Give the mathematical expression for coulomb's force if q1, q2 are the magnitude of charges and r is the distance between them.
F=K q1q2/r2
cornvet 500000grams in short form of using suitable prefix.
Answer:
0.5 mega grams
Explanation:
If the range of a projectile's trajectory is six times larger than the height of the trajectory, then what was the angle of launch with respect to the horizontal? (Assume a flat and horizontal landscape.)
Answer:
H = 1/2 g t^2 where t is time to fall a height H
H = 1/8 g T^2 where T is total time in air (2 t = T)
R = V T cos θ horizontal range
3/4 g T^2 = V T cos θ 6 H = R given in problem
cos θ = 3 g T / (4 V) (I)
Now t = V sin θ / g time for projectile to fall from max height
T = 2 V sin θ / g
T / V = 2 sin θ / g
cos θ = 3 g / 4 (T / V) from (I)
cos θ = 3 g / 4 * 2 sin V / g = 6 / 4 sin θ
tan θ = 2/3
θ = 33.7 deg
As a check- let V = 100 m/s
Vx = 100 cos 33.7 = 83,2
Vy = 100 sin 33,7 = 55.5
T = 2 * 55.5 / 9.8 = 11.3 sec
H = 1/2 * 9.8 * (11.3 / 2)^2 = 156
R = 83.2 * 11.3 = 932
R / H = 932 / 156 = 5.97 6 within rounding
The amount of energy released when 45 g of -175° C steam is cooled to 90° C is
A. 101,700 J
B. 317,781 J
C. 419,481 J
D. 417,600 J
Answer:
The answer should be choice B.
How much force is required to pull a spring 3.0 cm from its equilibrium position if the spring constant is 20 N/m?
[tex]\\ \rm\rightarrowtail F=-kx[/tex]
[tex]\\ \rm\rightarrowtail F=-20(0.03)[/tex]
[tex]\\ \rm\rightarrowtail F=-0.6N[/tex]
A source of light emits photons with a wavelength of 8.1 x 10-8 meters. What is the frequency of this light
Answer:
Explanation:
Speed of light v = 3 x 10⁸ m/s
wavelength λ = 8.1 x 10⁻⁸ m
frquency f = v/λ = 3.7 x 10¹⁵ Hz
If a source of light emits photons with a wavelength of 8.1 x 10⁻⁸ meters, then the frequency of the light would be 3.7 × 10¹⁵ Hz, as the wavelength and the frequency of the photon are inversely proportional to each other.
What is Wavelength?It can be understood in terms of the distance between any two similar successive points across any wave for example wavelength can be calculated by measuring the distance between any two successive crests.
C = λν
As given in the problem if a source of light emits photons with a wavelength of 8.1 x 10⁻⁸ meters, then we have to find out the frequency of the light,
The frequency of the light = 3 × 10⁸ / 8.1 x 10⁻⁸
=3.7 × 10¹⁵ Hz
Thus, the frequency of the light would be 3.7 × 10¹⁵ Hz
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A small block, with a mass of 250 g, starts from rest at the top of the apparatus shown above. It then slides without friction down the incline, around the loop, and then onto the final level section on the right. The maximum height of the incline is 80 cm, and the radius of the loop is 15 cm.
a.) Find the initial energy of the block.
b.) Find the velocity of the block at the bottom of the loop.
c.) Find the velocity of the block at the top of the loop.
(a) The initial energy of the block due to its position is 1.96 J.
(b) The velocity of the block at the bottom of the loop is 3.96 m/s.
(c) the velocity of the block at the top of the loop is 3.13 m/s.
Initial energy of the blockThe initial energy of the block due to its position is calculated as follows;
P.E = mgh
P.E = 0.25 X 9.8 X 0.8
P.E = 1.96 J
Conversation of the energyThe velocity of the block at the bottom of the loop is determined by applying the principle of conservation of energy as shown below;
P.Ei + P.Ef = K.Ei + K.Ef
1.96 + 0 = 0 + ¹/₂mvf²
vf² = 2(1.96)/m
vf² = (2 x 1.96) / (0.25)
vf² = 15.68
vf = √15.68
vf = 3.96 m/s
Velocity of the block at top of the loopThe velocity of the block at the top is calculated by applying principle of conservation of energy,
P.Ei + P.Ef = K.Ei + K.Ef
1.96 = mghf + ¹/₂mvf²
where;
hf is the position of the ball at the top of the loop = 2r = 2 x 15 cm = 30 cm = 0.31.96 = 0.25 x 9.8 x 0.3 + 0.5 x 0.25vf²
1.225 = 0.125vf²
vf² = 1.225/0.125
vf² = 9.8
vf = 3.13 m/s
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Alex (31kg) and Cassie (19Kg) sit on a 10kg metre-long see-saw at the local park. The pivot of the see-saw is in the middle of its length. If Cassie sits at one end of the see-saw, where relative to the other end must Alex sit so the net torque is balanced? (unit:metres)
Answer:
M1 g L1 = 19 kg * 9.8 m/s^2 * 5 m = counter clockwise torque - Cassie at left end
M1 g L1 = M2 g L2 for torques to balance
L2 = M1 L1 / M2 = 19 * 5 / 31 = 3.06 M
Alex should sit at 3.1 m from the fulcrum (at 5 m from each end)
1. What is the distance covered by a T-Rex that goes from 0 m/s to 9 m/s in 6.78 seconds? (10
points)
With the use of first and third equation of motion, the distance covered by a T-Rex is 30.51 m
Linear MotionWhen a body is in linear motion, the body is moving in a straight line. some of the parameters to consider are:
Distance coveredSpeedVelocityAccelerationE.T.CGiven that a T-Rex move from 0 m/s to 9 m/s in 6.78 seconds, the distance covered can be found by calculating the acceleration.
Let us use equation 1
V = U + at
9 = 0 + 6.78a
a = 9 / 6.78
a = 1.33 m/[tex]s^{2}[/tex]
Now let us use equation 3
[tex]v^{2}[/tex] = [tex]u^{2}[/tex] + 2as
[tex]9^{2}[/tex] = 2 x 1.33 x S
81 = 2.655S
S = 81/2.655
S = 30.51 m
Therefore, the distance covered by a T-Rex is 30.51 m.
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how r u
________________.
A wire is attached to the ceiling so that the current flows south to north. A student is standing directly below the wire facing north. What is the direction of the B-field (caused by the current in the wire) at this observation point
Answer:
If one wraps the fingers around the wire and points the thumb in the direction of the "conventional" current the fingers will point towards the North pole - the direction of the B-field.
In this case the B-field is pointed "West".
When resting, a person generates about 412005 joules of heat from the body. The person is submerged neck-deep into a tub containing 2124 kg of water at 20.9 °C. If the heat from the person goes only into the water, find the water temperature.
If a person generates about 412005 joules of heat from the body, the water temperature is mathematically given as
t=21.6296C
What is the water temperature.?Question Parameter(s):
The person is submerged neck-deep into a tub containing 2124 kg of water at 20.9 °C
Generally, the equation for the Heat is mathematically given as
Heat gained =Heat loess
Thereofore
mw*cw*(t-2160)=1.5*10^5
[tex]t=21.60+\frac{1.5*10^5}{mw*Cw}\\\\t=21.60+\frac{1.5*10^5}{1.2*10^3*4186}[/tex]
t=21.6296C
In conclusion, the tempreature
t=21.6296C
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What is the x component of a vector that is defined as
45m at -35°?
the x- component of the vector is 36.86 m.
What is a vector?Vectors are quantities that have both magnitude and direcion
To calculate the x-component of the vector, we use the formula below.
Formula:
dx = dcosθ.......... Equation 1Where;
dx = x-component of the vectord = vector between the x-y componentθ = Angle of the vector to the horizontal.From the question,
Given:
d = 45 mθ = -35°Substitute these values into equation 1
dx = 45cos(-35°)dx = 45×0.918dx = 36.86 m.Hence, the x- component of the vector is 36.86 m.
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A magnet gets demagnetized when it is heated.
Answer:
The delicate balance between temperature and magnetic domains is destabilized when a magnet is subjected to high temperatures. If a magnet is exposed to this temperature for an extended length of time or heated over its Curie temperature, it will lose its magnetism and become irreversibly demagnetized.
Explanation:
A train moves from rest to a speed of 22 m/s in 34.0 seconds. What is its acceleration?
Answer: a = 0.647 m/s^2
Explanation:
Acceleration = change in speed / time → a = 22 / 34 → a = .647 m/s^2
At an altitude of 1.3x10^7 m above the surface of the earth an incoming meteor mass of 1x10^6 kg has a speed of 6.5x10^3 m/s. What would be the speed just before impact with the surface of earth?Ignore air resistance.
Show all steps.
Answer:
Approximately [tex]1.1 \times 10^{4}\; {\rm m\cdot s^{-1}}[/tex] if air friction is negligible.
Explanation:
Let [tex]G[/tex] denote the gravitational cosntant. Let [tex]M[/tex] denote the mass of the earth. Lookup the value of both values: [tex]G \approx 6.67 \times 10^{-11}\; {\rm N\cdot m^{2}\cdot kg^{-2}}[/tex] while [tex]M \approx 5.697 \times 10^{24}\; {\rm kg}[/tex].
Let [tex]m[/tex] denote the mass of the meteor.
Let [tex]v_{0}[/tex] denote the initial velocity of the meteor. Let [tex]r_{0}[/tex] denote the initial distance between the meteor and the center of the earth.
Let [tex]r_{1}[/tex] denote the distance between the meteor and the center of the earth just before the meteor lands.
Let [tex]v_{1}[/tex] denote the velocity of the meteor just before landing.
The radius of planet earth is approximately [tex]6.371 \times 10^{6}\; {\rm m}[/tex]. Therefore:
At an altitude of [tex]1.3 \times 10^{7}\; {\rm m}[/tex] about the surface of the earth, the meteor would be approximately [tex]r_{0} \approx 6.371 \times 10^{6}\; {\rm m} + 1.3 \times 10^{7}\; {\rm m} \approx 1.9 \times 10^{7}\; {\rm m}[/tex] away from the surface of planet earth. The meteor would be only [tex]r_{1} \approx 6.371 \times 10^{6}\; {\rm m}[/tex] away from the center of planet earth just before landing.Note the significant difference between the two distances. Thus, the gravitational field strength (and hence acceleration of the meteor) would likely have changed significant during the descent. Thus, SUVAT equations would not be appropriate.
During the descent, gravitational potential energy ([tex]\text{GPE}[/tex]) of the meteor was turned into the kinetic energy ([tex]\text{KE}[/tex]) of the meteor. Make use of conservation of energy to find the velocity of the meteor just before landing.
Initial [tex]\text{KE}[/tex] of the meteor:
[tex]\displaystyle (\text{Initial KE}) = \frac{1}{2}\, m\, {v_{0}}^{2}[/tex].
Initial [tex]\text{GPE}[/tex] of the meteor:
[tex]\displaystyle (\text{Initial GPE}) &= -\frac{G\, M\, m}{r_{0}}[/tex].
(Note the negative sign in front of the fraction.)
Just before landing, the [tex]\text{KE}[/tex] and the [tex]\text{GPE}[/tex] of this meteor would be:
[tex]\displaystyle (\text{Final KE}) = \frac{1}{2}\, m\, {v_{1}}^{2}[/tex].
[tex]\displaystyle (\text{Final GPE}) &= -\frac{G\, M\, m}{r_{1}}[/tex].
If the air friction on this meteor is negligible, then by the conservation of mechanical energy:
[tex]\begin{aligned}& (\text{Initial KE}) + (\text{Initial GPE}) \\ =\; & (\text{Final KE}) + (\text{Final GPE})\end{aligned}[/tex].
[tex]\begin{aligned}& \frac{1}{2}\, m\, {v_{0}}^{2} - \frac{G\, M\, m}{r_{0}} \\ =\; & \frac{1}{2}\, m\, {v_{1}}^{2} - \frac{G\, M\, m}{r_{1}}\end{aligned}[/tex].
Rearrange and solve for [tex]v_{1}[/tex], the velocity of the meteor just before landing:
[tex]\begin{aligned}{v_{1}} &= \sqrt{\frac{\displaystyle \frac{1}{2}\, m\, {v_{0}}^{2} - \frac{G\, M\, m}{r_{0}} + \frac{G\, M\, m}{r_{1}}}{(1/2)\, m}} \\ &= \sqrt{{v_{0}}^{2} - \frac{G\, M}{r_{0}} + \frac{G\, M}{r_{1}}} \\ &= \sqrt{{v_{0}}^{2} - G\, M\, \left(\frac{1}{r_{1}} - \frac{1}{r_{0}}\right)}\end{aligned}[/tex].
Substitute in the values and evaluate:
[tex]\begin{aligned}v_{1} &= \sqrt{{v_{0}}^{2} - G\, M\, \left(\frac{1}{r_{1}} - \frac{1}{r_{0}}\right)} \\ &\approx \sqrt{\begin{aligned}(& 6.5 \times 10^{3}\; {\rm m \cdot s^{-1}}) \\ & - [6.67 \times 10^{-11}\; {\rm N \cdot {m}^{2}\cdot {kg}^{2} \times 5.697\; {\rm kg}}\\ &\quad\quad \times (1 / (6.371 \times 10^{6}\; {\rm m}) - 1 / (1.9371 \times 10^{7}\; {\rm m}))]\end{aligned}} \\ &\approx 1.1 \times 10^{4}\; {\rm m\cdot {s}^{-1}}\end{aligned}[/tex].
(Note that assuming a constant acceleration of [tex]g = 9.81\; {\rm m\cdot s^{-2}}[/tex] would give [tex]v_{1} \approx 1.7\times 10^{4}\; {\rm m\cdot s^{-1}}[/tex], an inaccurate approximation.
There are two space ships traveling next to each other. The first one is 500
Kg and the second one is 498 Kg. Since they are 35 meters apart, what is
the force of gravity between the two space ships?
This question involves the concept of Newton's law of gravitation.
The force of gravity between the two spaceships is "1355.78 N".
Newton's Law Of GravitationAccording to Newton's Law of Gravitation:
[tex]F=\frac{Gm_1m_2}{r^2}[/tex]
where,
F = force of gravity between ships = ?G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²m₁ = mass of first ship = 500 kgm₂ = mass of second ship = 498 kgr = distance between ships = 35 mTherefore,
[tex]F=\frac{(6.67\ x\ 10^{-11}\ N.m^2/kg^2)(500\ kg)(498\ kg)}{(35\ m)^2}\\\\[/tex]
F = 1355.78 N = 1.356 KN
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Two space ships traveling next to each other. The first one is 500 kg and the second one is 498 kg. They are 35 meters apart, the Force of gravity between the two spaceships is 1355.78 N.
It is given that the First spaceship's weight ([tex]m_{1}[/tex]) is 500 kg,
The second spaceship's weight ([tex]\rm m_{2}[/tex]) is 498 kg.
The distance between spaceships (r) is 35 meters.
It is required to find the Force of gravity between these spaceships.
What is Gravitational force?It is defined as the force which attracts any two masses in the universe.
By Newton's law of Gravitation:
[tex]\rm F= \frac{Gm_1m_2}{r^2}[/tex] , Where
[tex]\rm F = The\ force \ of \ gravity \ between \ the \ spaceships\\\rm G= Universal\ Gravitational \ Constant = 6.67 \times 10^{-11} N.m^2/kg^2[/tex]
Putting values in the above formula:
[tex]\rm F = \frac{(6.67\times 10^{-11} N.m^2/kg^2)(500kg)(498kg)}{(35m)^2}[/tex]
F = 1355.78 N = 1.356 KN
Thus, Two spaceships travel next to each other. The first one is 500 kg and the second one is 498 kg. They are 35 meters apart, the Force of gravity between the two spaceships is 1355.78 N.
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A bucket of mass m is attached to a rope that is wound around the outside of a solid sphere (I = 2/5 M^2) of radius R. When the bucket is allowed to fall from rest, it falls with an acceleration of a down. What is the mass of the sphere in terms of m, R, a, and g?
Answer:
[tex]\displaystyle \sqrt{\frac{(5/2)\, (g - a)\, m\, R^{2}}{M^{2}\, a}}[/tex], assuming that the tension in the rope is the only tangential force on the sphere ([tex]g[/tex] denote the gravitational acceleration.)
Explanation:
The forces on the bucket are:
Weight of the bucket: [tex]m\, g[/tex] (downward.)Tension in the rope (upward.)Since the weight of the bucket and the tension from the rope are in opposite directions, the magnitude of the net force would be:
[tex]\begin{aligned} \|\text{Net Force}\| =\; & \|\text{Weight}\| - \|\text{Tension}\| \end{aligned}[/tex].
The upward tension in the rope prevents the bucket from accelerating at [tex]g[/tex] (free fall.) Rather, the bucket is accelerating at an acceleration of only [tex]a[/tex]. The net force on the bucket would be thus [tex]m\, a[/tex].
Rearrange the equation for the net force on the bucket to find the magnitude of the tension in the rope would be:
[tex]\begin{aligned} & \|\text{Tension}\| \\ =\; & \|\text{Weight}\| - \|\text{Net Force}\| \\ =\; & m\, g - m\, a \\ =\; & (g - a)\, m\end{aligned}[/tex].
At a distance of [tex]R[/tex] from the center of the sphere, the tension in the rope [tex](g - a)\, m[/tex] would exert a torque of [tex](g - a)\, m\, R[/tex] on the sphere. If this tension is the only tangential force on this sphere, the net torque on the sphere would be [tex](g - a)\, m\, R\![/tex].
Let [tex]M[/tex] denote the mass of this sphere. The moment of inertia of this filled sphere would be [tex]I = (2/5)\, M^{2}[/tex].
Therefore, the magnitude of the angular acceleration of this sphere would be:
[tex]\begin{aligned}& \|\text{Angular Acceleration}\| \\ =\; & \frac{\|\text{Net Torque}\|}{(\text{Moment of Inertia})} \\ =\; & \frac{(g - a)\, m\, R}{(2/5)\, M^{2}} \end{aligned}[/tex].
The bucket is accelerating at a magnutide of [tex]a[/tex] downwards. The rope around the sphere need to unroll at an acceleration of the same magnitude, [tex]a\![/tex]. The tangential acceleration of the sphere at the surface would also need to be [tex]\! a[/tex].
Since the surface of the sphere is at a distance of [tex]R[/tex] from the center, the angular acceleration of this sphere would be [tex](a / R)[/tex].
Hence the equation:
[tex]\begin{aligned}& \frac{(g - a)\, m\, R^{2}}{(2/5)\, M^{2}} = \|\text{Angular Acceleration}\| = \frac{a}{R} \end{aligned}[/tex].
Solve this equation for [tex]M[/tex], the mass of this sphere:
[tex]\begin{aligned}& \frac{(g - a)\, m\, R^{2}}{(2/5)\, M^{2}} = \frac{a}{R} \end{aligned}[/tex].
[tex]\begin{aligned}M^{2} &= \frac{(g - a)\, m\, R^{2}}{(2/5)\, a} \\ &= \frac{(5/2)\, (g - a)\, m\, R^{2}}{a}\end{aligned}[/tex].
[tex]\begin{aligned}M&= \sqrt{\frac{(5/2)\, (g - a)\, m\, R^{2}}{a}}\end{aligned}[/tex].
Read the text below. Each sentence is about one, two or no energy at all. (5 points) Name the type (s) of energy for each sentence, or leave the space blank (if in the sentence no energy is mentioned). Artan decided to paint the house. He moved the furniture, climbed the stairs, and began work. After two hours he took a break, ate lunch and turned on the radio to listen to some music. When done, turn on a heater to allow the paint to dry as quickly as possible. At dinner everything had ended. a) ............................................................................................................................................ b) ............................................................................................................................................ c) ............................................................................................................................................ d) ............................................................................................................................................ e) ............................................................................................................................................
the radius of a ball is increasing at a rate of 2 mm per second. how fast is the volume of the ball increasing when the diameter is 40 mm
Step 1: Define an equation that relates the volume of a sphere to its radius.
V = 4/3*π*r3
Step 2: Take the derivative of each side with respect to time (we will define time as "t").
(d/dt)V = (d/dt)(4/3*π*r3)
dV/dt = 4πr2*dr/dt
Step 3: We are told in the problem statement that diameter is 100m, so therefore r = 50mm. We are also told the radius of the sphere is increasing at a rate of 2mm/s, so therefore dr/dt = 2mm/s. We are looking for how fast the volume of the sphere is increasing, or dV/dt.
dV/dt = 4π(50mm)2*(2mm/s)
dV/dt = 62,832 mm3/s
Someone please help me !!
Answer:25
Explanation: because higher means less kinetic energ
An electron with an initial speed of 700,000 m/s is brought to rest by an electric field. What was the potential difference that stopped the electron? What was the initial kinetic energy of the electron, in electron volts?
Answer:
See below.
Explanation:
According to the question, we know that,
work done is given by, [tex]W=qV[/tex]
and change in kinetic energy is, Δ [tex]KE=W=1=1/2[mv^{2} ][/tex]
therefore equating both the equations we get,
[tex]qV=1/2[mv^{2} ][/tex] ⇒ [tex]V=\frac{mv^{2} }{2q}[/tex]
m= mass of electron = [tex]9.1*10^{-31} kg[/tex]
q= charge on an electron = [tex]1.6*10^{-19} C[/tex]
v= speed of electron= 700000m/s
substituting the values in the above equation, we get
[tex]V=\frac{9.1*10^{-31} *(700000)^{2} }{2*1.6*10^{-19} } =1.39V[/tex]
(1). the potential difference that stopped the electron is 1.39 volts.
now the kinetic energy equation is : 2 ways[tex]KE=1/2[mv^{2} ]=\frac{9.1*10^{-31} *700000^{2} }{2} =2.22*10^{-19} J\\[/tex]
or [tex]KE=\frac{2.22*10^{-19} }{1.6*10^{-19} } =1.39eV[/tex]
(2). the initial kinetic energy of the electron is 1.39eV.
A 1980-kg car is traveling with a speed of 15.5 m/s. What is the magnitude of the horizontal net force that is required to bring the car to a halt in a distance of 39.2 m
Answer: 6067.5 N
Explanation:
Work = Change in Energy. To start, all of the energy is kinetic energy, so find the total KE using: KE = 1/2(m)(v^2). Plug in 1980 kg for m and 15.5 m/s for v and get KE = 237847.5 J.
Now, plug this in for work: Work = Force * Distance; so, divide work by distance to get 6067.5 N.
The weight of an object on the moon is less because the _______ on the moon is less.
A. mass
B. kilogram
C. newton
D. acceleration of gravity
E. weight
Answer:
acceleration of the gravity
Explanation:
The weight of an object on the moon is less because the acceleration of the gravity on the moon is less.
Which planet is the farthest?
can somebody please help
The amplitude of the wave on the given sinusoidal wave graph is 10 cm.
What is amplitude of wave?
The amplitude of a wave is the maximum displacement of a wave. This is the highest vertical position of the wave from the origin.
Amplitude of the wave is calculated as follows;
From the graph, the amplitude of the wave or maximum displacement of the wave is 10 cm.
Thus, the amplitude of the wave on the given sinusoidal wave graph is 10 cm.
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a 1. You found that the MCB was tied with a thread and the thread was fixed with a nail on the wall in your friend's house. i. Is it good idea to do this? ii. What could be the possible hazard of this? iii. What should have done to keep the circuit safe?
The miniature circuit breaker should rather be fastned to a wall using nails and other neccessary tools.
What is a miniature circuit breaker?A miniature circuit breaker is a circuit breaker that is used in homes as a means of guarding against damage to appliances due to a very high current.
This miniature circuit breaker is also harzardous in the sense that it could lead to an electrical fault related fire outbreak especially when it is being blown freely by wind as you tie it with a thread. Doing this a very bad idea because of the risk of a fire hazard.
The miniature circuit breaker should rather be fastned to a wall using nails and other neccessary tools.
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5 waves with a length of 4m hit the shore every 2 seconds, what is the frequency?
The frequency of the 5 waves with a length of 4m hit the shore every 2 seconds is 2.5 Hz.
What is frequency?This is the number of cycles completed by a wave in one second. The s.i
unit of frequency is Hert (Hz).
From the question, to calculate the frequency of 5 waves with length of 4 m that hit the shores every 2 seconds, we use the formula below.
Formula:
F = n/t........... Equation 1Where:
n = Number of waveF = Frequencyt = timeFrom the question,
Given:
n = 5 waves t = 2 secondsSubstitute these values into equation 1
F = 5/2F = 2.5 Hz.Hence, The frequency of the wave is 2.5 Hz.
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An object of mass m is oscillating with a period T. The position x of the object as a function of time is given by the equation x(t)=Acosωt . The maximum net force exerted on the object while it is oscillating has a magnitude F. Which of the following expressions is correct for the maximum speed of the object during its motion?
What the equation given?
x(t)=Acos[tex]\omega[/tex]tMaximum velocity occurs at the equilibrium position
So
x=0Now
x(0)=Acos0[/tex]x(0)=ANow
As we know the formula
[tex]\\ \rm\rightarrowtail V_max=A\omega[/tex]
These expressions can be used
The maximum speed of the oscillating object will be given by [tex]V_{max}=Aw[/tex]
What is oscillation?An oscillation is defined as the repitative periodic motion of any object about its mean or equilibrium position.
The given equation is as follows:
x(t)=Acost
Maximum velocity occurs at the equilibrium position x=0
x(0)=Acos0
x(0)=A
Hence the maximum velocity will be [tex]V_{max}=Aw[/tex]
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a 2.99 kg sphere makes a perfectly inelastic collision with a second sphere that is intially at rest. the composite moves with a speed equal to one third the original speed of the 2.99kg. what is the mass of the second sphere?
Answer:
5.98 kg
Explanation:
To solve this problem, let use the Linear Momentum Conservation Law:
Before collision: [tex]\sum p=p_{1}+p_{2}=m_{1}v_{1}+m_{2}v_{2}=(2.99v_{1})+0=2.99v_{1}[/tex]
After collision: [tex]\sum p'=p_{1}'+p_{2}'=(2.99+m_{2})(v_{1}/3)[/tex]
So, we obtain:
[tex]\sum p =\sum p' \rightarrow 2.99v_{1}=(2.99+m_{2})(v_{1}/3) \rightarrow 8.97 = 2.99 + m_{2}[/tex]
[tex]m_{2}=8.97-2.99=5.98 kg[/tex]
A 7 kg ball of clay traveling at 12 m/s collides with a 25 kg ball of clay traveling in the
same direction at 6 m/s. What is their combined speed if the two balls stick together
when they touch?
Answer:
Given:
m1 = 7 kg
V1 = 12 m/s
m2 = 25 kg
V2 = 6 m/s
To find:
Combined speed of two balls stick together after collision V = ?
Solution:
According to law of conservation of momentum,
m1V1 + m2V2 = (m1+m2)V
7×12 + 25×6 = (7+25)V
84 + 150 = 32V
V = 234/32
V = 7.31 m/s
Combined speed of two ball is 7.31 m/s
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