Answer:
a) a = 27.44 m / s², b) a = 5.39 m / s², c) a = 156.8 m / s², cabinet maximum acceleration does not change
Explanation:
a) In this exercise the wheels of the truck rotate to provide acceleration, but the contact point between the ground and the 2 wheels remains fixed, therefore the coefficient of friction for this point is static.
Let's apply Newton's second law
we set a regency hiss where the x axis is in the direction of movement of the truck
Y axis y
N- W = 0
N = W = m g
X axis
2fr = m a
the expression for the friction force is
fr = μ N
fr = μ m g
we substitute
2 μ m g = m /2 a
a = 4 μ g
a = 4 0.7 9.8
a = 27.44 m / s²
b) let's look for the maximum acceleration that can be applied to the cabinet
fr = m a
μ N = ma
μ m g = m a
a = μ g
a = 0.55 9.8
a = 5.39 m / s²
as the acceleration of the platform is greater than this acceleration the cabinet must slip
c) the friction force is in the four wheels as well
With when the truck had two-wheel Thracian the weight of distributed evenly between the wheels, in this case with 4-wheel Thracian the weight must be distributed among all
applying Newton's second law
4 fr = (m/4) a
16 mg = (m) a
a = 16 g
a = 16 9.8
a = 156.8 m / s²
cabinet maximum acceleration does not change
why does self inductance acts as electrical inertia?
Answer:
self-indulgence of coil is the property by virtue of wich is tends to maintain magnatic flux link with it and opposed any any change in the flux inducing current in it
Help me plssssssss cause I’m struggling
Answer:
I am pretty sure it is C
Explanation:
It can be found all over the universe
A solid uniform disk of diameter 3.20 m and mass 42 kg rolls without slipping to the bottom of a hill, starting from rest. If the angular speed of the disk is 4.27 rad/s at the bottom, how high did it start on the hill?
A) 3.57 m.
B) 4.28 m.
C) 3.14 m.
D) 2.68 m.
Answer:
A(3.56m)
Explanation:
We have a conservation of energy problem here as well. Potential energy is being converted into linear kinetic energy and rotational kinetic energy.
We are given ω= 4.27rad/s, so v = ωr, which is 6.832 m/s. Place your coordinate system at top of the hill so E initial is 0.
Ef= Ug+Klin+Krot= -mgh+1/2mv^2+1/2Iω^2
Since it is a solid uniform disk I= 1/2MR^2, so Krot will be 1/4Mv^2(r^2ω^2= v^2).
Ef= -mgh+3/4mv^2
Since Ef=Ei=0
Mgh=3/4mv^2
gh=3/4v^2
h=0.75v^2/g
plug in givens to get h= 3.57m
explain why sound wave travel faster in liquid than gas
Answer:
Because gas contains free molecules but not liquid.
Please mark as brainliast
A cylindrical specimen of aluminum having a diameter of 0.505 in. (12.8 mm) and a gauge length of 2.0 in. (50.8 mm) is pulled in tension. Use the load-elongation characteristics tabulated below to complete parts (a) through (f).
a. Plot the data as engineering stress versus engineering strain.
b. Compute the modulus of elasticity.
c. Determine the yield strength at a strain offset of 0.002.
d. Determine the tensile strength of this alloy.
e. What is the approximate ductility, in percent elongation?
f. Compute the strain energy density up to yielding (modulus of resilience).
( Load in N Load in lb Length in mm Length in in. 2.000 2.002 2.004 2.006 2.008 2.010 2.020 2.040 2.080 2.120 2.160 2.200 2.240 2.270 2.300 2.330 Fracture 50.800 7330 15,100 3400 23,100 5200 30,400 6850 34,400 7750 38,400 8650 41,3009300 44,800 10,100 46,200 10,400 53, 47,300 10,650 54.864 47,500 10,700 55.880 46,100 10,400 44,800 10,100 42,600 9600 3,400 8200 Fracture Fracture Fracture 50.851 50.902 50.952 51.003 51.054 1650 51.308 51.816 52.832 848 56.896 57.658 58.420 59.182
Answer:
A cylindrical specimen of aluminum having a diameter of 0.505 in. (12.8 mm) and a gauge length of 2.0 in. (50.8 mm) is pulled in tension. Use the load-elongation characteristics tabulated below to complete parts (a) through (f).
Una cuerda horizontal tiene una longitud de 5 m y masa de 0,00145 kg. Si sobre esta cuerda se da un pulso generando una longitud de onda de 0,6 m y una frecuencia de 120 Hz. La tensión a la cual está sometida la cuerda es:
a. 1,5 N
b. 15,0 N
c. 3,1 N
d. 5,2 N
Answer:
Option (A) is correct.
Explanation:
A horizontal rope has a length of 5 m and a mass of 0.00145 kg. If a pulse occurs on this string, generating a wavelength of 0.6 m and a frequency of 120 Hz. The tension to which the string is subjected is
mass of string, m = 0.00145 kg
Frequency, f = 120 Hz
wavelength = 0.6 m
Speed = frequency x wavelength
speed = 120 x 0.6 = 72 m/s
Let the tension is T.
Use the formula
[tex]v =\sqrt\frac{T L}{m}\\\\72 = \sqrt\frac{T\times 5}{0.00145}\\\\T = 1.5 N[/tex]
Option (A) is correct.
why is it wrong to leave our light on
Answer:
you will get huge electricity bills ............
Find the coefficient of kinetic friction between a 3.80-kg block and the horizontal surface on which it rests if an 87.0-N/m spring must be stretched by 6.50 cm to pull it with constant speed. Assume that the spring pulls in the horizontal direction.
Answer:
μ = 0.15
Explanation:
Let's start by using Hooke's law to find the force applied to the block
F = k x
F = 87.0 0.065
F = 5.655 N
Now we use the translational equilibrium relation since the block has no acceleration
∑ F = 0
F -fr = 0
F = fr
the expression for the friction force is
fr = μ N
if we write Newton's second law for the y-axis
N -W = 0
N = W = mg
we substitute
F = μ mg
μ = F / mg
μ = [tex]\frac{5.655}{3.8 \ 9.8}[/tex]
μ = 0.15
Two cars are moving. The first car has twice the mass of the second car but only half as much kinetic energy. When both cars increase their speed by 2.76 m/s, they then have the same kinetic energy. Calculate the original speeds of the two cars.
Let m be the mass of the second car, so the first car's mass is 2m.
Let K be the kinetic energy of the second car, so the first car's kinetic energy would be K/2.
Let u and v be the speeds of the first car and the second car, respectively. At the start,
• the first car has kinetic energy
K/2 = 1/2 (2m) u ² = mu ² ==> K = 2mu ²
• the second car starts with kinetic energy
K = 1/2 mv ²
It follows that
2mu ² = 1/2 mv ²
==> 4u ² = v ²
When their speeds are both increased by 2.76 m/s,
• the first car now has kinetic energy
1/2 (2m) (u + 2.76 m/s)² = m (u + 2.76 m/s)²
• the second car now has kinetic energy
1/2 m (v + 2.76 m/s)²
These two kinetic energies are equal, so
m (u + 2.76 m/s)² = 1/2 m (v + 2.76 m/s)²
==> 2 (u + 2.76 m/s)² = (v + 2.76 m/s)²
Solving the equations in bold gives u ≈ 1.95 m/s and v ≈ 3.90 m/s.
A particle of mass 1.2 mg is projected vertically upward from the ground with a velocity of 1.62 x 10 cm/h. Use the above information to answer the following four questions: 7. The kinetic energy of the particle at time t = 0 s is A. 1.215 x 10-3 J B. 2.430 J C. 1215 J D. 9.72 x 106 J E. OJ (2)
Answer:
K = 0 J
Explanation:
Given that,
The mass of the particle, m = 1.2 mg
The speed of the particle, [tex]v=1.62\times 10\ cm/h[/tex]
We need to find the kinetic energy of the particle at time t = 0 s.
At t = 0 s, the particle is at rest, v = 0
So,
[tex]K=\dfrac{1}{2}mv^2[/tex]
If v = 0,
[tex]K=0\ J[/tex]
So, the kinetic energy of the particle at time t = 0 s is 0 J.
A piston-cylinder device contains 5 kg of refrigerant-134a at 0.7 MPa and 60°C. The refrigerant is now cooled at constant pressure until it exists as a liquid at 24°C. If the surroundings are at 100 kPa and-24°C, determine:
(a) the exergy of the refrigerant at the initial and the final states and
(b) the exergy destroyed during this process.
Answer:
Yes sure, keep it going, and never give up because your dreams are so important
A) The exergy of the refrigerant at the initial and final states are :
Initial state = - 135.5285 kJ Final state = -51.96 kJB) The exergy destroyed during this process is : - 1048.4397 kJ
Given data :
Mass ( M ) = 5 kg
P1 = 0.7 Mpa = P2
T1 = 60°C = 333 k
To = 24°C = 297 k
P2 = 100 kPa
A) Determine the exergy at initial and final states
At initial state :
U = 274.01 kJ/Kg , V = 0.034875 m³/kg , S = 1.0256 KJ/kg.k
exergy ( Ф ) at initial state = M ( U + P₂V - T₀S )
= 5 ( 274.01 + 100* 10³ * 0.034875 - 297 * 1.0256)
≈ - 135.5285 kJ
At final state :
U = 84.44 kJ / kg , V = 0.0008261 m³/kg, S = 0.31958 kJ/kg.k
exergy ( ( Ф ) at final state = M ( U + P₂V - T₀S )
= -51.96 kJ
B) Determine the exergy destroyed
exergy destroyed = To * M ( S2 - S1 )
= 297 * 5 ( 0.31958 - 1.0256 )
= - 1048.4397 KJ
Hence we can conclude that A) The exergy of the refrigerant at the initial and final states are : Initial state = - 135.5285 kJ, Final state = -51.96 kJ and The exergy destroyed during this process is : - 1048.4397 kJ
Learn more about exergy : https://brainly.com/question/25534266
A nerve impulse travels along a myelinated neuron at 90.1 m/s.
What is this speed in mi/h?
Answer:
201.5537 mph
Explanation:
Given the following data;
Speed = 90.1 m/s
Speed can be defined as distance covered per unit time. Speed is a scalar quantity and as such it has magnitude but no direction.
Mathematically, speed is given by the formula;
Speed = distance/time
To convert this value into miles per hour;
Conversion;
1 meter = 0.000621 mile
90.1 meters = 90.1 * 0.000621 = 0.05595 miles
1 metre per second = 2.237 miles per hour
90.1 meters per seconds = 90.1 * 2.237 = 201.5537 miles per hour
90.1 m/s = 201.5537 mph
a student weighs 1200N they are standing in an elevator that is moving downwards at a constant speed of
Answer:
Elevator That Is Moving Downwards At A Constant Speed Of 4.9 M/S. What Is The Magnitude Of The Net Force Acing On The Student?
This problem has been solved!
This problem has been solved!See the answer
This problem has been solved!See the answerA student weighs 1200N. They are standing in an elevator that is moving downwards at a constant speed of 4.9 m/s. What is the magnitude of the net force acing on the student?
Explanation:
use this R= m(g-a), where R = reaction = weight, m= mass, a= acceleration and g= acceleration due to gravity
Question 3 of 10
Which statement describes the law of conservation of energy?
A. Air resistance has no effect on the energy of a system.
B. Energy cannot be created or destroyed.
C. The total energy in a system can only increase.
D. Energy cannot change forms.
هما
SUBMIT
Answer:
B . energy cannot be created or destroyed
If I could lift up to ten tons and I threw a ball the size of an orange but weighed a ton, to the ground, how big of an impact would it make? And could you also show me the equation to solve similar problems myself. Thank you.
Answer:
The impact force is 98000 N.
Explanation:
mass = 10 tons
The impact force is the weight of the object.
Weight =mass x gravity
W = 10 x 1000 x 9.8
W = 98000 N
The impact force is 98000 N.
A ball has a mass of 4.65 kg and approximates a ping pong ball of mass 0.060 kg that is at rest by striking it in an elastic collision. The initial velocity of the bowling ball is 5.00 m / s, determine the final velocities of both masses after the collision. Use equations 9.21 and 9.22 from the textbook. The book is on WebAssign.
Answer:
the final velocity of the ball is 4.87 m/s
the final velocity of the ping ball is 9.87 m/s
Explanation:
Given;
mass of the ball, m₁ = 4.65 kg
mass of the ping ball, m₂ = 0.06 kg
initial velocity of the ping ball, u₂ = 0
initial velocity of the ball, u₁ = 5 m/s
let the final velocity of the ball = v₁
let the final velocity of the ping ball, = v₂
Apply the principle of conservation of linear momentum for elastic collision;
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
4.65(5) + 0.06(0) = 4.65v₁ + 0.06v₂
23.25 + 0 = 4.65v₁ + 0.06v₂
23.25 = 4.65v₁ + 0.06v₂ ------ (1)
Apply one-directional velocity equation;
u₁ + v₁ = u₂ + v₂
5 + v₁ = 0 + v₂
5 + v₁ = v₂
v₁ = v₂ - 5 -------- (2)
substitute equation (2) into (1)
23.25 = 4.65(v₂ - 5) + 0.06v₂
23.25 = 4.65v₂ - 23.25 + 0.06v₂
46.5 = 4.71 v₂
v₂ = 46.5/4.71
v₂ = 9.87 m/s
v₁ = v₂ - 5
v₁ = 9.87 - 5
v₁ = 4.87 m/s
A simple pendulum takes 2.00 s to make one compete swing. If we now triple the length, how long will it take for one complete swing
Answer:
3.464 seconds.
Explanation:
We know that we can write the period (the time for a complete swing) of a pendulum as:
[tex]T = 2*\pi*\sqrt{\frac{L}{g} }[/tex]
Where:
[tex]\pi = 3.14[/tex]
L is the length of the pendulum
g is the gravitational acceleration:
g = 9.8m/s^2
We know that the original period is of 2.00 s, then:
T = 2.00s
We can solve that for L, the original length:
[tex]2.00s = 2*3.14*\sqrt{\frac{L}{9.8m/s^2} }\\\\\frac{2s}{2*3.14} = \sqrt{\frac{L}{9.8m/s^2}}\\\\(\frac{2s}{2*3.14})^2*9.8m/s^2 = L = 0.994m[/tex]
So if we triple the length of the pendulum, we will have:
L' = 3*0.994m = 2.982m
The new period will be:
[tex]T = 2*3.14*\sqrt{\frac{2.982m}{9.8 m/s^2} } = 3.464s[/tex]
The new period will be 3.464 seconds.
A magnetic field of 0.080 T is in the y-direction. The velocity of wire segment S has a magnitude of 78 m/s and components of 18 m/s in the x-direction, 24 m/s in the y-direction, and 72 m/s in the z-direction. The segment has length 0.50 m and is parallel to the z-axis as it moves.
Required:
a. Find the motional emf induced between the ends of the segment.
b. What would the motional emf be if the wire segment was parallel to the y-axis?
Answer:
Explanation:
From the information given:
The motional emf can be computed by using the formula:
[tex]E = L^{\to}*(V^\to*\beta^{\to})[/tex]
[tex]E = L^{\to}*((x+y+z)*\beta^{\to})[/tex]
[tex]E = 0.50*((18\hat i+24 \hat j +72 \hat k )*0.0800)[/tex]
[tex]E = 0.50*((18*0.800)\hat k +0j+(72*0.080) \hat -i ))[/tex]
[tex]E = 0.50*((18*0.800)[/tex]
E = 0.72 volts
According to the question, suppose the wire segment was parallel, there will no be any emf induced since the magnetic field is present along the y-axis.
As such, for any motional emf should be induced, the magnetic field, length, and velocity are required to be perpendicular to one another .
Then the motional emf will be:
[tex]E = 0.50 \hat j *((18*0.800)\hat k -(72*0.080) \hat i ))[/tex]
E = 0 (zero)
two point charges with charge q are initially separated by a distance d. if you double the charge on both charges, how far should the charges be separated for the potential energy between them to remain the same
Answer:
r ’= 4 r
Explanation:
Electric potential energy is
U = [tex]k \frac{q_1q_2}{r_{12}}[/tex]k q1q2 / r12
in this exercise
q₁ = q₂ = q
U = k q² / r
for when the charge change
U ’= k q’² / r’
indicate that
q ’= 2q
U ’= U
we substitute
U = k (2q) ² / r ’
U = 4 k q² / r ’
we substitute
[tex]k \ \frac{q^2}{r} = 4 k \ \frac{q^2}{r'}[/tex]k q² / r = 4 k q² / r ’
r ’= 4 r
Is the actual height the puck reached greater or less than your prediction? Offer a possible reason why this might be.
Answer:
Answer to the following question is as follows;
Explanation:
The puck's real altitude is lower than ones projection. That's because the mechanism may not be completely frictionless. Electricity is nevertheless wasted owing to particle interactions such as friction, which might explain why the present the results is lower than predicted.
Riley, a student, notices that the protractor tool does not measure the angle just as the ball leaves the surface. She sees that the ball must travel some distance before it crosses the protractor, so the direction of travel may have changed as the ball moves upwards. She says that this is the cause of the discrepancy between her predicted angle and the measured angle. Does this reasoning explain the discrepancy between your predicted angle and your measured angle. Use evidence to support your claim.
Answer:
Riley's reasoning is correct
Explanation:
Her reasoning is correct because as the ball moves upwards, the acceleration due to gravity would be vertical and in downward position. Therefore at all points as the ball moves, the velocity of the ball is going to change in magnitude as well as in direction. given that the direction keeps changing at certain points, the angle made by the initial velocity just as the ball left the surface would also have to continuously change.
If Riley has to wait for this ball to move some inches before she uses the protractor to measure the angle, the angle of travel would have to change.
Therefore there is going to be discrepancies between the measured angle and the predicted angle. The predicted is the angle of velocity with the horizontal just as this ball moves from the surface.
A horizontal, mass spring system undergoes simple harmonic motion. which of the following statements is correct reguarding the mass in the system when it is located at its maximum distance from the equilibrium position?
a. The acceleration of the mass is zero.
b. The potential energy of the spring attached to the mass is at a minimum.
c. The total mechanical energy of the mass is zero.
d. The kinetic energy of the mass is at a maximum.
e. The speed of the mass is zero.
Answer:
Option (e)
Explanation:
A body executing SHM moves to and fro or back and forth about its mean position.
When the particle is at mean position, its velocity is maximum and when it is at extreme position its velocity is zero.
So, when it is at maximum distance:
a.
The acceleration is maximum.
b.
The potential energy is maximum.
c.
The total mechanical energy is non zero.
d.
The kinetic energy is zero.
e. The speed is zero. Correct
Which one of the following is not an example of convection? An eagle soars on an updraft of wind. A person gets a suntan on a beach. An electric heater warms a room. Smoke rises above a fire. Spaghetti is cooked in water.
Answer: The statement that is not an example of convection is (A person gets a suntan on a beach).
Explanation:
There are different modes of heat energy transfer which includes:
--> conduction
--> Radiation and
--> Convection
CONVECTION is a process by which heat energy is transferred in a fluid or air by the actual movement of the heated molecules. The cooler portion of the air surrounding a warmer part exerts a buoyant force on it. As the warmer part of the air moves, it is replaced by cooler air that is subsequently warmed.
Convection in gases is very common and gas expands more than liquid when subjected to high temperature.
--> it is used in bringing about the circulation of fresh air in the room in a process known as ventilation.Here, cool air is constantly being replaced with denser air ( warm air).
-->An electric heater warms a room and Smoke rises above a fire are typical example of convection in gases.
-->Spaghetti is cooked in water: As the water close to the burner warms, it rises to the top and boils. At the same time, cooler water on top moves downward to replace the rising hot water.
--> also the eagle uses convection current to stay afloat in the sky without flapping its wings to conserve energy.
But the option (A person gets a suntan on a beach) is an example of heat transfer through radiation. This is because the sun emits it's rays from the sky down to earth without any material medium unlike others. Therefore, this option is the ODD one out.
1. An excited lithium atom emits a red light with wavelength a = 671nm. What is the corresponding photon energy? hc (6.63 x 10-34).S)(3.0 x 108m/s)
Answer:
E = 2,964 10⁻¹⁹ J
Explanation:
The energy of the photons is given by the Planck relation
E = h f
the speed of light is related to wavelength and frequency
c = λ f
we substitute
E = h c /λ
let's reduce the magnitude to the SI system
λ = 671 nm = 671 10⁻⁹ m
let's calculate
E = 6.63 10⁻³⁴ 3 10⁸ /671 10⁻⁹
E = 2,964 10⁻¹⁹ J
which is the correct formula for calculating the age of meteor right if using half life
Answer:
n × t_1/2
Exmplanation:
The age of meteorite is calculated by multiplying it's quantity n with the half life . This means that the formula is for age of this meteorite is;
Age of meteorite= n × t_1/2
where;
n = quantity of the meteorite
t_/2 = half life of the meteorite
Thus:
The correct formula is; n × t_1/2
Many collisions, like the collision of a bat with a baseball, appear to be instantaneous. Most people also would not imagine the bat and ball as bending or being compressed during the collision. Consider the following possibilities: The collision is instantaneous. The collision takes a finite amount of time, during which the ball and bat retain their shapes and remain in contact. The collision takes a finite amount of time, during which the ball and bat are bending or being compressed. How can two of these be ruled out based on energy or momentum considerations?
The collision is instantaneous.
The collision takes a finite amount of time, during which the ball and bat retain their shapes and remain in contact.
The collision takes a finite amount of time, during which the ball and bat are bending or being compressed.
How can two of these be ruled out based on energy or momentum considerations?
Answer:
The collision takes a finite amount of time, during which the ball and bat are bending or being compressed
Explanation:
These two conditions can be ruled out on the fact that :The collision takes a finite amount of time, during which the ball and bat are bending or being compressed
The rule of energy is been broken here because during the collision of objects energy and momentum is conserved. i.e. the change in shape of the ball when hit by the bat should not be noticed because the compression and returning to normal shape happens instantaneously
A child with a weight of 230 N swings on a playground swing attached to 2.20-m-long chains. What is the gravitational potential energy of the child-Earth system relative to the child's lowest position at the following times?
(a) when the chains are horizontal (in J)
(b) when the chains make an angle of 33.0° with respect to the vertical (in J)
(c) when the child is at his lowest position (in J)
Answer:
a) U = 506 J, b) U = 37.11 J, c) U = 0
Explanation:
The gravitational power energy is given by the expression
U = m g (y -y₀)
In general, a reference system is set that allows the expression to be simplified, in this case let's assume the reference system at the child's lowest point, therefore y₀ = 0
Let's use trigonometry to find the child's height
h = y = L - L cos θ
we substitute
U = m g L (1 - cos θ)
a) when the chain is horizontal θ = 90 and cos 90 = 0
U = mg L
weight and mass are related
W = mg
m = W / g
U = 230 2.20
U = 506 J
b) θ = 33.0º
cos 33 = 0.83867
U = 230 (1 - 0.83867)
U = 37.11 J
c) in this case θ = 0 cos 0 = 1
U = 0
As it pulls itself up to a branch, a chimpanzee accelerates upward at 2.4 m/s2 at the instant it exerts a 260-N force downward on the branch.Find the magnitude of the force the chimpanzee exerts on the Earth.
Answer:
[tex]F=208.83N[/tex]
Explanation:
From the question we are told that:
Acceleration [tex]a=2.4m/s^2[/tex]
Force of Branch [tex]F=260N[/tex]
Generally the Newton's equation second law for Force is mathematically given by
[tex]ma=F-mg[/tex]
[tex]m=\frac{260}{2.4+9.8}[/tex]
[tex]m=21.31kg[/tex]
Therefore
[tex]F=mg[/tex]
[tex]F=(21.31)(9.8)[/tex]
[tex]F=208.83N[/tex]
If one lawn mower causes an 80-dB sound level at a point nearby, four lawnmowers together would cause a sound level of ____________ at that point. a.92 dB b.84 dB c.86 dB d.none of the above
Answer:
The intensity of 4 lawn movers is 86 dB.
Explanation:
Intensity of one lawnmower = 80 dB
Let the intensity is I.
Use the formula of intensity
[tex]dB = 10 log\left ( \frac{I}{Io} \right )\\\\80=10log\left ( \frac{I}{Io} \right )\\\\10^8 = \frac{I}{10^{-12}}\\\\I = 10^{-4} W/m^2[/tex]
Now the intensity of 4 lawn movers is
[tex]dB = 10 log\left ( \frac{4I}{Io} \right )\\\\dB=10log\left ( \frac{4\times10^{-4}}{10^{-12}} \right )\\\\dB = 86 dB\\[/tex]
A 0.033-kg bullet is fired vertically at 222 m/s into a 0.15-kg baseball that is initially at rest. How high does the combined bullet and baseball rise after the collision, assuming the bullet embeds itself in the ball
Answer:
The maximum height risen by the bullet-baseball system after the collision is 81.76 m.
Explanation:
Given;
mass of the bullet, m₁ = 0.033 kg
mass of the baseball, m₂ = 0.15 kg
initial velocity of the bullet, u₁ = 222 m/s
initial velocity of the baseball, u₂ = 0
let the common final velocity of the system after collision = v
Apply the principle of conservation of linear momentum to determine the common final velocity.
m₁u₁ + m₂u₂ = v(m₁ + m₂)
0.033 x 222 + 0.15 x 0 = v(0.033 + 0.15)
7.326 = v(0.183)
v = 7.326 / 0.183
v = 40.03 m/s
Let the height risen by the system after collision = h
Initial velocity of the system after collision = Vi = 40.03 m/s
At maximum height, the final velocity, Vf = 0
acceleration due to gravity for upward motion, g = -9.8 m/s²
[tex]v_f^2 = v_i^2 +2gh\\\\0 = 40.03^2 - (2\times 9.8)h\\\\19.6h = 1602.4\\\\h = \frac{1602.4}{19.6} \\\\h = 81.76 \ m[/tex]
Therefore, the maximum height risen by the bullet-baseball system after the collision is 81.76 m.
