Answer:
Rotational speed increases
Explanation:
The formula for rotational speed is:
Rotational speed = rotations / time.
When you approach the center the number of rotations increases, because the radius of the circle decreases, and this increases rotational speed.
If you walk from the edge to the center of a rotating carousel, the carousel will rotate faster. This is due to the conservation of angular momentum. As you move closer to the center, the carousel's moment of inertia decreases and to conserve total angular momentum, its rotational speed increases.
Explanation:Your position on the carousel affects its rotational speed. If you stand on the edge of a rotating carousel and then you walk toward the center, you change your rotational inertia, which according to the law of conservation of angular momentum, makes the carousel rotate faster.
This happens because when you move closer to the center, the rotational inertia or the resistance to the rotation of the carousel decreases and thus, the rotational speed or the speed at which the carousel is turning increases to conserve the total angular momentum.
The conservation of angular momentum is the principle which tells that the total angular momentum remains constant in a system without any external torques. As we see in the case of the carousel, when you are at the edge of the spinning carousel, there is a large radius, leading to a high moment of inertia.
But as you move towards the center, the radius decreases, which in turn decreases the moment of inertia. Therefore, to keep the angular momentum constant, the angular velocity, in this case, the speed of the carousel, must increase.
The same concept can be applied to figure skaters. When a figure skater pulls in her arms and spins, she reduces her rotational inertia and therefore spins faster to conserve angular momentum.
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A ball is thrown straight up. What are the velocity and acceleration of the ball at the highest point in its path?
a. v = 9.8 m/s up, a = 0.
b. v = 0, a = 9.8 m/s2 down.
c. v = 0, a = 9.8 m/s2 up.
d. v = 9.8 m/s down, a = 0.
e. v = 0, a = 0.
Answer:
b. v = 0, a = 9.8 m/s² down.
Explanation:
Hi there!
The acceleration of gravity is always directed to the ground (down) and, near the surface of the earth, has a constant value of 9.8 m/s². Since the answer "b" is the only option with an acceleration of 9.8 m/s² directed downwards, that would solve the exercise. But why is the velocity zero at the highest point?
Let´s take a look at the height function:
h(t) = h0 + v0 · t + 1/2 g · t²
Where
h0 = initial height
v0 = initial velocity
t = time
g = acceleration due to gravity
Notice that the function is a negative parabola if we consider downward as negative (in that case "g" would be negative). Then, the function has a maximum (the highest point) at the vertex of the parabola. At the maximum point, the slope of the tangent line to the function is zero, because the tangent line is horizontal at a maximum point. The slope of the tangent line to the function is the rate of change of height with respect to time, i.e, the velocity. Then, the velocity is zero at the maximum height.
Another way to see it (without calculus):
When the ball is going up, the velocity vector points up and the velocity is positive. After reaching the maximum height, the velocity vector points down and is negative (the ball starts to fall). At the maximum height, the velocity vector changed its direction from positive to negative, then at that point, the velocity vector has to be zero.
The buoyant force on a ball of weight W, dropped in a container of water is F. If the same ball is dropped into a container of liquid with half the density of water, the buoyant force on the ball will be
Answer:[tex]\frac{F}{2}[/tex]
Explanation:
Given
weight of Object is W
Buoyant Force [tex]F_b=F[/tex]
let the density of First Liquid is [tex]\rho [/tex]
volume of object [tex]=\frac{W}{\rho _0g}[/tex]
where [tex]\rho _0[/tex]=object density
thus buoyant Force in first Liquid is
[tex]F=\rho \times \frac{W}{\rho _0g}\times g=\frac{W\rho }{\rho _0}[/tex]
For Second Liquid
[tex]F'=\frac{\rho }{2}\times \frac{W}{\rho _0g}\times g=\frac{W\rho }{2\rho _0}[/tex]
[tex]F'=\frac{F}{2}[/tex]
A car speedometer that is supposed to read the linear speed of the car uses a device that actually measures the angular speed of the tires. If larger-diameter tires are mounted on the car instead, how will that affect the speedometer reading? The speedometer
a. will still read the speed accurately.
b. will read low.
c. will read high.
A speedometer often known as aspeed meter do act as a boundary or gauge that helps us to take note, measures and displays the speed of a moving automobile.
The speedometer will read low.An increase in the radius of the tires does not increase the speed measured by a speedometer.
An increase in the size of the tires will cause the car to move faster than it would with smaller tires. This will leaf tothe speed of the car will be much more than what the speedometer is reading.
Conclusively we can say that option b is the best option that explains what the statement means
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Investigators are exploring ways to treat milk for longer shelf life by using pulsed electric fields to destroy bacterial contamination. One system uses 8.0-cm-diameter circular plates separated by 0.95 cm. The space between the plates is filled with milk, which has the same dielectric constant as that of water. The plates are briefly charged to 30,000 V. What is the capacitance of the system? How much charge is on each plate when they are fully charged?
Answer:
C = 3.77*10⁻¹⁰ F = 377 pF
Q = 1.13*10⁻⁵ C
Explanation:
Given
D = 8.0 cm = 0.08 m
d = 0.95 cm = 0.95*10⁻² m
k = 80.4 (dielectric constant of the milk)
V = 30000 V
C = ?
Q = ?
We can get the capacitance of the system applying the formula
C = k*ε₀*A / d
where
ε₀ = 8.854*10⁻¹² F/m
and A = π*D²/4 = π*(0.08 m)²/4
⇒ A = 0.00502655 m²
then
C = (80.4)*(8.854*10⁻¹² F/m)*(0.00502655 m²) / (0.95*10⁻² m)
⇒ C = 3.77*10⁻¹⁰ F = 377 pF
Now, we use the following equation in order to obtain the charge on each plate when they are fully charged
Q = C*V
⇒ Q = (3.77*10⁻¹⁰ F)*(30000 V)
⇒ Q = 1.13*10⁻⁵ C
The capacitance of the system is calculated by applying the physics of parallel-plate capacitors, using characteristics of the plates and the milk as the dielectric constant. The charge on each plate when fully charged is then deduced from the calculated capacitance and the applied voltage.
Explanation:The concept this question revolves around is the electrical mechanism in a parallel-plate capacitor. This system uses the property of the dielectric constant (the same as water for milk, in this case) of the substance between the plates.
The capacitance of such a system may be calculated as C = ε * (A/d), where ε denotes the permittivity of the substance (milk), A is the area of one of the plates, and d is the distance separating the plates. For a circular plate, the area can be computed as A = π * (D/2)^2 where D is the diameter of the plate.
Here, ε = 8.854*10^-12 F/m (for water), D = 8 cm = 0.08 m (converted into meters for unit consistency), and d = 0.95 cm = 0.0095 m. Substituting these values, we compute the capacitance C which should then be used to calculate the charge on each plate when fully charged using the formula, Q = C * V, where V is the voltage applied.
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An object is attached to a vertical spring and slowly lowered to its equilibrium position. This stretches the spring by an amount d. If the same object is attached to the same vertical spring but permitted to fall instead, through what maximum distance does it stretch the spring? Please show all work. Step by step.
Answer:
2d
Explanation:
For any instance equivalent force acting on the body is
[tex]mg-kd= m\frac{d}{dt}\frac{dx}{dt}[/tex]
Where
m is the mass of the object
k is the force constant of the spring
d is the extension in the spring
and
d/dt(dx/dt)= is the acceleration of the object
solving the above equation we get
[tex]x= Asin\omega t +d[/tex]
where
[tex]\omega= \sqrt{\frac{k}{m} } = \frac{2\pi}{T}[/tex]
A is the amplitude of oscillation from the mean position.
k= spring constant , T= time period
Here we are assuming that at t=T/4
x= 0 since, no extension in the spring
then
A=- d
Hence
x=- d sin wt + d
now, x is maximum when sin wt=- 1
Therefore,
x(maximum)=2d
Final answer:
When an object falls onto a spring, it will stretch the spring further than if it were slowly lowered due to kinetic energy turning into additional elastic potential energy. The maximum stretch can be calculated using energy conservation, resulting in a stretch that is √(2mgd/k) where d is the equilibrium stretch, m is the mass, g is gravity, and k is the spring constant.
Explanation:
Maximum Stretch of a Spring with a Falling Object
When an object of mass m is attached to a vertical spring and slowly lowered to its equilibrium position, the spring stretches by a distance d, producing potential energy in the spring equal to k·d²/2. Meanwhile, the work done by gravity equals mg·d. If instead the object is allowed to fall and attaches to the spring, the spring will stretch further due to the kinetic energy acquired during the fall, turning into additional elastic potential energy at maximum stretch.
If we assume no energy losses due to non-conservative forces, the energy conservation equation just before the object attaches to the spring would be:
Kinetic energy + gravitational potential energy = elastic potential energy at maximum extension.
Since the object starts from rest at a height d above the equilibrium position:
1/2·mv² + mgd = 1/2·k·x² (with x being the total distance the spring stretches).
At the point of maximum stretch, all kinetic energy has been converted into elastic potential energy, and the object is momentarily at rest, so:
mgd = 1/2·k·x²
In the special case where there are no other forms of energy conversion (like heat due to air resistance), the spring will stretch maximum distance x satisfying:
x = √(2·mg·d / k)
Using the initial condition that the spring stretches by d when the object is slowly lowered, we can deduce that the object in free fall will stretch the spring more than d due to extra energy from its fall. The exact factor would depend on the stiffness of the spring (expressed by k) and the mass of the object (m).
A 3.00-m rod is pivoted about its left end. A force of 7.80 N is applied perpendicular to the rod at a distance of 1.60 m from the pivot causing a ccw torque, and a force of 2.60 N is applied at the end of the rod 3.00 m from the pivot. The 2.60-N force is at an angle of 30.0o to the rod and causes a cw torque. What is the net torque about the pivot? (Take ccw as positive.)
a. 26.4 N·m
b. 4.68 N·m
c. 8.58 N·m
d. -8.58 N·m
e. -16.4 N·m
Answer:
The net torque about the pivot is and the answer is 'c'
c. [tex]T_{net}=8.58[/tex]
Explanation:
[tex]T=F*d[/tex]
The torque is the force apply in a distance so it is the moment so depends on the way to be put it the signs so:
[tex]T_1=F_1*d_1[/tex]
[tex]T_1=7.8N*1.6m=12.48N*m[/tex]
[tex]T_2=F_2*d_2[/tex]
[tex]T_2=2.60N**cos(30)*3.0m[/tex]
[tex]T_2= - 3.9 N*m[/tex]
Now to find the net Torque is the summation of both torques
[tex]T_{net}=T_1+T_2[/tex]
[tex]T_{net}=12.48N-3.9N=8.58N[/tex]
The net torque about the pivot is calculated by finding the individual torques caused by each force and then subtracting the clockwise torque from the counterclockwise torque, resulting in a positive net torque of 8.58 N·m. So the correct option is c.
Explanation:The student is asking to calculate the net torque about the pivot on a rod being acted upon by two forces causing opposing torques. To find the net torque, we need to calculate each torque and then subtract the clockwise (cw) torque from the counterclockwise (ccw) torque. Torque (τ) can be calculated using the formula τ = rFsin(θ), where r is the distance from the pivot, F is the force applied, and θ is the angle at which the force is applied relative to the rod.
Step-by-step calculation:
Since the counterclockwise (ccw) torque is the positive direction, the net torque of 8.58 N·m is positive, giving us option c) 8.58 N·m as the correct answer.
The vapor pressures (in torr) of solid and liquid chlorine are given by ln ( P s ) = 24.320 − 3777 K T ln ( P l ) = 17.892 − 2669 K T where T is the absolute temperature. Calculate the temperature and pressure at the triple point of chlorine.
Answer:
[tex]T=172.37\ K[/tex]
[tex]P_s=11.1095\ Pa[/tex]
Explanation:
Triple point is that temperature and pressure for a substance at which the chemical substance coexists in liquid, solid and gaseous state with mutual equilibrium.
At this point the equilibrium vapor pressure of solid, liquid and gas phases are mutually equal.Given expressions of vapor pressure:
[tex]ln (P_s)=24.320 -\frac{3777 K}{T}[/tex] ........................................(1)
[tex]ln (P_l)=17.892 -\frac{2669 K}{T}[/tex] ..........................................(2)
Equating the vapor pressure of two phases of chlorine:
[tex]24.320 -\frac{3777 K}{T}=17.892 -\frac{2669 K}{T}[/tex]
[tex]T=172.37\ K[/tex]
The vapor pressure at this temperature:
[tex]ln (P_s)=24.320 -\frac{3777 }{172.37}[/tex]
[tex]P_s=11.1095\ Pa[/tex]
To find the triple point of chlorine, we set the vapor pressure equations of solid and liquid chlorine equal and solve for temperature, then use either equation to find the vapor pressure. However, numerical values for the vapor pressure equations are required to compute the specific temperature and pressure. The Clausius-Clapeyron equation can also be relevant in such calculations.
Explanation:The triple point of chlorine is determined by finding the temperature and pressure at which the vapor pressures of solid and liquid chlorine are equal. Given the vapor pressure equations for solid and liquid chlorine, both can be set equal to each other and solved for temperature (T), as both will have the same pressure at the triple point.
Unfortunately, without the numerical expressions for ln(Ps) and ln(Pl), it is impossible to provide a numerical answer. Typically, this would involve algebraic manipulation to find T, and then either expression can be used to solve for the vapor pressure at the triple point.
For example, the Clausius-Clapeyron equation is used to relate the temperatures and pressures for phase changes, such as moving from solid to gas (sublimation) at the triple point. However, more information is necessary to calculate the temperature and pressure explicitly.
Children have about 1016 synapses that can transfer signals between neurons in the brain and between neurons and muscle cells. Suppose that these synapses simultaneously transmit a signal, sending 1000 calcium ions (Ca2+) across the membrane at each synaptic ending. Determine the total electric charge transfer in coulombs during that short time interval. By comparison, a lightning flash involves about 5C of charge transfer. (Note: This is a fictional scenario. All human neurons do not simultaneously produce signals.) e=1.6×10−19C.
Answer:
3.2 C
Explanation:
e = Charge of electron = [tex]1.6\times 10^{-19}\ C[/tex]
Number of synapses = [tex]10^{16}[/tex]
Signals transferred by each synapse 1000 calcium ions [tex]Ca^{2+}[/tex]
Each calcium ion carries 2 charges
Number of charges would be [tex]2\times 1000=2000[/tex]
Total charges would be
[tex]N=10^{16}\times 2000=2\times 10^{19}[/tex]
Charge is given by
[tex]Q=Ne\\\Rightarrow Q=2\times 10^{19}\times 1.6\times 10^{-19}\\\Rightarrow Q=3.2\ C[/tex]
The total electric charge transfer in coulombs during that short time interval is 3.2 C
Final answer:
In a fictional scenario where 10¹⁶ synapses simultaneously transmit signals using 1000 calcium ions each, the total electric charge transferred is 3.20 coulombs, which is significant but still less than the 5 coulombs typically involved in a lightning flash.
Explanation:
Children have about 1016 synapses that can transfer signals between neurons in the brain and between neurons and muscle cells. If these synapses simultaneously transmit a signal, sending 1000 calcium ions (Ca²⁺) across the membrane at each synaptic ending, and given that the charge of each Ca²⁺ is twice that of the electron charge (e = 1.60 × 10⁻¹⁹ C), we can calculate the total electric charge transfer in coulombs during that time interval.
The total number of calcium ions transferred is 1016 synapses × 1000 ions/synapse = 10¹⁹ ions. Since each Ca²⁺ carries twice the charge of an electron, the charge per ion is 2 × 1.60 × 10⁻¹⁹ C = 3.20 × 10⁻¹⁹ C. Therefore, the total charge transferred is 1019 ions × 3.20 × 10⁻¹⁹ C/ion = 3.20 coulombs.
Comparing this to a lightning flash, which involves about 5 coulombs of charge transfer, we can see that the total charge transfer in this hypothetical scenario is significant, yet smaller than that of a lightning flash.
A stone with heat capacity C = 1.2 J/K is left outside on a cold day to reach a temperature of 273.15 K. The stone is then brought inside where the air temperature is 293.15 K. The stone is used as the cold side of a reversible engine. (The air is the hot side.) What is the maximum work that can be accomplished ?
To solve the problem it is necessary to apply the concepts related to Helmholtz free energy. By definition in a thermodynamic system the Helmholtz energy is defined as
[tex]\Delta F = \Delta U - T\Delta S[/tex]
Where,
[tex]\Delta U[/tex] is the internal energy equivalent to
[tex]\Delta U = C \Delta T[/tex]
And [tex]\Delta S[/tex] means the change in entropy represented as
[tex]\Delta S = C ln \frac{T_2}{T_1}[/tex]
Note: C means heat capacity.
Replacing in the general equation we have to
[tex]\Delta F = C \Delta T - T C ln \frac{T_2}{T_1}[/tex]
The work done of a thermodynamic system is related by Helmholtz free energy as,
[tex]W = - \Delta F[/tex]
[tex]W = -(C \Delta T - T C ln \frac{T_2}{T_1})[/tex]
[tex]W = T C ln \frac{\T_2}{T_1}-C \Delta T[/tex]
Replacing with our values we have,
[tex]W = (293.15)(1.2)ln(\frac{293.15}{273.15})-(1.2)(20)[/tex]
[tex]W = 0.858 J[/tex]
Therefore the maximum work that can be accomplished is 0.858J
The maximum work that can be accomplished by the stone is 24 J.
Explanation:To find the maximum work that can be accomplished by the stone as the cold side of a reversible engine, we can use the formula for the maximum work of a Carnot engine:
Wmax = Qh - Qc
where Qh is the heat transferred from the hot reservoir and Qc is the heat transferred to the cold reservoir.
Since the stone is the cold side, Qc is the negative of the heat capacity times the temperature difference:
Qc = -C(Tin - Tcold)
Substituting the given values, we have:
Qc = -(1.2 J/K)(293.15 K - 273.15 K) = -24 J
Since Qh is the negative of Qc, we have:
Qh = -Qc = -(-24 J) = 24 J
Therefore, the maximum work that can be accomplished by the stone is:
Wmax = 24 J
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A 2.3 g spider is hanging at the end of a silk thread. You can make the spider bounce up and down on the thread by tapping lightly on his feet with a pencil. You soon discover that you can give the spider the largest amplitude on his little bungee cord if you tap exactly once every 3 seconds.What is the spring constant of the silk thread?
Answer:[tex]k=10.091\times 10^{-3} N/m[/tex]
Explanation:
Given
mass of spider [tex]m=2.3 gm[/tex]
Largest amplitude can be obtained by Tapping after every 3 second
i.e. Time period of oscillation is [tex]T= 3 s[/tex]
considering spider to execute Simple harmonic motion
time of oscillation is given by
[tex]T=2\pi\sqrt{\frac{m}{k}}[/tex]
substituting values
[tex]3=2\pi \sqrt{\frac{2.3\times 10^{-3}}{k}}[/tex]
[tex]\frac{3}{2\pi }=\sqrt{\frac{2.3\times 10^{-3}}{k}}[/tex]
[tex]0.477=\sqrt{\frac{2.3\times 10^{-3}}{k}}[/tex]
[tex](0.477)^2=\frac{2.3\times 10^{-3}}{k}[/tex]
[tex]\frac{2.3\times 10^{-3}}{k}=0.228[/tex]
[tex]k=10.091\times 10^{-3} N/m[/tex]
A satellite beams microwave radiation with a power of 13 toward the Earth's surface, 550 away. When thebeam strikes Earth, its circular diameter is about 1000. Find the rms electric field strength of the beam at the surface of the Earth.
Answer:
[tex]E_{rms} = 2.27 V/m[/tex]
Explanation:
Power of microwave radiation = 13 kW = 13 x 10³ W
Diameter = 1000 m
Radius = R = 1000/2 = 500 m
distance, r = 500 Km
r = 500 x 10³ m
Electric strength of beam of light is given as
[tex]S = \dfrac{P}{A} = c\epsilon_0 E_{rms}^2[/tex]
Now,
[tex]E_{rms} =\sqrt{\dfrac{P}{Ac\epsilon_0}}[/tex]
c is speed of light
ε₀ is permittivity of free space
A is the area of circle = π r² = π x (500 x 10³)² = 7.853 x 10¹¹ m²
[tex]E_{rms} =\sqrt{\dfrac{13\times 10^3}{7.853 \times 10^{11}\times 3 \times 10^8 \times 8.85\times 10^{-12}}}[/tex]
[tex]E_{rms} =\sqrt{\dfrac{13\times 10^3}{7.853 \times 10^{11}\times 3 \times 10^8 \times 8.85\times 10^{-12}}}[/tex]
[tex]E_{rms} = 2.27 V/m[/tex]
A truck going 15 km/h has a head-on collision with a small car going 30 km/h. Which statement best describes the situation?
a. The truck has the greater change of momentum because it has the greater mass.
b. None of the above is necessarily true.
c. Neither the car nor the truck changes its momentum in the collision because momentum is conserved.
d. They both have the same change in magnitude of momentum because momentum is conserved.
e. The car has the greater change of momentum because it has the greater speed.
Answer:
d. They both have the same change in magnitude of momentum because momentum is conserved.
Explanation:
We know that if there is no any external force on the system the linear momentum of the system remain conserve.Or in the other other words change in the linear momentum is zero.
Here the external force is zero ,that is why initial and the final momentum will be conserved.
change in the momentum for both car as well as truck will be same.Because momentum should be conserved.
Therefore the answer is -d
The statement that best describes the situation is that they both have the same change in magnitude of momentum because momentum is conserved. Option D is correct
Momentum has to do with the strength an object has when moving with speed.
For the truck and the small car given in the question, there was no external force acting on the system.
Since there was no force acting on the system, we can conclude that the linear momentum of the system is conserved that is there is no change in the linear momentum of the system.
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Which is true about the inductance of a solenoid?
O It depends on the number of turns per unit length but not on the area of each turn.
O It depends on the area of each turn but not on the number of turns per unit length.
O It depends on the number of turns per unit length and the area of each turn.
Answer:
It depends on the number of turns per unit length and the area of each turn.Explanation:
The inductance of a long solenoid can be approximated by:
[tex]L=\mu \frac{N^2A}{L}[/tex]
Where:
[tex]N=Number\hspace{3}of\hspace{3}turns\\A=Cross-sectional\hspace{3}area\\L=Length\hspace{3}of\hspace{3}the\hspace{3}coil\\\mu=Magnetic\hspace{3}permeability[/tex]
Therefore, according to this, we can conclude that the inductance of a solenoid depends on the number of turns per unit length and the area of each turn.
One at your summer lunar space camp activities is to launch n 1170-kg rocket from the surface of the Moon. You are a serious space camper and you launch a serious rocket: II reaches an altitude of 211 km. What gain m gravitational potential memo does the launch accomplish? The mass and radios of the Moon are 7.36 x 10^22 kg and 1740 km, respectively.
To solve this problem it is necessary to apply the concepts related to gravitational potential energy.
The change in gravitational potential energy is given by,
[tex]\Delta PE = PE_f - PE_i[/tex]
Where,
[tex]PE = \frac{GMm}{R}[/tex]
Here,
G = Gravitational Universal Constant
M = Mass of Earth
m = Mass of Object
R = Radius
Replacing we have that
[tex]\Delta PE = \frac{GMm}{R+h} -\frac{GMm}{R}[/tex]
Note that h is the height for this object. Then replacing with our values we have,
[tex]\Delta PE = \frac{GMm}{R+h} -\frac{GMm}{R}[/tex]
[tex]\Delta PE = GMm(\frac{1}{R} -\frac{1}{R+h})[/tex]
[tex]\Delta PE = (6.65*10^{-11})(7.36*10^{22})(1170)(\frac{1}{1740*10^3} -\frac{1}{211*10^3+1740*10^3})[/tex]
[tex]\Delta PE = 57264.48*10^{11}(5.1255*10^{-7}-5.747*10^{-7})[/tex]
[tex]\Delta PE = 3.56*10^8J[/tex]
Therefore the gravitational potential is [tex]3.56*10^8J[/tex]
A hoop, a solid cylinder, a solid sphere, and a thin, spherical shell each have the same mass of 4.01 kg and the same radius of 0.260 m. (a) What is the moment of inertia (in kg · m2) for each object as it rotates about its axis
Final answer:
The moment of inertia for each object can be calculated using specific formulas. The formulas are: For a thin hoop: I = mR², for a solid cylinder: I = (1/2)MR², for a solid sphere: I = (2/5)MR², and for a thin spherical shell: I = (2/3)MR².
Explanation:
The moment of inertia for each object as it rotates about its axis can be calculated using the formulas for moment of inertia for different objects:
1. Hoop: The moment of inertia for a thin hoop rotating about an axis perpendicular to its plane is given by the formula I = mR², where m is the mass and R is the radius of the hoop. In this case, the mass is 4.01 kg and the radius is 0.260 m, so the moment of inertia for the hoop is (4.01 kg)(0.260 m)².
2. Solid Cylinder: The moment of inertia for a solid cylinder rotating about its central axis is given by the formula I = (1/2)MR², where M is the mass and R is the radius of the cylinder. In this case, the mass is 4.01 kg and the radius is 0.260 m, so the moment of inertia for the solid cylinder is (1/2)(4.01 kg)(0.260 m)².
3. Solid Sphere: The moment of inertia for a solid sphere rotating about its central axis is given by the formula I = (2/5)MR², where M is the mass and R is the radius of the sphere. In this case, the mass is 4.01 kg and the radius is 0.260 m, so the moment of inertia for the solid sphere is (2/5)(4.01 kg)(0.260 m)².
4. Thin Spherical Shell: The moment of inertia for a thin spherical shell rotating about its central axis is given by the formula I = (2/3)MR², where M is the mass and R is the radius of the shell. In this case, the mass is 4.01 kg and the radius is 0.260 m, so the moment of inertia for the thin spherical shell is (2/3)(4.01 kg)(0.260 m)².
A bat strikes a 0.145-kg baseball. Just before impact, the ball is traveling horizontally to the right at and it leaves the bat traveling to the left at an angle of above horizontal with a speed of If the ball and bat are in contact for 1.75 ms, find the horizontal and vertical components of the average force on the ball.
Answer
given,
mass of the baseball = 0.145 Kg
Assuming the horizontal velocity of ball = V x = 50 m/s
ball left at an angle of = 30°
At the speed of 65 m/s
time of contact = 1.75 ms
velocity of time along horizontal direction
v_{horizontal} = -65 cos 30° - 50
= -106.29 m/s
Impulse = force x time
impulse is equal to change in momentum
now,
force x time = m v
[tex]F = \dfrac{mv}{t}[/tex]
[tex]F = \dfrac{0.145 \times (-106.29)}{1.75 \times 10^{-3}}[/tex]
F_{horizontal} = - 8.806 kN
now velocity in vertical component
v_{vertical} = 65 sin 30°
= 32.5 m/s
[tex]F = \dfrac{mv}{t}[/tex]
[tex]F = \dfrac{0.145 \times (32.5)}{1.75 \times 10^{-3}}[/tex]
F_{vertical} = 2.692 kN
(a) The horizontal component of the average force on the ball is 7,633.63 N.
(b) The vertical component of the average force on the ball is 2,929 N.
The given parameters;
mass of the bat, m₁ = 0.145 kgtime, t = 1.75 ms initial velocity of the ball to right, v₀ = 50 m/sfinal velocity of the ball to left v = 55 m/s at 40⁰The horizontal component of the ball's acceleration is calculated as follows;
[tex]-v_f_x = v_0_x - a_xt\\\\a_xt = v_0_x + v_f_x\\\\a_x = \frac{v_0_x + v_f_x}{t} \\\\a_x = \frac{50 \ + \ 55cos(40)}{0.00175} \\\\a_x = 52,645.7 \ m/s^2[/tex]
The horizontal component of the average force on the ball is calculated as follows;
[tex]F_x = ma_x\\\\F_x = 0.145 \times 52,645.7 \\\\F_x = 7,633.63 \ N[/tex]
The vertical component of the ball's acceleration is calculated as follows;
[tex]v_f_y = v_0y + a_yt\\\\v_f_y = 0+ a_yt\\\\a_y = \frac{v_f_y}{t} \\\\a_y = \frac{55 \times sin(40)}{0.00175} \\\\a_y = 20,200 \ m/s^2[/tex]
The vertical component of the average force on the ball is calculated as follows;
[tex]F_y = ma_y\\\\F_y = 0.145 \times 20,200\\\\F_y = 2,929 \ N[/tex]
"Your question is not complete, it seems to be missing the following information";
A bat strikes a 0.145-kg baseball. Just before impact, the ball is traveling horizontally to the right at 50.0 m/s , and it leaves the bat traveling to the left at an angle of 40 ∘ above horizontal with a speed of 55.0 m/s . The ball and bat are in contact for 1.75 ms .
find the horizontal and vertical components of the average force on the ball.
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andy is entering early adolescence. although he used to be a big help to his mother by doing household chores, he now refuses, saying that the chores are "women's jobs." he also has a new interest in competitive sports and is being assertive in his interactions with friends. andy is most likely experiencing
a. gender intensification.
b. gender segregation.
c. gender typing.
d. the timing of puberty effect
Answer:
a. gender intensification
Explanation:
This theory states that in the adolescence both girls and boys are pressured to follow gender roles established culturally.
Two identical masses are connected to two different flywheels that are initially stationary. Flywheel A is larger and has more mass, but has hexagonal sections where material has been removed. The attached masses are released from rest and allowed to fall a height h.Which of the following statements about their angular accelerations is true? a. The angular acceleration of the two flywheels is different but it is impossible to tell which is greater. b. The angular acceleration of flywheel A is greater The angular acceleration of flywheel B is greater. c. Not enough information is provided to determine. d. The angular accelerations of the two flywheels are equal.
The angular acceleration of the two flywheels will be different due to their differing moments of inertia, which depend on both the mass and its distribution relative to the rotational axis. The removed hexagonal sections on flywheel A alters its moment of inertia.
Explanation:The answer to your question is: The angular acceleration of the two flywheels is different but it is impossible to tell which is greater.
The reason for this is tied to the concept known as moment of inertia, an object's resistance to changes in its state of rotation. The moment of inertia depends not only on the mass of an object, but also on its distribution of mass relative to the axis around which it rotates. The hexagonal sections removed from flywheel A affect its distribution of mass, potentially lowering its moment of inertia.
Angular acceleration is the rate of change of angular velocity over time, and is given by the equation: torque = moment of inertia x angular acceleration. The torque on both flywheels is the same, since the same force is exerted on both through the identical masses falling from height h. Therefore, if the moments of inertia of the two flywheels are different, then their angular accelerations must also be different as they both satisfy the above equation.
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The angular acceleration of the two flywheels is different but it is impossible to tell which is greater.
Explanation:The angular acceleration of the two flywheels is different but it is impossible to tell which is greater. The angular acceleration of a flywheel depends on both its mass and its moment of inertia, which takes into account the distribution of mass relative to the axis of rotation.
In this case, the larger mass of flywheel A would tend to result in a smaller angular acceleration, but the removal of material in hexagonal sections could change the moment of inertia and potentially increase the angular acceleration. Without specific information about the moment of inertia for each flywheel, it is impossible to determine which one would have a greater angular acceleration.
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Romeo (85.0 kg) entertains Juliet (59.0 kg) by playing his guitar from the rear of their boat at rest in still water, 2.70 m away from Juliet, who is in the front of the boat. After the serenade, Juliet carefully moves to the rear of the boat (away from shore) to plant a kiss on Romeo's cheek. (a) How far (in m) does the 75.0 kg boat move toward the shore it is facing? m
Answer:
0.7274 m
Explanation:
X denotes the distance from the reference point
M denotes the mass of the point
The center of mass of a system is given by
[tex]X_{CM}=\sum_{n=1}^n\frac{X_n\times M_n}{M_n}[/tex]
Taking the position of Juliet as the reference point
[tex]X_{CM}=\frac{59\times 0+85\times 2.7+75\times 1.35}{59+85+75}\\\Rightarrow X_{CM}=1.51027\ m[/tex]
After the boat has moved the distance between the center of mass of the boat and the shore remains the same. The change is observed relative to the rear of the boat
[tex]X_{CM}=\frac{75\times 1.35+(85+59)\times 2.7}{59+85+75}\\\Rightarrow X_{CM}=2.23767\ m[/tex]
The displacement of the boat towards the shore is
[tex]2.23767-1.51027=0.7274\ m[/tex]
By applying the conservation of momentum for a system with no external forces, the distance the boat moves when Juliet goes to the rear to kiss Romeo is 2.124 meters toward the shore.
The question involves the concept of conservation of momentum, particularly in systems with no external forces. Here, we consider a scenario where Juliet transitions from the front to the rear of the boat, causing the boat to move toward shore. To solve for the movement of the boat toward the shore, we need to realize that the total momentum of the system (Juliet, Romeo, and the boat) before and after Juliet's movement must remain constant, as there is no external force in the horizontal direction.
Initially, both Juliet and the boat are at rest, so the total initial momentum is zero. When Juliet, having mass 59.0 kg, moves a distance of 2.70 m towards Romeo, the boat will move in the opposite direction to conserve momentum.
To find the distance the boat moves, we use the conservation of momentum equation: (mass of Juliet) x (distance Juliet moves) = (mass of boat) x (distance boat moves). Rearranging gives us the distance the boat moves: distance boat moves = (mass of Juliet x distance Juliet moves) / mass of boat.
By substituting the given values, we get:
distance boat moves = (59.0 kg x 2.70 m) / 75.0 kg
distance boat moves = 2.124 m
Therefore, the boat moves 2.124 meters toward the shore when Juliet moves to the rear to kiss Romeo.
The Young�s modulus for steel is 20.7*1010 N/m2. How much will a 2.5 mm diameter wire 12 cm long, be strained when it supports a load of 450 N?
Answer:
ΔL = 53.14*10⁻⁶ m
Explanation:
Given
E = 20.7*10¹⁰ N/m²
D = 2.5 mm = 2.5*10⁻³ m
L = 12 cm = 0.12 m
P = 450 N
ΔL = ?
We can use the formula
ΔL = P*L / (A*E)
where A = π*D² / 4 = π*(2.5*10⁻³ m)² / 4
⇒ A = 4.908*10⁻⁶ m²
then
ΔL = (450 N)*(0.12 m) / (4.908*10⁻⁶ m²*20.7*10¹⁰ N/m²) = 53.14*10⁻⁶ m
Metal surfaces on spacecraft in bright sunlight develop a net electric charge.
Do they develop a negative or a positive charge?
Answer:
Positive.
Explanation:
As a consequence of the photoelectric effect, electrons that will get hit by sufficiently energetic photons will abandon the metal surfaces exposed to bright sunlight. This decreases the negative charge of the surface, thus causing it to develop a positive net charge.
An inductor L = 0.0345 H and a 30.5 Ω resistor are connected in series to a 3.20 volt battery and a switch. At t = 0 the switch is closed to complete the circuit. (a) What is the potential difference across the resistor immediately after the switch is closed?
Answer:
0 V
Explanation:
given,
Inductance = L = 0.0345 H
resistor of resistance = R = 30.5 Ω
connected in series with battery of voltage = 3.20 V
at t= 0 switch is closed
potential difference when switch is closed = ?
The potential difference when the switch is just closed across the resistor will be equal to Zero.
Because at t =0 the capacitor will act as a short circuit.
hence, the potential will be 0 V
A balloon has an initial radius of 0.075 m. A circle is painted on the balloon using silver metal paint. When the paint dries, the circle is a very good electrical conductor. With the balloon oriented such that a 1.5-T magnetic field is oriented perpendicular to the plane of the circle, air is blown into the balloon so that it expands uniformly. The silver circle expands to a radius 0.125 m in 1.5 s. Determine the induced emf for this silver circle during this period of expansion.
Answer:
The induced emf for this silver circle during this period of expansion is 0.0314 V.
Explanation:
Given that,
Initial radius = 0.075 m
Magnetic field = 1.5 T
Radius = 0.125 m
Time t =1.5 s
We need to calculate the induced emf for this silver circle
Using formula of emf
[tex]\epsilon=\dfrac{d(BA)}{dt}[/tex]
[tex]\epsilon=\dfrac{B(A_{f}-A_{i})}{t}[/tex]
[tex]\epsilon=\dfrac{B(\pi r_{f}^2-\pi r_{i}^2)}{t}[/tex]
Put the value ino the formula
[tex]\epsilon=\dfrac{1.5(\pi\times(0.125)^2-\pi(0.075)^2)}{1.5}[/tex]
[tex]\epsilon=0.0314\ V[/tex]
Hence, The induced emf for this silver circle during this period of expansion is 0.0314 V.
The induced emf for the silver circle during this period of expansion is approximately [tex]\(\boxed{0.0314 \, \text{V}}\).[/tex]
The induced emf in the silver circle during the expansion can be calculated using Faraday's law of induction, which states that the induced emf in a closed loop is equal to the negative rate of change of the magnetic flux through the loop. Mathematically, this is expressed as:
[tex]\[ \varepsilon = -\frac{d\Phi_B}{dt} \][/tex]
The initial area [tex]\( A_i \)[/tex] when the radius [tex]\( r_i \[/tex]) is 0.075 m is:
[tex]\[ A_i = \pi r_i^2 = \pi (0.075 \, \text{m})^2 \][/tex]
The final area [tex]\( A_f \)[/tex] when the radius [tex]\( r_f \)[/tex] is 0.125 m is:
[tex]\[ A_f = \pi r_f^2 = \pi (0.125 \, \text{m})^2 \][/tex]
The change in area [tex]\( \Delta A \)[/tex] is:
[tex]\[ \Delta A = A_f - A_i = \pi (0.125 \, \text{m})^2 - \pi (0.075 \, \text{m})^2 \][/tex]
The time taken for this change [tex]\( \Delta t \)[/tex] is 1.5 s.
The induced emf [tex]\( \varepsilon \)[/tex] is then:
[tex]\[ \varepsilon = -\frac{d\Phi_B}{dt} = -B \frac{\Delta A}{\Delta t} \][/tex]
[tex]\[ \varepsilon = -1.5 \, \text{T} \times \frac{\pi ((0.125 \, \text{m})^2 - (0.075 \, \text{m})^2)}{1.5 \, \text{s}} \][/tex]
Now, let's calculate the numerical value:
[tex]\[ \varepsilon = -1.5 \times \pi \times \frac{(0.125^2 - 0.075^2)}{1.5} \][/tex]
[tex]\[ \varepsilon = -1.5 \times \pi \times \frac{(0.015625 - 0.005625)}{1.5} \][/tex]
[tex]\[ \varepsilon = -1.5 \times \pi \times \frac{0.01}{1.5} \][/tex]
[tex]\[ \varepsilon = -1.5 \times \pi \times 0.00666\ldots \][/tex]
[tex]\[ \varepsilon \approx -0.0314 \, \text{V} \][/tex]
The negative sign indicates the direction of the induced emf according to Lenz's law, which is opposite to the direction of the change in magnetic flux. However, the magnitude of the induced emf is approximately 0.0314 V.
Therefore, the induced emf for the silver circle during this period of expansion is approximately [tex]\(\boxed{0.0314 \, \text{V}}\).[/tex]
The universe is filled with photons left over from the Big Bang that today have an average energy of about 2 × 10−4 eV (corresponding to a temperature of 2.7 K). As derived in lecture, the number of available energy states per unit volume for photons is ????(????)????????
Answer:
The number of available energy states per unit volume is [tex]4.01\times10^{48}[/tex]
Explanation:
Given that,
Average energy [tex]E=2\times10^{-4}\ eV[/tex]
Photon = [tex]4\times10^{-5}\ eV[/tex]
We need to calculate the number of available energy states per unit volume
Using formula of energy
[tex]g(\epsilon)d\epsilon=\dfrac{8\pi E^2dE}{(hc)^3}[/tex]
Where, E = energy
h = Planck constant
c = speed of light
Put the value into the formula
[tex]g(\epsilon)d\epsilon=\dfrac{8\times\pi\times2\times10^{-4}\times4\times10^{-5}\times1.6\times10^{-19}}{(6.67\times10^{-34}\times3\times10^{8})^3}[/tex]
[tex]g(\epsilon)d\epsilon=4.01\times10^{48}[/tex]
Hence, The number of available energy states per unit volume is [tex]4.01\times10^{48}[/tex]
You are in a room in a basement with a smooth concrete floor (friction force equals 40 N) and a nice rug (friction force equals 55 N) that is 3 m by 4 m. However, you have to push a very heavy box from one corner of the rug to the opposite corner of the rug. Will you do more work against friction going around the floor or across the rug, and how much extra?
Answer:
I will do more work against friction going around the floor.
ΔW = 5 J
Explanation:
Given
Ff₁ = 40 N
Ff₂ = 55 N
a = 3 m
b = 4 m
We have to get the distance d₁ as follows
d₁ = a + b = 3 m + 4 m = 7 m
And d₂:
d₂ = √(a² + b²) = √((3 m)² + (4 m)²)
⇒ d₂ = 5 m
then we use the equations
W₁ = Ff₁*d₁ = (40 N)(7 m) = 280 J (work against friction going around the floor)
W₂ = Ff₂*d₂ = (55 N)(5 m) = 275 J (work against friction going across the rug)
since W₂ < W₁ I will do more work against friction going around the floor.
Now, we apply the formula
ΔW = W₁ - W₂ = 280 J - 275 J = 5 J
When pushing the heavy box from one corner of the rug to the opposite corner, you will do more work against friction going across the rug than going around the floor. The extra work against friction across the rug is calculated to be 375 J.
Explanation:When pushing the heavy box from one corner of the rug to the opposite corner, you will do more work against friction going across the rug. The force of friction on the rug is greater than the force of friction on the smooth concrete floor. To calculate the extra work, you can use the formula:
Extra work = (force of friction on the rug - force of friction on the floor) x distance
Extra work = (55 N - 40 N) x (diagonal distance of the rug)
Since the rug is 3 m by 4 m, the diagonal distance can be calculated using the Pythagorean theorem:
diagonal distance = √(3^2 + 4^2) = √(9 + 16) = √25 = 5 m
Therefore, the extra work against friction going across the rug is (55 N - 40 N) x 5 m = 75 N x 5 m = 375 J.
For fully developed laminar pipe flow in a circular pipe, the velocity profile is given by u(r) = 2 (1 - r2/R2) in m/s, where R is the inner radius of the pipe. Assuming that the pipe diameter is 3.1 cm, find the (a) maximum and (b) average velocities in the pipe as well as (c) the volume flow rate.
Answer:
a)Uo= 2 m/s
b)[tex]u_{avg}=1 \ m/s[/tex]
c)Q=7.54 x 10⁻⁴ m³/s
Explanation:
Given that
[tex]u(r)=2\left(1-\dfrac{r^2}{R^2}\right)[/tex]
Diameter ,D= 3.1 cm
Radius ,R= 1.55 cm
We know that in the pipe flow the general equation for laminar fully developed flow given as
[tex]u(r)=U_o\left(1-\dfrac{r^2}{R^2}\right)[/tex]
Uo=Maximum velocity
Therefore maximum velocity
Uo= 2 m/s
The average velocity
[tex]u_{avg}=\dfrac{U_o}{2}[/tex]
[tex]u_{avg}=\dfrac{2}{2}\ m/s[/tex]
[tex]u_{avg}=1 \ m/s[/tex]
The volume flow rate
[tex]Q=u_{avg}. A[/tex]
[tex]Q=\pi R^2\times u_{avg}\ m^3/s[/tex]
[tex]Q=\pi \times (1.55\times 10^{-2})^2\times 1\ m^3/s[/tex]
Q=0.000754 m³/s
Q=7.54 x 10⁻⁴ m³/s
A uniform, thin rod of length h and mass M is held vertically with its lower end resting on a frictionless horizontal surface. The rod is then released to fall freely. (a) What is the speed of its center of mass just before it hits the horizontal surface? (Use any variable or symbol stated above along with the following as necessary: g for the acceleration of gravity.)
The speed of the center of mass just before it hits the horizontal surface can be found using the principle of conservation of energy.
Explanation:To find the speed of the center of mass just before it hits the horizontal surface, we can use the principle of conservation of energy. When the rod falls freely, it gains gravitational potential energy which is converted into kinetic energy. At the lowest point, where the center of mass hits the horizontal surface, all the gravitational potential energy is converted into kinetic energy. Therefore, we can equate the gravitational potential energy at the top to the kinetic energy at the bottom:
mgh = (1/2)mv^2
Where m is the mass of the rod, g is the acceleration due to gravity, h is the height of the rod, and v is the speed of the center of mass. Solving for v:
v^2 = 2gh
v = sqrt(2gh)
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. An elastic bar (Young’s modulus E) of initial length L is fixed at one end and is axially loaded at the other end with a force P (Fig. 1a). (a) Derive the expression to obtain the displacement ∆ of the loaded end of the bar. (b) Calculate the displacement if E = 70 GPa, L = 100 mm, and P = 1 kN.
Answer:
ΔL = 1.43 10 -9 m
Explanation:
a) Let's start from Newton's second law, the force in a spring is elastic
F = - k Δx
Let's divide the two sides by the area
F / A = -k Δx / A
In general area is long by wide, the formulated pressure is
P = F / A
P = - k Δx / l x
P = (-k / l) Δx/x
Call us at Young's constant module
P = E Δx / x
Let's change x for L
E = P / (ΔL/L)
b) we cleared
ΔL = P L / E
Let's reduce the magnitudes to the SI system
E = 70 GPa = 70 109 Pa
L = 100mm (1m / 1000mm) = 0.100m
P = 1 kN = 1 103 N
ΔL = 1 103 0.100/ 70 109
ΔL = 1.43 10 -9 m
(a) What is the length of a simple pendulum that oscillates with a period of 3.2 s on Earth, where the acceleration due to gravity is 9.80 m/s2, and on Mars, where the acceleration due to gravity is 3.70 m/s2?
Length (Earth)=?
Length (Mars)=?
(b) What mass would you need to suspend from a spring with a force constant of 20 N/m in order for the mass-spring system to oscillate with a period of 3.2 s on Earth, where the acceleration due to gravity is 9.80 m/s2, and on Mars, where the acceleration due to gravity is 3.70 m/s2?
Mass (Earth)=?
Mass (Mars)=?
Answer:
Explanation:
The expression relating length and time period
T = 2π [tex]\sqrt{\frac{l}{g} }[/tex]
3.2 = [tex]2\pi \sqrt{\frac{l}{9.8} }[/tex]
l = 2.54 m
On Mars g = 3.7
[tex]3.2 = 2\pi \sqrt{\frac{L}{3.7} }[/tex]
L = .96 m
b )
Expression for elastic constant and time period is as follows
[tex]T = 2\pi \sqrt{\frac{m}{k} }[/tex]
[tex]3.2=2\pi \sqrt{\frac{m}{20} }[/tex]
m = 5.19 N/s
Time period of oscillation due to spring is not dependent on g , so same time period will be found on Mars as that on the earth.
A body is traveling at 5.0 m/s along the positive direction of an x axis; no net force acts on the body. An internal explosion separates the body into two parts, each of mass 4 kg, and increases the total kinetic energy by 100 J. The forward part continues to move in the original direction of motion. (a) What is the speed of the rear part? (b) What is the speed of the forward part?
To solve this problem it is necessary to apply the concepts related to the conservation of momentum and conservation of kinetic energy.
By definition kinetic energy is defined as
[tex]KE = \frac{1}{2} mv^2[/tex]
Where,
m = Mass
v = Velocity
On the other hand we have the conservation of the moment, which for this case would be defined as
[tex]m*V_i = m_1V_1+m_2V_2[/tex]
Here,
m = Total mass (8Kg at this case)
[tex]m_1=m_2 =[/tex] Mass each part
[tex]V_i =[/tex] Initial velocity
[tex]V_2 =[/tex] Final velocity particle 2
[tex]V_1 =[/tex] Final velocity particle 1
The initial kinetic energy would be given by,
[tex]KE_i=\frac{1}{2}mv^2[/tex]
[tex]KE_i = \frac{1}{2}8*5^2[/tex]
[tex]KE_i = 100J[/tex]
In the end the energy increased 100J, that is,
[tex]KE_f = KE_i KE_{increased}[/tex]
[tex]KE_f = 100+100 = 200J[/tex]
By conservation of the moment then,
[tex]m*V_i = m_1V_1+m_2V_2[/tex]
Replacing we have,
[tex](8)*5 = 4*V_1+4*V_2[/tex]
[tex]40 = 4(V_1+V_2)[/tex]
[tex]V_1+V_2 = 10[/tex]
[tex]V_2 = 10-V_1[/tex](1)
In the final point the cinematic energy of EACH particle would be given by
[tex]KE_f = \frac{1}{2}mv^2[/tex]
[tex]KE_f = \frac{1}{2}4*(V_1^2+V_2^2)[/tex]
[tex]200J=\frac{1}{2}4*(V_1^2+V_2^2)[/tex](2)
So we have a system of 2x2 equations
[tex]V_2 = 10-V_1[/tex]
[tex]200J=\frac{1}{2}4*(V_1^2+V_2^2)[/tex]
Replacing (1) in (2) and solving we have to,
200J=\frac{1}{2}4*(V_1^2+(10-V_1)^2)
PART A: [tex]V_1 = 10m/s[/tex]
Then replacing in (1) we have that
PART B: [tex]V_2 = 0m/s[/tex]