (0+ 10)/2 = 5 m/s
then to cover 50 m will take 50 m / 5 m/s = 10 seconds
change in velocity/ change in time = acceleration = 10/10 = 1 m/s^2
Determine the speed at which the medicine leaves the needle
The speed at which the medicine leaves the needle is 2.462 m/s
What is Bernoulli's theorem?When an incompressible, ideal fluid is flowing through a tube or pipe, the total energy remains constant.
p₁ /ρg + v₁²/2g +z₁ = p₂ /ρg + v₂²/2g +z₂
Where, p/ρg = pressure energy
v²/2g = kinetic energy
z = potential energy
Given is during an injection, pressure in the barrel of syringe is 1.03 atm while pressure in the needle section is 1.00 atm. Assuming the syringe lays horizontally and mass density of the liquid medicine ρ =1000 kg/m³.
The fluid is initially at rest.
Using the Bernoulli's equation, we have
v₁² - v₂²= 2 x (p₂ -p₁) / ρ
Substituting the values, we get
0 - v₂² = 2 x (1.00 -1.03) x 1.01 x 10⁵ /1000
v₂² = 6.06
v₂ = 2.462 m/s
Thus, the speed at which the medicine leaves the needle is 2.462 m/s
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A projectile is fired at an upward angle of 45.0º from the top of a 265-m cliff with a speed of .sm 185 What will be its speed when it strikes the ground below? (Use conservation of energy.)
The final velocity of the projectile when it strikes the ground below is 198.51 m/s.
Time of motion of the projectileThe time taken for the projectile to fall to the ground is calculated as follows;
h = vt + ¹/₂gt²
where;
h is height of the cliffv is velocityt is time of motion265 = (185 x sin45)t + (0.5)(9.8)t²
265 = 130.8t + 4.9t²
4.9t² + 130.8t - 265 = 0
solve the quadratic equation using formula method,
t = 1.89 s
Final velocity of the projectilevyf = vyi + gt
where;
vyf is the final vertical velocityvyi is initial vertical velocityvyf = (185 x sin45) + (9.8 x 1.89)
vyf = 149.322 m/s
vxf = vxi
where;
vxf is the final horizontal velocityvxi is the initial horizontal velocityvxf = 185 x cos(45)
vxf = 130.8 m/s
vf = √(vyf² + vxf²)
where;
vf is the speed of the projectile when it strikes the ground belowvf = √(149.322² + 130.8²)
vf = 198.51 m/s
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What is the wavelength of a compression wave with a speed of 500 m/s and a frequency of 250 Hz?
Use this formula: λ = s/f
Select one:
a. 2 m
b. 1.25 m
c. 250 m
d. 750 m
Answer:
a (2m)
Explanation:
according to wavelenght = speed / frequency,
the wavelength is 2 m
Erbium-165 has a half-life of 10.4 hours. If you start with 1,000 grams of
erbium-165, how much time will it take to have 125 grams of erbium-165 left
in the sample?
A. 41.6 hours
OB. 20.8 hours
C. 31.2 hours
D. 10.4 hours
The amount of time it will take to have 125 grams of erbium-165 left
in the sample is; C: 31.2 hours
How to calculate decay time?The formula for amount of substance after decay is;
N(t) = N₀(¹/₂)^(t/t_¹/₂)
We are given;
Half life; t_¹/₂ = 10.4 hours
Initial amount; N₀ = 1000 g
Amount left; N(t) = 125 g
Thus;
125 = 1000 * (¹/₂)^(t/10.4)
125/1000 = (¹/₂)^(t/10.4)
In 0.125 = (t/10.4) In 0.5
-2.08 = -0.693(t/10.4)
t/10.4 = -2.08/-0.693
t = 10.4 * 3
t = 31.2 hours
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A force of 20N is applied to the end of a wire of length 5m to produce an extension of 0.20mm
calculate: the stress on the wire.
The stress on the wire is determined as 6.37 x 10⁶ N/m².
Stress on the wireThe stress on the wire is calculated as follows;
σ = F/A
where;
F is the force applied on the wireA is area of the wireLet the diameter of the wire = 2 mm
Area = πr²
Area = π(1 x 10⁻³)²
Area = 3.142 x 10⁻⁶ m²
Stress = 20/(3.142 x 10⁻⁶)
Stress = 6.37 x 10⁶ N/m²
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Which of the following can be contracted from contact with bloodborne pathogens?
HIV can be contracted from contact with bloodborne pathogens.
Other bloodborne diseases are HBV, malaria, syphilis and brucellosis
What are bloodborne pathogens?Bloodborne pathogens can be defined as those microorganisms or pathogenic organisms that cause disease and are present in human blood.
Blood borne pathogens can also be contacted through the following means
Se- xual contactNeedle contactIn conclusion; HIV can be contracted from contact with bloodborne pathogens.
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Aude slipped on ice going 7.6 m/s .What was her velocity before she hit the wal?
Answer:
Define universal gravitational constant. Why is newton's law of gravitation called universal law?
An arrow is shot horizontally from the top of a building and it lands 200m from the foot of the building after 10s.Assuming air resistance is negligible,calculate the initial velocity of the arrow and the height of the building?
Need an answer urgently please
The initial velocity of the arrow and the height of the building will be 29.05 m/s² and 781 m respectively.
What is velocity?The change of distance with respect to time is defined as speed. Speed is a scalar quantity. It is a time-based component. Its unit is m/sec.
The given data in the problem is;
u is the initial velocity of fall = ? m/sec
h is the distance of fall = 200 m
g is the acceleration of free fall = 9.81 m/sec²
H is the height of the building
t is the time period = 10 second
According to Newton's second equation of motion,
[tex]\rm h= ut+\frac{1}{2} gt^2 \\\\\ h-\frac{1}{2} gt^2 =ut \\\\ 200 - 0.5 \times(9.81) \times 10^2 = 10 u \\\\ u = - 29.05 \ m/sec[/tex]
- ve shows the direction is downward.The magnitude of the initial velocity is found as;
u = 29.05 m/sec
The height of the building
[tex]\rm H= ut+\frac{1}{2} gt^2 \\\\\ H = 29.05 \times 10 + 0.5 \times 9.81 \times 10^2 \\\\ H = 781 \ m[/tex]
Hence the initial velocity of the arrow and the height of the building will be 29.05 m/s² and 781 m respectively.
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Taking the density of air to be 1.29 kg/m3, what is the magnitude of the angular momentum (in kg · m2/s) of a cubic meter of air moving with a wind speed of 73.0 mi/h in a hurricane? Assume the air is 51.2 km from the center of the hurricane "eye."
The magnitude of the angular momentum of air will be 4.128 x 10^(-3) kg·m^2/s
The rotating equivalent of linear momentum in physics is called angular momentum. Because it is a conserved quantity—the total angular momentum of a closed system stays constant—it is significant in physics. Both the direction and the amplitude of angular momentum are preserved.
Given the density of air to be 1.29 kg/m3 and a wind speed of 73.0 mi/h
We have to find the magnitude of the angular momentum
Let,
ρ = Density of air = 1.29 kg/m^3
v = Speed of wind = 73.0 mi/h = 0.032 km/s
M = angular momentum of air
Let the volume of air be 1 m^3
Mass = Volume x ρ = 1 x 1.29 = 1.29 kg
Momentum = M = mass x velocity
Momentum = 1.29 x 0.0032
Momentum = 4.128 x 10^(-3) kg·m^2/s
Hence the magnitude of the angular momentum of air will be 4.128 x 10^(-3) kg·m^2/s
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Question 3 of 15
Which statement describes an advantage of using nuclear fission to produce
electricity?
A. It uses a fuel made of atoms with large nuclei that will be
destroyed.
B. It is much more efficient than any other process used to generate
power.
C. It happens in large buildings that are expensive to build.
D. It produces radioactive waste products that must be stored safely.
Answer:
it is much more efficient than any other process to generate power
What is the gravitational potential energy of a 15.0kg object that is 5.00m above the ground relative to a point 8.00m above the ground?
-441J
735J
-735J
441J
Explanation:
an object's gravitational potential energy Eg is m×g×h where:
m=mass
g=9.8m/s²
h=height relative to the closest object below it (because it cannot potentially fall through it
so Eg = 15×9.8×5=735J
differentiate e^x - 8x +7
please answer the questions asap
Answer:
eˣ-8.
Explanation:
(eˣ-8x+7)'=eˣ-8.
the temprerture of an object is 40°c.what is this tempreture in farhenhite scale?
This is an exercise of thermal scales.
Data:T = 40 °C
T = °F ¿?
To convert degrees Celsius to degrees Fahrenheit, we have the formula:
[tex]\bf{^{\circ}F=\dfrac{9}{5} \ ^{\circ}C+32 \ \ \to \ \ \ Formula }[/tex]
We substitute our data into the formula.
[tex]\bf{^{\circ}F=\dfrac{9}{5} \times 40+32}[/tex]
First we divide 9/5 multiplied by 40.
[tex]\bf{=75+32}[/tex]
Add 75 plus 32
[tex]\bf{=104 \ ^{\circ}F}[/tex]
Therefore, the temperature change of 40 °C on the °F scale is equal to 104.
[tex]\huge \red{\boxed{\green{\boxed{\boldsymbol{\purple{Piscis04}}}}}}[/tex]
According to the __________perspective, people who suffer from PTSD are classically conditioned.
According to the pavlovian perspective, people who suffer from PTSD are classically conditioned.
What is PSTD?
This is referred to as post traumatic stress disorder which is a mental disorder from a terrifying incident.
The pavlovian perspective however view this disorder as being a type of classical conditioned one.
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During World War II, it was found that r, the radius of the shockwave produced during an atomic bomb explosion, depends on the energy released, W, the elapsed time, t, and the air density, .
The atomic bomb explosion, depends on the energy released, W, the elapsed time, t, and the air density which altogether lead to the vaporisation of the contents within the mushroom cloud created.
What is an atomic bomb?An atomic bomb is type of nuclear weapon that releases a vast amount of energy upon explosion in the form of fission reaction.
During an explosion of atomic bomb, a cloud of mushroom fire is formed which vaporises anything found within it. The extent of destruction depends on:
energy released, W, the elapsed time, t, and the air density.Therefore, the atomic bomb explosion, depends on the energy released, W, the elapsed time, t, and the air density which altogether lead to the vaporisation of the contents within the mushroom cloud created.
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Part A
Calculate the spring constant.
• Look at the graph of x vs. t graph. (You may want to double-click on
it to examine it in the Data Tool view.)
. From the graph, determine the period T.
• The first video frame lists the mass.
• Use the equation below to solve for the spring constant k. Show
your work below.
m
T = 2π√√T
Answer:
Using 3 periods to get an accurate reading:
3T = (6.40 S - 0.90 s) = 5.50 S So, T = 1.83 s
m = 0.250 kg
Using algebra:
k= [tex]\frac{4\pi ^{2}m }{T^{2} }[/tex]=[tex]\frac{4\pi ^{2}0.250 }{1.90^{2} }[/tex]=2.95kg/sec
Using Hooke's law, what happens when the displacement is tripled?
If the displacement is tripled then the spring force also is tripled.
What is Hooke's law?The amount of force required to stretch or compress the spring is known as the spring force. Its unit is Newton(N).
Froom the hooks law;
F = kx
Where,
F is the applied force
x is the displacement in case 1
Case 2;
x'=3x
F'=kx'
F'=3KX
F'=3x'
Where,
x' is the displacement when the force is tripled
F' is the force in the case2
Hence, if the displacement is tripled then the spring force also be tripled.
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Resolve the vector shown below into its components.
Answer:
D. s=3x+4y
Explanation:
The line is at the 3rd column in the 4th row.
The value of acceleration due to gravity (g) on Pluto is about 0.61 meters/second2. How much will an object that weighs 250 newtons on Earth weigh on Pluto? Note that the value of acceleration due to gravity on Earth is 9.8 meters/second2.
The weight of an object on pluto will be 15.56. Mass multiplied by the gravitational acceleration gives weight.
What is gravitation?Gravitation is a natural law by which all things with all matter are attracted towards one another. Gravity is responsible for large-scale structures present in the Universe.
By dividing the object's weight on Earth by 9.8 m/s², as illustrated below, one may calculate the object's mass.
m = 250 N / 9.8 m/s²
m = 25.51 kg
Multiply the acquired mass by Pluto's gravitational acceleration (g);
W = (25.51 kg) x (0.61 m/s²)
W = 15.56 N
Thus, the item will only be 15.56 N in weight.
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Answer: B.
15.6 newtons
Explanation: edmentum
A small sphere of mass 10 kg
is released from rest at a height of
15.0 m above the ground level.
The sphere experiences a constant
resistive force (due to air
resistance) of magnitude R = 10.0
N.
a) Calculate the speed of the
sphere after it has fallen
through a distance of 5.00 m
bCalculate the speed of the ball just before a it hits the gound.
Answer:
Approximately [tex]9.39 \; {\rm m\cdot s^{-1}}[/tex] after the sphere has travelled a distance of [tex]5\; {\rm m}[/tex].
Approximately [tex]16.3\; {\rm m\cdot s^{-1}}[/tex] right before touching the ground (a distance of [tex]15\; {\rm m}[/tex].)
Assumption: [tex]g = 9.81\; {\rm N\cdot kg^{-1}}[/tex].
Explanation:
Weight of the sphere: [tex]m\, g = 9.81\; {\rm N \cdot kg^{-1}} \times 10\; {\rm kg} = 98.1\; {\rm N}[/tex], downwards.
Drag on the sphere: [tex]10.0\; {\rm N}[/tex] upwards.
Net force on the sphere: [tex]98.1\; {\rm N} - 10\; {\rm N} = 88.1\; {\rm N}[/tex] downwards.
Acceleration of the sphere: [tex]a = F_\text{net} / m = 88.1\; {\rm N} / (10\; {\rm kg}) = 8.81\; {\rm m\cdot s^{-2}}[/tex].
Apply the SUVAT equation [tex]v^{2} - u^{2} = 2\, a\, x[/tex], where [tex]v[/tex] is the final velocity, [tex]u[/tex] is the initial velocity ([tex]0[/tex] in this case, as the sphere was released from rest,) and [tex]x[/tex] is the distance (displacement) that the sphere has travelled so far.
Rearrange this equation to obtain an expression for [tex]v[/tex]:
[tex]\displaystyle v = \sqrt{2\, a\, x + u^{2}}[/tex].
For example, after the ball travelled a distance of [tex]5.00\; {\rm m}[/tex], [tex]x = 5.00 \; {\rm m}[/tex]:
[tex]\begin{aligned} v &= \sqrt{2\, a\, x + u^{2}} \\ &= \sqrt{2 \times 8.81\; {\rm m\cdot s^{-2}} \times 5.0\; {\rm m} + 0} \\ &\approx 9.39\; {\rm m\cdot s^{-1}}\end{aligned}[/tex].
Similarly, [tex]x = 15.0\; {\rm m}[/tex] right before landing, such that:
[tex]\begin{aligned} v &= \sqrt{2\, a\, x + u^{2}} \\ &= \sqrt{2 \times 8.81\; {\rm m\cdot s^{-2}} \times 15.0\; {\rm m} + 0} \\ &\approx 16.3\; {\rm m\cdot s^{-1}}\end{aligned}[/tex].
A 3.42 kg steel ball strikes a massive wall at 14.3 m/s at an angle of α = 61.5° with respect to the plane of the wall. It bounces off with the same speed and angle as seen in the figure.
If the ball is in contact with the wall for 0.207 s, what is the average force exerted on the ball by the wall?
The magnitude of the average force exerted on the ball by the wall is 225.469 N.
What is impulse?The change in momentum is equal to the product of impact force applied while colliding and time for that impact.
Impulse F. t = m (Vf -Vi)
where, Vf is the final velocity and Vi is the initial velocity.
Given is a 3.42 kg steel ball strikes a massive wall at 14.3 m/s at an angle of α = 61.5° with respect to the plane of the wall. It bounces off with the same speed and angle. The ball is in contact with the wall for 0.207 s
Substitute the values into the expression, we get
Impulse = 2 x mvcosθ
Impulse= 2 x 3.42 x 14.3 x cos 61.5°
Impulse = 46.672 kg.m/s
The impact force can be written as
F.t = I
Put the given values, we have
F = 46.672 / 0.207
F = 225.469 N
Thus, the magnitude of force exerted by the wall is 225.469 N
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The action force is the ballon pushing the air out. What is the magnitude of the reaction force of the air pushing the ballon
The magnitude of the reaction force of the air pushing the balloon would be equal and opposite.
How is the reaction force equal and opposite?A push or a pull that an object experiences as a result of interacting with another item is known as a reaction force.
Newton's third law is officially expressed as follows: There is an equal and opposite reaction to every action. The implication of the statement is that there are always two forces acting on the two interacting objects. The force acting on the first object is equal in size to the force acting on the second. The force acting on the first object is acting in the opposite direction to the force acting on the second object. Force pairs—equal and opposing action-reaction force pairs—always exist in pairs.
Knowing that everything has an equal and opposite reaction according to Newton's second law. The preasured air that a balloon had to push out into the free air acts as the reaction force.
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