Answer:
a) v₀ = 44.27 m / s, b) stone A v = 44.276 m / s, stone B v = 0.006 m / s
Explanation:
a) This is a kinematics exercise, let's start by finding the time it takes for stone A to reach half the height of the building y = 100 m
y = y₀ + v₀ t - ½ gt²
as the stone is released its initial velocity is zero
y- y₀ = 0 - ½ g t²
t = [tex]\sqrt{ -2(y-y_o)/g}[/tex]
t = [tex]\sqrt{ -2(100-200)/9.8}[/tex]
t = 4.518 s
now we can find the initial velocity of stone B to reach this height at the same time
y = y₀ + v₀ t - ½ g t²
stone B leaves the floor so its initial height is zero
100 = 0 + v₀ 4.518 - ½ 9.8 4.518²
100 = 4.518 v₀ - 100.02
v₀ = [tex]\frac{100-100.02}{4.518}[/tex]
v₀ = 44.27 m / s
b) the speed of the two stones at the meeting point
stone A
v = v₀ - gt
v = 0 - 9.8 4.518
v = 44.276 m / s
stone B
v = v₀ -g t
v = 44.27 - 9.8 4.518
v = 0.006 m / s
Which phase of matter makes up stars?
O liquid
O gas
O plasma
Answer:
The answer to this question is plasma
Answer:
Plasma
Explanation:
Where would the normal force exerted on the rover when it rests on the surface of the planet be greater
Answer:
Normal force exerted on the rover would be greater at a point on the surface of the planet where the weight of the rover is experienced to be greater.
Explanation:
Since weight is a vector quantity, it can vary with position. Weight is the amount of force the planet exerts on the rover centered towards the planet.
Such a force is the result of gravitational pull and is quantified as:
[tex]F=G\times \frac{M.m}{R^2}[/tex]
and [tex]M=\rho\times \frac{4\pi.r^3}{3}[/tex]
where:
R = distance between the center of mass of the two bodies (here planet & rover)
G = universal gravitational constant
M = mass of the planet
m = mass of the rover
This gravitational pull varies from place to place on the planet because the planet may not be perfectly spherical so the distance R varies from place to place and also the density of the planet may not be uniform hence there is variation in weight.
Weight is basically a force that a mass on the surface of the planet experiences.
According to Newton's third law the there is an equal and opposite reaction force on the body (here rover) which is the normal force.
Complete the following statement: The electromotive force is:______.a. the force that accelerates electrons through a wire when a battery is connected to it.b. the maximum potential difference between the terminals of a battery. c. the force that accelerates protons through a wire when a battery is connected to it.d. the maximum capacitance between the terminals of a battery.e. the potential difference between the terminals of a battery when the battery is not in use.
Answer:
The electromotive force is the maximum potential difference between the terminals of a battery.
The electromotive force is the maximum potential difference between the terminals of a battery. The correct option is b.
What is electromotive force?The electromotive force also called as EMF, is the force which causes current to flow from the positive to negative terminal of the battery.
The electromotive force is the maximum potential difference between the terminals of a battery.
Thus, the correct option is b.
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93 cm3 liquid has a mass of 77 g. When calculating its density what is the appropriate number of significant figures
Answer:
828 kg/m³ or 0.828 g/cm³
Explanation:
Applying,
D = m/V............. Equation 1
Where D = density of the liquid, m = mass of the liquid, V = volume of the liquid.
From the question,
Given: m = 77 g , V = 93 cm³
Substitute these values into equation 1
D = 77/93
D = 0.828 g/cm³
Converting to kg/m³
D = 828 kg/m³
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
A uniform steel rod of length 0.9 m and mass 3.8 kg has two point masses of 2.3 kg each at the two ends. Calculate the moment of inertia of the system about an axis perpendicular to the rod, and passing through its center.
Answer: [tex]2.4705\ kg.m^2[/tex]
Explanation:
Given
length of the rod is L=0.9 m
Mass of the rod m=3.8 kg
Point masses has mass of m=2.3 kg
Moment of Inertia of the rod about the center is
[tex]\Rightarrow I_o=\dfrac{1}{12}ML^2[/tex]
Moment of inertia of combined system is the sum of rod and two point masses.
[tex]\Rightarrow I=I_o+2mr^2[/tex]
[tex]\Rightarrow I=\dfrac{1}{12}3.8\times 0.9^2+2\times 2.3\times \left(\dfrac{0.9}{2}\right)^2\\\\\Rightarrow I=1.539+0.9315\\\Rightarrow I=2.4705\ kg-m^2[/tex]
A 2090-kg test rocket is launched vertically from the launch pad. Its fuel (of negligible mass) provides a thrust force so that its vertical velocity as a function of time is given by v(t) =At+Bt^2 , where A and B are constants and time is measured from the instant the fuel is ignited. The rocket has an upward acceleration of 1.50m/s 2 at the instant of ignition and, 1.00 s later, an upward velocity of 2.00 m/s. (a) Determine A and B , including their SI units. (b) At 4.00 s after fuel ignition, what is the acceleration of the rocket, and (c) what thrust force does the burning fuel exert on it, assuming no air resistance? Express the thrust in newtons and as a multiple of the rocket’s weight. (d) What was the initial thrust due to the fuel?
Answer:
a) A = 1.50 m / s², B = 1.33 m/s³, b) a = 12.1667 m / s²,
c) I = M (1.5 t + 1.333 t²) , d) ΔI = M 2.833 N
Explanation:
In this exercise give the expression for the speed of the rocket
v (t) = A t + B t²
and the initial conditions
a = 1.50 m / s² for t = 0 s
v = 2.00 m / s for t = 1.00 s
a) it is asked to determine the constants.
Let's look for acceleration with its definition
a = [tex]\frac{dv}{dt}[/tex]
a = A + 2B t
we apply the first condition t = 0 s
a = A
A = 1.50 m / s²
we apply the second condition t = 1.00 s
v = 1.5 1 + B 1²
2 = 1.5 + B
B = 2 / 1.5
B = 1.33 m/s³
the equation remains
v = 1.50 t + 1.333 t²
b) the acceleration for t = 4.00 s
a = 1.50 + 1.333 2t
a = 1.50 + 2.666 4
a = 12.1667 m / s²
c) The thrust
I = ∫ F dt = p_f - p₀
Newton's second law
F = M a
F = M (1.5 + 2 1.333 t) dt
we replace and integrate
I = M ∫ (1.5 + 2.666 t) dt
I = 1.5 t + 2.666 t²/2
I = M (1.5 t + 1.333 t²) + cte
in general the initial rockets with velocity v = 0 for t = 0, where we can calculate the constant
cte = 0
I = M (1.5 t + 1.333 t²)
d) the initial push
For this we must assume some small time interval, for example between
t = 0 s and t = 1 s
ΔI = I_f - I₀
ΔI = M (1.5 1 + 1.333 1²)
ΔI = M 2.833 N
If you change the motor in your vehicle you need to notify the DMV within ____,
days of this change.
-20
-25
-10
-15
when you change your motor on your vehicle you need to notify the DMV within 10 days
If you change the motor in your vehicle you need to notify the DMV within 10 days of this change.
An engine or motorAn engine or motor is a machine designed to convert one or more forms of energy into mechanical energy.
Available energy sources include potential energy (e.g. energy of the Earth's gravitational field as exploited in hydroelectric power generation), heat energy (e.g. geothermal), chemical energy, electric potential, and nuclear energy (from nuclear fission or nuclear fusion). Many of these processes generate heat as an intermediate energy form, so heat engines have special importance. Some natural processes, such as atmospheric convection cells convert environmental heat into motion (e.g. in the form of rising air currents). Mechanical energy is of particular importance in transportation but also plays a role in many industrial processes such as cutting, grinding, crushing, and mixing.
Mechanical heat engines convert heat into work via various thermodynamic processes. The internal combustion engine is perhaps the most common example of a mechanical heat engine, in which heat from the combustion of fuel causes rapid pressurization of the gaseous combustion products in the combustion chamber, causing them to expand and drive a piston, which turns a crankshaft. Unlike internal combustion engines, a reaction engine (such as a jet engine) produces thrust by expelling reaction mass, by Newton's third law of motion.
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A boy throws a ball upward with a velocity of 4.50 m/s at 60.0o. What is the maximum height reached by the ball?
Answer:
3.1m
Explanation:
Since we only care about the y direction we only need to find vy. Once u draw your vector you will realize that vy= 4.5sin60=3.897m/s.
use vf²=v²+2a(y)
At the maximum height the velocity is 0 and since the object is in freefall, a=-g
Plug in all values
0=15.1875-2*9.8(y)
solve for y
-15.1875*2/-9.8=y
y=3.1m
Answer:
0.774m
Explanation:
The formula for maximum height is given by:
hmax = ∨₀² ₓ Sin (α)² / 2 × g
where;
∨₀ = initial velocity
Sin (α) = angle of launch
g = gravitational acceleration which is equal to 9.8m/s²
Plugging in our values, we will have:
hmax = (4.50m/s)² × (Sin 60.0)² / 2 × 9.8m/s²
hmax= 20.25m/s × 0.75 / 19.8m/s²
hmax = 15.1875 / 19.8
hmax = 0.774m
A popular car stereo has four speakers, each rated at 60 W. In answering the following questions, assume that the speakers produce sound at their maximum power.
A) Find the intensity I of the sound waves produced by one 60-Wspeaker at a distance of 1.0 m.
B) Find the intensity I of the sound waves produced by one 60-Wspeaker at a distance of 1.5 m.
C) Find the intensity I of the sound waves produced by four 60-Wspeakers as heard by the driver. Assume that the driver is located 1.0 m from each of the two front speakers and 1.5 m from each of the two rear speakers.
D)The threshold of hearing is defined as the minimum discernible intensity of the sound. It is approximately 10^(-12) W/m2. Find the distance dfrom the car at which the sound from the stereo can still be discerned. Assume that the windows are rolled down and that each speaker actually produces 0.06 W of sound, as suggested in the last follow-up comment.
Answer:
Explanation:
Intensity of sound = sound energy emitted by source / 4 π d² , where d is distance of the source .
A )
Intensity of sound at 1 m distance = 60 /4 π d²
d = 1 m
Intensity of sound at 1 m distance = 60 /(4 π 1²)
= 4.78 W m⁻² s⁻¹
B )
Intensity of sound at 1.5 m distance = 60 /4 π d²
d = 1.5 m
Intensity of sound at 1 m distance = 60 /(4 π 1.5²)
= 2.12 W m⁻² s⁻¹
C )
Intensity of sound due to 4 speakers at 1.5 m distance = 4 x 60 /4 π d²
d = 1.5 m
= 4 x 60 /(4 π 1.5²)
= 8.48 W m⁻² s⁻¹
D )
Intensity of sound due to .06 W speaker must be 10⁻¹² W s ⁻² . Let the distance be d .
.06 /4 π d² = 10⁻¹²
d² = .06 /4 π 10⁻¹²
d = 6.9 x 10⁴ m .
TIME REMAINING
45:43
What are possible units for impulse? Check all that apply.
kg • m
kg • meters per second
N • s
N • m
StartFraction Newtons times meters per second EndFraction
Answer:
n.m maby
Explanation:
i think or its kg m/s
Answer:
answer (B) & (C)
Explanation:
kg • /N • s
What is the temperature of a system in thermal equilibrium with another system made up of water and steam at one atmosphere of pressure
Full Question:
What is the temperature of a system in thermal equilibrium with another system made up of water and steam at one atmosphere of pressure?
A) 0°F
B) 273 K
C) 0 K
D) 100°C
E) 273°C
Answer:
The correction Option is D) 100°C
Explanation:
The temperature above is referred to as the critical point.
it is the highest temperature and pressure at which water (which has three phases - liquid, solid, and gas) can exist in vapor/liquid equilibrium. If the temperature goes higher than 100 degrees celsius, it cannot remain is liquid form regardless of what the pressure is at that point.
There is also a condition under which water can exist in its three forms: that is
- Ice (solid)
- Liquid (fluid)
- Gas (vapor)
That state is called triple point. The conditions necessary for that to occur are:
273.1600 K (0.0100 °C; 32.0180 °F) as temperature and611.657 pascals (6.11657 mbar; 0.00603659 atm) as pressureCheers
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In an exciting game, a baseball player manages to safely slide into second base. The mass of the baseball player is 88.9 kg and the coefficient of kinetic friction between the ground and the player is 0.53. (a) Find the magnitude of the frictional force in newtons. N (b) It takes the player 1.7 s to come to rest. What was his initial velocity (in m/s)
Answer:
Look at explanation
Explanation:
a) Kinetic Friction= μmg
μmg=0.53*88.9*9.8=461.75N
b) -461.75N=ma
a= -5.19m/s^2
v=v0+at
5.19*1.7=v0
v0=8.81m/s^2
(a) The magnitude of the frictional force will be 461.75N
(b)The initial velocity will be 8.81 m/s.
What is kinetic friction?A force that acts among sliding parts is referred to as kinetic friction. A body moving on the surface is subjected to a force that opposes its progressive motion
The size of the force will be determined by the kinetic friction coefficient between the two materials.
The given data in the problem is;
μ is the coefficient of kinetic friction= 0.53.
m is the mass = 88.9 kg
g is the acceleration due to gravity= 9.81 m/s²
v is the speed =?
The formula for friction force is;
[tex]\rm F= \mu R \\\\ R=mg \\\\ F= \mu mg \\\\\ F=0.53 \times 88.9 \times 9.81 \\\\ F= 461.75 \ N[/tex]
Mechanical force is found as;
F=ma
-461.75=(88.9)a
(-ve shows the -ve work done)
a=-5.19 m/s
From the Newton's first equation of motion;
v=u+at
0=u+at
u=-at
u=(- (-5.19)(1.7)
u=8.81 m/s²
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2.
Select the correct answer.
Erica is working in the lab. She wants to remove the fine dust particles suspended in a sample of oil. Which method is she most likely to use?
Answer:
Reverse Osmosis
Explanation:
Reverse osmosis is a type of filtration that involves passing a solvent through a semipermeable membrane in the opposite direction that natural osmosis does. Separation is always enforced through the use of pressure in this process. Ions, fine dust particles, molecules, and larger particles are typically removed from solvents using this method. The technique is particularly popular in the treatment and purification of water.
Answer:
filtration is used to separate
Calculate the volume of 10g of helium ( M= 4kg/kmol) at 25C and 600 mmHg
Answer:
T=273+25=298 K
n= m/M = 10/ 4 = 2.5
R=0.08206 L.atm /mol/k
760mmHg = 1 atm therefore
600mmHg = X atm
760 X = 600mmHg
X = 600/760 = 0.789 atm
P = 0.789 atm
V= ?
PV= nRT
0.789 V = 2.5 × 0.08206 × 298
V= 2.5 × 0.08206 ×298 / 0.789
V= 77.48 L
I hope I helped you ^_^
2. g A spring extends by 20 cm when a force of 2 N is applied. What is the value of the spring constant in N/m
10N/m
Explanation:
f=kx
k=f/x
k=20N/0.2m
k=10N/m
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
Find a parametric representation for the surface. The plane through the origin that contains the vectors i - j and j - k
Answer:
parametric representation: x = u, y = v - u , z = - v
Explanation:
Given vectors :
i - j , j - k
represent the vector equation of the plane as:
r ( u, v ) = r₀ + ua + vb
where: r₀ = position vector
u and v = real numbers
a and b = nonparallel vectors
expressing the nonparallel vectors as :
a = i -j , b = j - k , r = ( x,y,z ) and r₀ = ( x₀, y₀, z₀ )
hence we can express vector equation of the plane as
r(u,v) = ( x₀ + u, y₀ - u + v, z₀ - v )
Finally the parametric representation of the surface through (0,0,0) i.e. origin = 0
( x, y , z ) = ( x₀ + u, y₀ - u + v, z₀ - v )
x = 0 + u ,
y = 0 - u + v
z = 0 - v
∴ parametric representation: x = u, y = v - u , z = - v
A wheel 30 cm in diameter accelerates uniformly from 245 rpm to 380 rpm in 6.1 s . Part A How far will a point on the edge of the wheel have traveled in this time
Answer:
A point on the edge of the wheel will travel 199.563 radians at the given time.
Explanation:
Given;
initial angular velocity of the wheel; [tex]\omega _i = 245 \ rev/\min = 245\ \frac{rev}{\min} \times \frac{2\pi}{1\ rev} \times \frac{1 \ \min}{60 \ s} = 25.66 \ rad/s[/tex]
final angular velocity of the wheel;
[tex]\omega _f = 380 \ rev/\min = 380 \ \frac{rev}{\min} \times \frac{2\pi}{1\ rev} \times \frac{1 \ \min}{60 \ s} = 39.80 \ rad/s[/tex]
radius of the wheel, d/2 = (30 cm ) / 2 = 15 cm = 0.15 m
time of motion, t = 6.1 s
The angular distance traveled by the edge of the wheel is calculated as;
[tex]\theta = (\frac{\omega_f + \omega_i}{2} )t\\\\\theta = (\frac{39.8 + 25.66}{2} )\times 6.1\\\\\theta = 199.653 \ radian[/tex]
Therefore, a point on the edge of the wheel will travel 199.563 radians at the given time.
A 36.0 kg child slides down a long slide in a playground. She starts from rest at a height h1 of 24.00 m. When she is partway down the slide, at a height h2 of 11.00 m, she is moving at a speed of 7.80 m/s.
Calculate the mechanical energy lost due to friction (as heat, etc.).
Answer:
E = 3495.96 J
Explanation:
From the law of conservation of energy:
Total Mechanical Energy at h1 = Total Mechanical Energy at h2
Kinetic energy at h1 + potential energy at h1 = Kinetic energy at h2 + potential energy at h2 + Mechanical Energy Lost due to Friction
[tex]K.E_{h1}+P.E_{h1} = K.E_{h2}+P.E_{h2} + E\\\\\frac{1}{2}mv_1^2\ J + mgh_1 = \frac{1}{2}mv_2^2 + mgh_2 + E\\\\\frac{1}{2}(36\ kg)(0\ m/s)_1^2\ J + (36\ kg)(9.81\ m/s^2)(24\ m)_1 = \frac{1}{2}(36\ kg)(7.8\ m/s)_2^2 + (36\ kg)(9.81\ m/s^2)(11\ m)_2 + E\\\\0\ J + 8475.84\ J = 1095.12\ J + 3884.76\ J + E\\E = 8475.84\ J - 1095.12\ J - 3884.76\ J\\[/tex]
E = 3495.96 J
Stationary waves are
A) transverse waves
B) longitudinal waves
C) mechanical waves
Answer:
stationary waves are transverse waves
А pressure gauge with a measurement range of 0-10 bar has a quoted inaccuracy of £1.0% f.s. (+1% of full-scale reading). (a) What is the maximum measurement error expected for this instrument? (b) What is the likely measurement error expressed as a percentage of the or reading if this pressure gauge is measuring a pressure of 1 bar?
Answer:
I am not able to answer this question please don't mind...Explanation:
please marks me as brainliests...Help pleaseeeeeeeeeeeeee
Answer:
hello the answer is 47m/s
explain why sound wave travel faster in liquid than gas
Answer:
Because gas contains free molecules but not liquid.
Please mark as brainliast
Which of the following statements is correct about the magnitude of the static friction force between an object and a surface?
a. Static friction depends on the mass of the object.
b. Static friction depends on the shape of the object.
c. Static friction depends on what the object is made of but not what the surface is made of.
d. None of the above is correct.
Answer:
Static friction depends on the mass of the object.
Explanation:
Friction is the force between two surfaces in contact. The force of friction between two surfaces in contact depends on;
1) nature of the object and the surface(how rough or smooth the surfaces are)
2)surface area of the object and the surface
3) mass of the object
Since;
F=μmg
Where;
μ= coefficient of static friction
m= mass of the object
g= acceleration due to gravity
Hence, as the mass of the object increases, the magnitude of static friction force between an object and a surface increases and vice versa.
NEED HELP ASAP- Please show work
The angular position of an object is given by θ = 4t3 +10t −40 , where θ is in radians and t is in seconds what is:
(a) (5 points) The angular velocity at t = 2 s?
(b) (5 points) The angular acceleration at t = 2 s?
Answer:
Look at work
Explanation:
Θ= 4t^3+10t-40
a) In order to find ω, we need to find displacement so plug in t=2 to find Θ.
Θ= 4*8+20-40=12
use ω=Θ/t
Plug in values
ω=6 rad/s
b) In order to find α we use ω/t.
Plug in values
α=6/2= 3 rad/s^2
An AC power source has an rms voltage of 120 V and operates at a frequency of 60.0 Hz. If a purely inductive circuit is made from the power source and a 47.2 H inductor, determine the inductive reactance and the rms current through the inductor.
Answer:
The inductance is 17784.96 ohm and rms current is 4.77 mA.
Explanation:
Voltage, V = 120 V
frequency, f = 60 Hz
Inductance, L = 47.2 H
The rms voltage is
[tex]V_{rms}=\frac{V_o}{\sqrt 2}\\\\V_{rms}=\frac{120}{\sqrt 2}\\\\V_{rms} = 84.87 V[/tex]
The reactance is given by
[tex]X_L = 2\pi f L\\\\X_L = 2\times 3.14\times 60\times 47.2 \\\\X_L = 17784.96 ohm[/tex]
The rms current is
[tex]I_{rms} =\frac{V_{rms}}{X_L}\\\\I_{rms}=\frac{84.87}{17784.96}\\\\I_{rms} = 4.77\times 10^{-3} A = 4.77 mA[/tex]
A solid piece of clear transparent material has an index of refraction of 1.61. If you place it into a clear transparent solution and it seems to disappear, approximately what is the index of refraction of the solution
Answer:
1.61
Explanation:
According to Oxford dictionary, refractive index is, ''the ratio of the velocity of light in a vacuum to its velocity in a specified medium.''
If the clear transparent solid disappears when dipped into the liquid, it means that the index of refraction of the solid and liquid are equal.
Hence, when a transparent solid is immersed in a liquid having the same refractive index, there is no refraction at the boundary between the two media. As long as there is no refraction between the two media, the solid can not be seen because the solid and liquid will appear to the eye as one material.
The cavity within a copper [β = 51 × 10-6 (C°)-1] sphere has a volume of 1.180 × 10-3 m3. Into this cavity is placed 1.100 × 10-3 m3 of benzene [β = 1240 × 10-6 (C°)-1]. Both the copper and the benzene have the same temperature. By what amount ΔT should the temperature of the sphere and the benzene within it be increased, so that the liquid just begins to spill out?
Answer:
The answer is "[tex]60.74^{\circ}[/tex]".
Explanation:
Cavity and benzene should be extended in equal quantities.
[tex]\to 1.18 \times 10^{-3}\times (1+ \Delta T \times 0.000051) = 1.1\times 10^{-3} \times (1+ \Delta T \times 0.00124)\\\\\to (\frac{1.18}{1.1})\times (1+ \Delta T \times 0.000051) = 1+ \Delta T \times 0.00124\\\\ \to 1.072\times (1+ \Delta T \times 0.000051) = 1+ \Delta T \times 0.00124\\\\ \to 1.072+ \Delta T \times 0.000054672 = 1+ \Delta T \times 0.00124\\\\ \to 1.072+ \Delta T \times 0.000054672 - 1- \Delta T \times 0.00124=0\\\\[/tex]
[tex]\to 0.072+ \Delta T \times 0.000054672 - \Delta T \times 0.00124=0\\\\ \to 0.072+ \Delta T ( 0.000054672 -0.00124)=0\\\\ \to \Delta T ( 0.000054672 -0.00124)= -0.072\\\\ \to \Delta T = -\frac{0.072}{( 0.000054672 -0.00124)}\\\\ \to \Delta T = -\frac{0.072}{-0.001185328 }\\[/tex]
[tex]\to \Delta T = \frac{0.072}{0.001185328 }\\\\ \to \Delta T = 60.74^{\circ}\\[/tex]
A cannon and a supply of cannonballs are inside a sealed railroad car of length L, as in Fig. 7-33. The cannon fires to the right; the car recoils to the left. The cannonballs remain in the car after hitting the far wall. (a) After all the cannonballs have been fired, what is the greatest distance the car can have moved from its original position
Answer:
Initially let n cannonballs with a total mass of m be to the left of the center of mass at L /2 and the mass of the car at L/2
x1 = [-m / (m + M)] * L / 2 is the original position of the CM
x2 = (m (x + L/2) + M x) / (m + M) * L/2 final position of CM with all cannon balls to the right
[-m x - m L / 2 + m x - M x] / (M + m) * L/2
= - ( m L / 2 + M x) / (m + M) * L/2 = Xcm
Check the math, but maximum distance occurs when the cannonballs of mass m move from -L/2 to L/2 and the car of mass M moves from zero to -x
