The cannon ball will travel the highest distance when the angle of projection is 70 degrees.
What is the maximum height of a projectile?The maximum height reached by a projectile is calculated using the following formula.
H = u²sin²θ/2g
where;
u is the initial velocity of the projectile θ is the angle of pojectiong is acceleration due to gravitywhen the angle of projection is 60 degrees;
H = (15² (sin 60)²) / (2 x 9.8)
H = 8.6 m
when the angle of projection is 70degrees;
H = (15² (sin 70)²) / (2 x 9.8)
H = 10.14 m
Thus, the cannon ball will travel the highest distance when the angle of projection is 70 degrees.
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Suppose the coefficient of kinetic friction between mA and the plane in the figure(Figure 1) is μk = 0.15, and that mA=mB=2.7kg. part aAs mB moves down, determine the magnitude of the acceleration of mA and mB, given θ = 30 ∘ .part b What smallest value of μk will keep the system from accelerating?
We are asked to determine the acceleration of the system. To do that we will consider the two masses to be a single system and we will use the following free-body diagram.
Where:
[tex]\begin{gathered} W_b=\text{ weight of B} \\ W_a=\text{ weight of A} \\ W_{ah}=\text{ component of the weight of A in the horizontal direction} \\ W_{av}=\text{ component of the weight of B in the vertical direction} \\ N=\text{ normal force} \\ F_f=\text{ force of friction} \end{gathered}[/tex]To determine the components of the weight of "A" we use the following triangle:
Therefore, the vertical component is given by:
[tex]W_{av}=W_a\cos30[/tex]Since the weight is the product of the mass and the acceleration we have;
[tex]W_{av}=m_ag\cos30[/tex]The horizontal component is given by:
[tex]W_{ah}=W_a\sin30[/tex]Substituting the formula for weight:
[tex]W_{ah}=m_ag\sin30[/tex]The friction force is given by:
[tex]F_f=\mu N[/tex]The normal force "N" is equivalent to the vertical component of the weight since there is no acceleration in that direction. Therefore, we have:
[tex]F_f=\mu W_{av}[/tex]Substituting the values:
[tex]F_f=\mu m_ag\cos30[/tex]Now, the net force is equivalent to the forces acting on the system. Therefore, the net force is:
[tex]F_{net}=m_bg-F_f-W_{ah}[/tex]Substituting we get:
[tex]F_{net}=m_bg-\mu m_ag\cos30-m_ag\sin30[/tex]According to Newton's second law we have that the net force is equal to the product of the mass and the acceleration:
[tex]F_{net}=ma[/tex]The total mass is the sum of the masses "A" and "B".
[tex]F_{net}=(m_a+m_b)a[/tex]Substituting the net force:
[tex]m_bg-\mu m_ag\cos30-m_ag\sin30=(m_a+m_b)a[/tex]Now, we divide both sides by the total mass:
[tex]\frac{m_bg-\mu m_ag\cos30-m_ag\sin30}{m_a+m_b}=a[/tex]Since we have that the masses are equal:
[tex]m_a=m_b=m[/tex]Substituting we get:
[tex]\frac{mg-\mu mg\cos(30)-mg\sin(30)}{m+m}=a[/tex]Adding the masses in the denominator:
[tex]\frac{mg-\mu mg\cos(30)-mg\sin(30)}{2m}=a[/tex]We can cancel out the "m":
[tex]\frac{g-\mu g\cos(30)-g\sin(30)}{2}=a[/tex]Now, we plug in the values:
[tex]\frac{9.8\frac{m}{s^2}-(0.15)(9.8\frac{m}{s^2})\cos30-9.8\frac{m}{s^2}\sin30}{2}=a[/tex]Solving the operations:
[tex]1.81\frac{m}{s^2}=a[/tex]Therefore, the acceleration is 1.81 meters per second square.
Part B. To determine the value of the coefficient of friction that will keep the system from accelerating we must go back to the formula for the acceleration:
[tex]\frac{g-\mu g\cos(30)- g\sin(30)}{2}= a[/tex]Now, we will set the acceleration to zero:
[tex]\frac{g-\mu g\cos(30)- g\sin(30)}{2}=0[/tex]Multiplying both sides by 2:
[tex]g-\mu g\cos(30)-g\sin(30)=0[/tex]Now, we can divide both sides by "g":
[tex]1-\mu\cos30-\sin30=0[/tex]Now, we solve for the coefficient of friction. We subtract 1 to both sides:
[tex]-\mu\cos(30)-\sin(30)=-1[/tex]Now, we add sin(30) to both sides:
[tex]-\mu\cos(30)=-1+\sin(30)[/tex]Now, we divide both sides by -cos(30):
[tex]\mu=\frac{-1+\sin(30)}{-\cos(30)}[/tex]Now, we solve the operations:
[tex]\mu=0.58[/tex]Therefore, the friction coefficient is 0.58
If a spring is stretched 4m from its starting length when 20n of force is applied, then how much work (in joules) is done by the spring when it is stretched 10 m?
ANSWER:
250 J
STEP-BY-STEP EXPLANATION:
F = 20N is required to stretch the spring by 4 meters
We know that the force is equal to:
[tex]F=k\cdot x[/tex]We solve for k (spring constant):
[tex]k=\frac{F}{x}=\frac{20}{4}=5\text{ N/m}[/tex]The work done in stretching the spring is given by the following equation (in this case the stretch is 10 meters:
[tex]\begin{gathered} W=\frac{1}{2}k\cdot x^2 \\ \text{ Replacing} \\ W=\frac{1}{2}\cdot5\cdot10^2 \\ W=250\text{ J} \end{gathered}[/tex]The work required is 250 joules.
ASAPPPPPPPPPPPPPPPPPPPPPPPP!!!!!!
Answer:
its A
Explanation:
Answer:
it A bc it makes the most sense
Explanation:
Find a model for simple harmonic motion satisfying the specified the conditions.
displacement (t=0) 0 centimeters
amplitude 8 centimeters
period 2 seconds
Answer: d = 8 sin (πt
Explanation:
when displacement is 0m in t = 0s, then it's a sine equation
∴ d = A sin (ωt + Ф)
when t = 0, then there's no initial phase
Period ⇒ 2π/ω = 2 ⇒ ω = π
∴ d = 8 sin (πt)
Anton applies a force on a shopping cart and makes it move forward. What can be said about the forces acting on the shopping cart at the moment Anton applies the force? The forces acting on the shopping cart are...
Given that, Anton is applying a force on a cart making it move around.
The forces acting on the cart are gravitational force, normal force, frictional force, and the force applied by Anton.
The gravitational force will be acting downwards. From Newton's third law there will be a reaction force to the gravitational force and that is called the normal force. And the formal force will be acting upwards. These two forces are equal to each other in magnitude. And thus they will cancel each other out making the net force on the cart in the vertical direction zero.
The force applied by Anton will be acting forward and the frictional force will be opposing the movement of the cart. The net force in the horizontal direction will be directed forward.
From Newton's second law, the net force is equal to the product of the mass of the cart and the acceleration produced by the net force.
The Big Dipper is an example of an observed pattern of stars called a ___________.
Answer:
The answer is the asterism.
Explanation:
An asterism is recognizable pattern of stars . some other commonly recognized asterisms are the little dipper ,orion's belt , and the teapot
Where do you find valence electrons?
Answer:Valence electrons are the electrons in the outermost shell, or energy level, of an atom. For example, oxygen has six valence electrons, two in the 2s subshell and four in the 2p subshell.
Explanation:
I need help with the answer
Answer:
Yes, of course
Explanation:
Ask the questionI'll answer itWhat is the index of refraction of a refractive medium if the angle of incidence in air (n=1.0003) is 87 degrees and the angle of refraction is 10 degrees?
Given:
The refractive index of the air, n₁=1.0003
The angle of incidence, θ₁=87°
The angle of refraction, θ₂=10°
To find:
The refractive index of the medium.
Explanation:
From the snell's law,
[tex]n_1\sin\theta_1=n_2\sin\theta_2[/tex]Where n₂ is the refractive index of the medium.
On substituting the known values,
[tex]\begin{gathered} 1.0003\sin87\degree=n_2\sin10\degree \\ \implies n_2=\frac{1.0003\times\sin87\degree}{\sin10\degree} \\ =5.75 \end{gathered}[/tex]Final answer:
Thus the refractive index of the medium is 5.75
What effect would that have on the force resulting in accordance to Newton's Third law ?
According to the Newton's Third law, the force sent back with equal and opposite in the reaction.
What is meant by Newton's Third law?According to Newton's third law, two objects exert a force of equal magnitude and opposite direction. This law represents a certain symmetry in nature: forces always occur in pairs, and one body cannot exert a force on another without experiencing a force on it. Newton's 3rd law of motion states that action and reaction are always equal but opposite in direction. Newton's third law that for every action in nature there is an equal and opposite reaction. If object A exerts a force on object B, object B also uses an alike and across-form force on object A.
So we can conclude that the force is in equal in magnitude but in opposite direction according to Newton's Third law.
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Which of the following methods can demonstrate cause and effect?
correlational
experimental
naturalistic observation
survey method
How is Newton's Third Law of Motion used in the sport Jokgu?
Newton's third law of motion states that every force has an equal and opposite reaction force. That is when two bodies interact with each other, they apply a force on each other which are equal in magnitude and opposite in direction.
In Jokgu, when you hit the ball it applies an equal force on the leg. And when a player presses his leg against the ground, the ground exerts an equal force on the player which helps him to move around.
A ball with a mass of 1.50 kg traveling +2.00 m/s collides with a stationery ball with a mass of 1.00 kg. After the collision, the velocity of the 1.50-kg ball is +0.40 m/s. What is the velocity of the 1.00-kg ball after the collision?
The velocity of the 1.00-Kg ball after the collision with the 1.50 Kg ball is 2.4m/s.
The mass M₁ of first ball is 1.5 Kg. The initial velocity U₁ of the first ball is 2m/s.
The mass M₂ of the second ball is 1 Kg. The initial velocity U₂ if the second ball is 0m/s.
After collision, the final velocity V₁ of the first ball is +0.40m/s.
Let us say that the final velocity of the second ball after collision is V₂.
As there is no external force,
We can apply the concept of conservation of linear momentum,
According to which, in a system when no external force is in effect, the total linear momentum of the system is conserved.
We can assume both balls to be a single system,
Initial momentum = Final momentum
M₁U₁+M₂U₂ = M₁V₁+M₂V₂
Putting all the values,
(1.5x2)+(1x0) = (1.5x0.4)+(1V₂)
V₂ = 1.5(2-0.4)
V₂ = 2.4m/s.
So, the final velocity of the 1.00kg ball is 2.4m.s.
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Suppose the ball is thrown from the same height as in the PRACTICE IT problem at an angle of 29.0° below the horizontal. If it strikes the ground 47.8 m away, find the following. (Hint: For part (a), use the equation for the x-displacement to eliminate v0t from the equation for the y-displacement.)
The time that the ball took to take off at an angle of 29.0° is 1.13 s
When two lines meet at a point, they form an angle. The word "angle" refers to the length of the "opening" between these two rays. The symbol is used to denote it.
Radians, a unit of circularity or rotation, and degrees are the two most common units used to measure angles. In daily life, angles are present. For the design of highways, structures, and sports venues, engineers and architects use angles.
The motion along the x-direction is always uniform, so the position along the x-direction is given by
x(t) = (v₀cos∅)t
where,
v₀ is the initial speed,
cos∅ is the angle of launching,
t is the time.
We can re-write the formula as
v₀t = x(t)/cos∅ ...1
Along the y-direction the motion is accelerated (free-fall), so the equation for the vertical position is
y(t) = h = v₀sin∅t - 1/2gt² ...2
where,
h = 45.0 m is the height of the building
g = 9.8m/s² is the acceleration of gravity.
Substituting (1) into (2),
y(t) = h = x(t)tan∅ - 1/2gt²
And when this happens, the displacement along the x-axis is x = 47.8 m. Solving the equation for t and substituting the numbers, we find the time of flight:
[tex]t = \sqrt{\frac{2(h-xtan\theta)}{g} }[/tex]
[tex]t = \sqrt{\frac{2(45-47.8tan39^{\circ})}{9.8} }[/tex]
t = 1.13 s
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Full question
Suppose the ball is thrown from a 45m height building as in the PRACTICE IT problem at an angle of 29.0° below the horizontal. If it strikes the ground 47.8 m away, find the following. (Hint: For part (a), use the equation for the x-displacement to eliminate v0t from the equation for the y-displacement.)
a) the time of flight s
A current of 17.0A is maintained in a single circular loop of 2.00m circumference. A magnetic field of 0.800T is directed parallel to the plane of the loop. (a) Calculate the magnetic moment of the loop. [5] (b) What is the magnitude of the torque exerted on the loop by the magnetic field?
A. The moment of the loop is 5.4 x [tex]10^{-3} Am^{2}[/tex]
B. The torque exerted on the loop by the magnetic field is 4.32 [tex]10^{-3}[/tex] Nm.
Given that the current I in the circular loop is 17.0 mA and the circumference of the loop is 2.00 m.
The radius of the circular loop is calculated as given below.
2πr = 2
r = 2/2 x 3.14
r = 0.18m
The moment in the circular loop is calculated as given below.
M = in
M = 17 x [tex]10^{-3}[/tex] x 3.14 x 0.318²
M = 5.4 x [tex]10^{-3} Am^{2}[/tex].
The torque exerted on the loop by the magnetic field is given below:
T = MB
T = 5.4 x [tex]10^{-3}[/tex] x 0.800
T = 4.32 x [tex]10^{-3} Nm[/tex].
This is the torque of the current-carrying loop in a uniform magnetic field. This formula can be shown to be valid for loops of any shape. The loop carries a current I, has N turns, and the plane A and the normal to the loop each make an angle θ with the field B. The net force on the loop is zero. A magnetic field exerts a force on a straight wire through which an electric current flows. Apply torque to the loop of the wire carrying the current. A torque rotates an object around a fixed axis.
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Describe the relationship between the state of matter of a pure substance (gas, liquid, solid) and the motion of the particles.
I will brainliest you!
Answer:
Gas is one of the state of pure substance and it have no shape and also it is non comprsable
Liquid is state of pure substance and it occupies the shape of its body and its slightly comrisable
Solid is the other physical state and it have definite volume and shape and also it is highly comrisable
Drawing of circuit in series, with 6 lamps , 3 electrical equipment, 1 power source, and is opened
The circuit is drawn given below
When the key is 'on' then the current is flowing and all the lamps will glow
Resistance is used to resist the flow of current.
Ammeter is used to measure the current flowing in the circuit
A stone is projected upwards to an angle of 30 degrees to the horizontal from the top of a tower of height 100m and it hits the ground at a point Q. If the initial velocity of projection is 100m/s. Calculate the maximum height of the stone above the ground.
The maximum height of the stone above the ground is 227.55 m.
What is the maximum height reached by the stone?
The maximum height reached by the stone is calculated by applying the following kinematic equation.
H = u²sin²θ/2g
where;
u is the initial velocity of the stoneθ is the angle of projection of the stoneg is acceleration due to gravitySubstitute the given parameters and solve for the maximum height reached by the stone.
H = (100² x (sin 30)²)/(2 x 9.8)
H = 127.55 m
Height of the stone above the ground = 127.55 m + 100 m = 227.55 m
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What will happen to the velocity of light ray when it passes from glass to air?
Answer:
So, when light passes from air to glass, its speed decreases. Q.
Explanation:
ray from the ghostbusters
A 32.3 kg mass ( ) is suspended by the cable assembly as shown in the figure. The cables have no mass of their own. The cable to the left ( 1 ) of the mass makes an angle of 0.00∘ with the horizontal, and the cable to the right ( 2 ) makes an angle ( 2 ) of 38.5∘ . If the mass is at rest, what is the tension in each of the cables, 1 and 2 ? The acceleration due to gravity is =9.81 m/s2 .
The tension in each of the wires, number 1 and number 2, respectively
T_1=398.35
This is further explained below.
What is tension ?Generally, At equilibrium, in direction [tex]$\Sigma F_y=0$[/tex]
[tex]\Rightarrow T_2 \sin \theta_2-m g=0 \text {. }[/tex]
[tex]$\Rightarrow \quad T_2 \sin \theta_2=m g$[/tex]
tension, [tex]$T_2=\frac{m g}{\sin \theta_2}$[/tex]
[tex]T_2=\frac{32.3 \times 9.81}{\sin (38.5}^{\circ})} \\\\T_2= {509.00} \mathrm{~N}$.[/tex]
[tex]In x-dirn; $\quad \Sigma F_x=0 \Rightarrow T_2 \cos \theta_2-T_1=0$.[/tex]
[tex]\Rightarrow T_1 &=T_2 \cos \theta_2 \\[/tex]
[tex]T_1 &=509.00\times \cos (38.5^{\circ}\right) \\\\[/tex]
T_1=398.35
In conclusion, the tension in each of the cables, 1 and 2
T_1=398.35
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in some places, insect "zappers", with their blue lights are a familiar sight on a summer's night. these devices use a high voltage to electrocute insects. one such device uses an ac voltage of 4320V, which is obtained from a standard 120,0V outlet by means of a transformer. if the primary coil has 21 turns, how many turns are in the secondary coil?
For a device that electrocutes insects with an AC voltage of 4320V which is obtained from a standard 1200V transformer, If the primary coil has 21 turns, the number of turns in the secondary coil is 76 turns.
Number of turns and output voltage of the primary coil and secondary coil in a transformer are related as the ratio of number of turns in the primary and secondary coil is equal to the ratio of output voltage in the primary and secondary coil respectively.
It can be stated as N₁/N₂ = V₁/V₂, where N₁ and N₂ are the number of turns in the primary and secondary coil respectively, and V₁ and V₂ are the output voltage of primary and secondary coil respectively.
According to the question,
21/N₂ = 1200/4320
N₂ = 75.6
Hence the number of turns should be an integer value, the number of turns in secondary coil are 76 turns.
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The terminal side contains the point (-6, -8). Find sin θ.Question 1 options:.8.6-.8-.6
Answer:
The first option: 0.8
Explanation:
Let us draw the angle.
Now,
[tex]\sin \theta=\frac{opposite}{\text{hypotenuse}}[/tex]Now, using the Pythagoras theorem we find that
[tex]hypotenuse=\sqrt[]{6^2+8^2}[/tex][tex]\begin{gathered} hypotenuse=\sqrt[]{36+64^{}} \\ \Rightarrow hypotenuse=10 \end{gathered}[/tex]Therefore,
[tex]\sin \theta=\frac{opposite}{hypotenuse}=\frac{opposite}{10}[/tex]since for our angle, opposite = 8, we have
[tex]\sin \theta=\frac{8}{10}[/tex][tex]\boxed{\sin \theta=0.8.}[/tex]Therefore, the first choice (0.8) is the correct answer.
draw a sample displacement verses time graph for a car that drives away at a constant speed from a point and has no initial speed. Label any known values.
On a graph of distance vs time, a straight line indicates that a car speed is constant.
Velocity is the pace and direction of an object's movement, whereas speed is the time rate at which an object is travelling along a path. In other words, velocity is a vector, whereas speed is a scalar value.On a speed-time graph, a horizontal line denotes a constant speed. On a speed-time graph, an inclining line denotes an acceleration. The sloping line indicates that the object's speed is varying. Either the item is accelerating or decelerating. Distance and displacement appear to have the same meaning, they actually have very different definitions and implications. Displacement is the measurement of how far an object is out of place, whereas distance refers to ,how much ground an object has covered during its motion.Therefore, when a car moves with constant speed the the distance time graph is in straight line.
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A 100-kg box is sliding down a frictional surface with an acceleration of -2.0m/s².
Determine the magnitude of the friction force on the object.
Answer: 1180 N
Explanation:
F - Fs = ma
⇒ (100 × 9.8) - Fs = 100 × (-2)
⇒ Fs = 1180 N
Light hits a mirror at a 45° angle. It will be reflected at an angle _____.
equal to 45°
greater than 45°
less than 45°
Light hits a mirror at a 45° angle. It will be reflected at an angle equal to 45°
According to law of reflection, when a light ray falls on a surface that is smooth, the angle of reflection will be equal to the angle of incidence. The incident ray, the reflected ray and the normal to the surface all will lie in the same plane.
Reflection is the process in which a light ray gets bounced back after falling on a surface. The angle between in the incident ray and the normal to the surface is known as angle of incidence. The angle between in the reflected ray and the normal to the surface is known as angle of reflection.
Therefore, light hits a mirror at a 45° angle. It will be reflected at an angle equal to 45°
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Situation: A .71-kg billiard ball moving at 2.5 m/s in the x-direction strikes a stationary ball of the same mass. After the collision, the first ball moves at 2.17 m/s, at an angle of 30.0° with respect to the original line of motion. Assuming an elastic collision (and ignoring friction and rotational motion), answer the following questions to find the struck ball's momentum after the collision.1. Calculate the x-component of the first ball's final momentum after the collision in kg*m/s.2. Using the conservation of momentum in the x-direction, find the struck ball's x-component of momentum.3. Calculate the y-component of the first ball's final momentum after the collision in kg*m/s.4. Using the conservation of momentum in the y-direction, find the struck ball's y-component of momentum.
We are given the following situation:
Ball 1 strikes ball 2 and after the collision forms an angle of 30 degrees. To determine the x-component of the final momentum of the first ball we use the following formula:
[tex]P_{1fx}=m_1v_{1f}\cos\theta_1[/tex]Where:
[tex]\begin{gathered} P_{1fx}=final\text{ x-component of the momentum of ball 1} \\ m_1=\text{ mass of ball 1} \\ v_{1f}=\text{ final velocity of ball 1} \\ \theta_1=\text{ angle of ball 1} \end{gathered}[/tex]Now, we substitute the values:
[tex]P_{1fx}=(0.71kg)(2.17\frac{m}{s})\cos(30)[/tex]Solving the operations we get:
[tex]P_{1fx}=1.33kg\frac{m}{s}[/tex]Therefore, the x-components of the momentum of the first ball is 1.33 kgm/s.
Part 2. To determine the x-component of the second ball we will do a balance of momentum in the x-direction:
[tex]P_{10x}+P_{20x}=P_{1fx}+P_{2fx}[/tex]Where:
[tex]\begin{gathered} P_{10x},P_{20x}=\text{ initial momentum in the x-direction of ball 1 and 2} \\ P_{2fx},P_{20x}=\text{ final momentum in the x-direction of ball 1 and 2} \end{gathered}[/tex]Since the second ball starts from rest we have that its initial momentum is zero, therefore:
[tex]P_{10x}=P_{1fx}+P_{2fx}[/tex]Now, we solve for the x-component of the momentum of the second ball:
[tex]P_{10x}-P_{1fx}=P_{2fx}[/tex]The initial momentum of the first ball is the product of its mass and velocity:
[tex]m_1v_{01}-P_{1fx}=P_{2fx}[/tex]Now, we plug in the values:
[tex](0.71kg)(2.5\frac{m}{s})-1.33kg\frac{m}{s}=P_{2fx}[/tex]Solving the operations:
[tex]0.44kg\frac{m}{s}=P_{2fx}[/tex]Part 3. To calculate the y-component of the first ball after the collision we will use the following formula:
[tex]P_{1fy}=m_1v_{1f}\sin\theta[/tex]Now, we plug in the values:
[tex]P_{1fy}=(0.71kg)(2.17\frac{m}{s})\sin(30)[/tex]Solving the operations we get:
[tex]P_{1fy}=0.77kg\frac{m}{s}[/tex]Therefore, the y-component of the momentum of the first ball is 0.77 kgm/s.
Part 4. To determine the y-component of the second ball we use a balance of momentum of the y-components of the balls:
[tex]P_{10y}+P_{20y}=P_{1fy}+P_{2fy}[/tex]Since the first ball is not moving in the y-direction this means that its y-component of the momentum is 0. Since ball 2 is not moving initially this means that its momentum in the y-direction is zero.
[tex]0=P_{1fy}+P_{2fy}[/tex]Now we solve for the y-component of the second ball, and we get:
[tex]-P_{1fy}=P_{2fy}[/tex]Therefore, the y-component of the second ball is:
[tex]-0.77kg\frac{m}{s}=P_{2fy}[/tex]Therefore, the y-component of the second ball is -0.77 kgm/s.
7. Which wave property increases as the energy of a wave increases? *O periodfrequencywavelengthamplitude
Energy is directly proportional to the amplitude.
Thus, the amplitude of the wave increases as the energy of the wave increases.
A ballerina begins a tour jeté (Fig. 11-20a) with angular speed w; and a rotational inertia consisting of two parts:
Ileg = 1.44 kg m² for her leg extended outward at angle
0 = 90.0° to her body and Itrunk = 0.660 kg m² for the rest of
her body (primarily her trunk). Near her maximum height
she holds both legs at angle 0 = 30.0° to her body and has angular speed wf (Fig. 11-20b). Assuming that trunk has not
changed, what is the ratio wf/w?
The term "moment of inertia" refers to the quantity that describes how a body resists angular acceleration. It is calculated by multiplying each particle's mass by its square of the distance from the axis of rotation.
A rigid body's moment of inertia—also referred to as its mass moment of inertia, angular mass, second moment of mass, or—most precisely—rotational inertia—determines the torque.
"I = L / W"
Let Ja be the moment of inertia of the legs.
When leg at 90 degrees,
Ja1 = 1.44 kgm² = m*1²...m = 1.44/1 = 1.44
When both legs at 26.5°
sin 26.5°= 0.45
Ja2 = 2*m*(sen26.5)² = 2.88*.45² = 0.58 kgm²
Jt1 = Ja1 + Jt = 1.44 + 0.76 = 2.15 kgm²
Jt2 = Ja2+Jt = 0.58+0.706 = 1.29 kgm²
wf/w = Jt1/Jt2 = 2.15/1.29 = 1.67
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Using m1v1i + m2v2i = m1v1f + m2v2f how does one describe the concept of left or right mathematically?
The concept of left or right can be described mathematically by using ( m₁*v₁)initial + (m₂*v₂)initial = (m₁*v₁)final + (m₂*v₂)final because as per the conservation of the momentum the, momentum before and after the collision will remain conserved.
What is momentum?It can be defined as the product of the mass and the speed of the particle, it represents the combined effect of mass and the speed of any particle,
P = mv
As given in the problem we have described the concept of left or right mathematically by using( m₁*v₁)initial + (m₂*v₂)initial = (m₁*v₁)final + (m₂*v₂)final
The total momentum of the object before the collision would be equal to the total momentum of the objects after the collision.
To learn more about momentum from here, refer to the link;
brainly.com/question/17662202
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What happens to balls dropped on opposite sides of earth
Answer:
It reaches its top speed at the very center of the globe and continues rocketing toward the other side
Explanation:
So what happens? Well, as the ball drops through the hole it picks up speed - that's the acceleration due to gravity. It reaches its top speed at the very center of the globe and continues rocketing toward the other side.