Answer: 56.44°
Explanation:
Given:
Let u represent the current speed of the plane, 1.2 MachConverting to SI Units (m/s):
= (1.2 mach)(340 ms^-1 / 1 Mach)
u = 408 m/s
Speed of sound in air, v = 340 m/sFind:
Angle the wave front of the shock wave relative to the plane's direction of motion, θWe have, sinθ = speed of sound / speed of object
sinθ = v / u
θ = sin^-1 (v / u)
= sin^-1 (340 / 408)
θ = 56.44°
Physics occurs all the time but often goes unnoticed. Here is your chance to reflect on physics in action. Other than the examples used in this lesson, think of a time where you witnessed the conservation of angular momentum. Describe the objects that had angular momentum and how angular momentum was conserved. You may also create an example if you cannot recall one in your personal experience.
Angular momentum is conserved in the above examples such as the ice skater, the torque or the rotating effect of the force is almost equal to zero because there is negligible friction between the skates and the ice.
What is principle of conservation of angular momentum?The principle of conservation of angular momentum states that the total angular momentum acting on an object is constant, provided there is no external torque acting on the object.
Angular momentum of a system is conserved as long as there is no net external torque acting on the system.
Examples of conservation of angular momentumthe spinning ice skatersomeone spinning in an office chaira child spinning on roller coasterThus, angular momentum is conserved in the above examples such as the ice skater, the torque or the rotating effect of the force is almost equal to zero because there is negligible friction between the skates and the ice.
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Friction typically __________ objects. A. Speeds up B. Slows down C. Doesn't affect D. Destroys
Answer:
B
Explanation:
friction opposes motion
Friction typically Slows down objects
Option "B"How does friction affect speed?Friction and Speed
While this is almost true for a wide range of low speeds, as speed increases and air friction is reckoned with, it has been found that friction depends not only on speed, but also on speed squared and sometimes on higher powers of friction. speed.
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A converging lens with a focal length of 4.0 cm is to the left of a second identical lens. When a feather is placed 12 cm to the left of the first lens, the final image is the same size and orientation as the feather itself. What is the separation between the lenses?
The distance of separation between the lenses is determined as 12 cm.
Distance of image formed by the first lens
The distance of the image formed by the first lens is calculated as follows;
1/d₁ = 1/f - 1/di₀
1/di₁ = 1/4 - 1/12
1/di₁ = 2/12
1/di₁ = 1/6
di₁ = 6 cm
Find the final position of the imageUse magnification formula as shown below;
M = (di₁di₂)/(d₀₁d₀₂)
1 = (di₁di₂)/(d₀₁d₀₂)
di₂ = (d₀₁d₀₂)/(di₁)
di₂ = (12 x d₀₂)/(6)
di₂ = 2d₀₂
1/f = 1/ d₀₂ + 1/di₂
1/f = 1/ d₀₂ + 1/2d₀₂
1/f = 1.5/d₀₂
d₀₂ = 1.5f
d₀₂ = 1.5(4 cm) = 6 cm
Distance between the two converging lensD = di₁ + d₀₂
D = 6 cm + 6 cm
D = 12 cm
Thus, the distance of separation between the lenses is determined as 12 cm.
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A ball is released from the top of a tower of height h meters. it takes T seconds to reach the ground . what is the position of the ball in T/3 seconds ?
Answer:the
8/9 h
Explanation:
Height = 1/2 a T^2 now change to T/3
now height = 1/2 a (T/3)^2 = 1/9 1/2 a T^2 <===== it is 1/9 of the way down or 8/9 h
A simple pendulum suspended in a rocket ship has a period [tex]T_{0}[/tex]. Assume that the rocket ship is near the earth in a uniform gravitational field.
True or false:
1. If the ship moves upward with a constant velocity, the period increases.
2. If the length of the pendulum is doubled, the new period will be: square root of 2 times [tex]T_{0}[/tex].
3. If the ship accelerates upward, the period increases.
4. If the ship accelerates downward at 9.81 [tex]m/s^{2}[/tex], the pendulum will no longer oscillate.
5. If the mass of the pendulum is halved, the period decreases.
(1) If the ship moves upward with a constant velocity, the period increases, false.
(2) If the length of the pendulum is doubled, the new period will be: square root of 2 times T₀. true.
(3) If the ship accelerates upward, the period increases, true.
(4) If the ship accelerates downward at 9.81 m/s² , the pendulum will no longer oscillate, false.
(5) If the mass of the pendulum is halved, the period decreases, false.
Period of a simple pendulum
The period of a simple pendulum is given as;
T = 2π√(L/g) ------- (1)
where;
L is length of the pendulumg is acceleration due to gravityWhen the ship moves upward with a constant velocity, the period will not change.
When the length of the pendulum is doubledT ∝√L
when, L = 2L₀
T = T₀√2
Thus, if the length of the pendulum is doubled, the new period will be: square root of 2 times T₀.
when ship accelerates upwardt = √2h/g
when the height increases, time or period increaseswhen mass of the pendulum is halvedT = √I/mgd
where;
m is mass of the pendulumwhen the mass is halved, the period increases.
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accelerates uniformly at 2.0 ms2 for 10.0s. Calculate its final velocity
Answer:
The distance is
=
7
m
Explanation:
Apply the equation of motion
s
(
t
)
=
u
t
+
1
2
a
t
2
The initial velocity is
u
=
0
m
s
−
1
The acceleration is
a
=
2
m
s
−
2
Therefore, when
t
=
3
s
, we get
s
(
3
)
=
0
+
1
2
⋅
2
⋅
3
2
=
9
m
and when
t
=
4
s
s
(
4
)
=
0
+
1
2
⋅
2
⋅
4
2
=
16
m
Therefore,
The distance travelled in the fourth second is
d
=
s
(
4
)
−
s
(
3
)
=
16
−
9
=
7
m
what is barometer ???
Answer:
an instrument measuring atmospheric pressure, used especially in forecasting the weather and determining altitude.
Explanation:
hope it helps ya
1. The diagram shows a satellite traveling in uniform circular motion around the Earth.
(a) Give the relation between radius of the orbit and the velocity of the satellite.
(b ) The satellite is kept in orbit by a force. On the diagram draw an arrow to show the
direction of this force.
Answer:
M V R = constant angular momentum is constant because no forces act in the direction of V
Since M (mass) = constant
V R = constant
The force is directed along the gravitational force vector (towards the center of rotation)
You are designing a delivery ramp for crates containing exercise equipment. The 1390 N crates will move at 1.8 m/s at the top of a ramp that slopes downward at 22.0∘. The ramp exerts a 515 N kinetic friction force on each crate, and the maximum static friction force also has this value. Each crate will compress a spring at the bottom of the ramp and will come to rest after traveling a total distance of 5.0 m along the ramp. Once stopped, a crate must not rebound back up the ramp.
The maximum force constant of the spring Kmax is 2337.9 N/m.
What is force constant of a spring?The force constant or spring constant is defined as the force required to stretch or compress a spring such that the displacement in the spring is 1 meter.
Force constant is denoted by K and its unit is N/m.
Force = K * xWhere;
K = spring constant
x = displacement
The work done by the spring is given below as follows:
Work done = Fx/2
Kinetic Energy = mv²/2
Force on an inclined plane = mgsinθ
Total force, F = mgsinθ + frictional force
F = 1390 * sin 22° + 515
F = 1035.7 N
Work done = change in KE
Fx/2 = mv²/2
Fx = mv²
m = 1390/9.81 = 141.692
Solving for x;
x = mv²/F
x = 141.692 * 1.8²/1035.7
x = 0.443 m
The maximum force constant of the spring Kmax = 1035.7/0.443
Kmax = 2337.9 N/m
In conclusion, the maximum force constant of the spring is the ratio of the total force and displacement.
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Note that the complete question is given below:
You are designing a delivery ramp for crates containing exercise equipment. The crates of weight 1490 N will move with speed 2.0 m/s at the top of a ramp that slopes downward at an angle 21.0 ∘. The ramp will exert a 533 N force of kinetic friction on each crate, and the maximum force of static friction also has this value. At the bottom of the ramp, each crate will come to rest after compressing a spring a distance x. Each crate will move a total distance of 8.0 m along the ramp; this distance includes x. Once stopped, a crate must not rebound back up the ramp. Calculate the maximum force constant of the spring Kmax that can be used in order to meet the design criteria
An object with an initial velocity of
0.12rad accelerates at 0.11rad over a
distance of 0.25 radians. What is the
final angular velocity of the object?
rad
S
Answer:
100rad because it angular velocity
a trolley and a sandbag have a combined mass of 4 kg. a bullet with a mass of 150 gram is fired towards the trolley and is lodged in the sand bag immediately after the Collision, on the trolley and sandbag in combination with the bullet, move backwards at 53 metre per second calculate the kinetic energy of the trolley , with the sandbag and bullet ,directly after the Collision
The correct answer is 5828.675 J.
Given combined mass 4kg and mass of bullet 150gm=0.150kg.
Total mass= 4+0.150=4.150kg
Velocity=53 m/s
Kinetic energy = [tex]\frac{1}{2} *m*v^{2}[/tex] =0.5*4.150*[tex]53^{2}[/tex] =5828.675 J
Kinetic energyKinetic energy is a type of power that an item or particle possesses as a result of motion. When an item undergoes work—the transfer of energy—by being subjected to a net force, it accelerates and consequently obtains kinetic energy. A moving object or particle's kinetic energy, which depends on both mass and speed, is one of its characteristics. Any combination of motions, including translation (or travel along a path from one location to another), rotation about an axis, and vibration, may be used as the type of motion.
A body's translational kinetic energy is equal to [tex]\frac{1}{2} *m*v^{2}[/tex] , or one-half of the product of its mass, m, and square of its velocity, v.
a trolley and a sandbag have a combined mass of 4 kg. a bullet with a mass of 150 gram is fired towards the trolley and is lodged in the sand bag immediately after the Collision, on the trolley and sandbag in combination with the bullet, move backwards at 53 metre per second calculate the kinetic energy of the trolley , with the sandbag and bullet ,directly after the Collision
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8- A bottle of water weighs 25 pounds. If the bottle weighs 5 pounds, how many gal does it contain ?
A bottle of water weighs 25 pounds. If the bottle weighs 5 pounds, then it contains 0.599132136584 gal .
A bottle weighs 25 pounds ,
we know that 1 Gallon = 8.34 Lb
so, 25 lb = 2.995660682922 gal
and if the bottle weighs 5 pounds then ,
5 lb = 0.599132136584 gal
hence, 2.4 gal less.
Define pound.
a. a unit of weight equivalent to l6 ounces avoirdupois (453.59237 grams), the fundamental unit of weight in the FPS system;
b. a unit of weight equal to 12 ounces troy or 12 ounces apothecaries' (373.2418 grams).
It is an imperial unit of mass or weight measurement.
Define gallons.
In both imperial and US customary units, the gallon is a unit of volume.
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An object is thrown upward with initial velocity of 30m/s at angle of 30 degree to the horizontal. calculate the components of the initial velocity.
Answer:
Vo (vertical) = Vo sin 30 = 30 m/s / 2 = 15 m/s
Vo (horizontal) = Vo cos 30 = .866 30 m/s = 26 m/s
A 2.0-kg projectile is fired with initial velocity components v0x = 30 m/s and v0y = 40 m/s from a point on the Earth's surface. Neglect any effects due to air resistance.
a) What is the kinetic energy of the projectile when it reaches the highest point in its trajectory?
b) How much work was done in firing the projectile?
Answer:
Kinetic energy of the projectile at the vertex of the trajectory: [tex]900\; {\rm J}[/tex].
Work done when firing this projectile: [tex]2500\; {\rm J}[/tex].
Explanation:
Since the drag on this projectile is negligible, the horizontal velocity [tex]v_{x}[/tex] of this projectile would stay the same (at [tex]30\; {\rm m\cdot s^{-1}}[/tex]) throughout the flight.
The vertical velocity [tex]v_{y}[/tex] of this projectile would be [tex]0\; {\rm m\cdot s^{-1}}[/tex] at the vertex (highest point) of its trajectory. (Otherwise, if [tex]v_{y} > 0[/tex], this projectile would continue moving up and reach an even higher point. If [tex]v_{y} < 0[/tex], the projectile would be moving downwards, meaning that its previous location was higher than the current one.)
Overall, the velocity of this projectile would be [tex]v = 30\; {\rm m\cdot s^{-1}}\![/tex] when it is at the top of the trajectory. The kinetic energy [tex]\text{KE}[/tex] of this projectile (mass [tex]m = 2.0\; {\rm kg}[/tex]) at the vertex of its trajectory would be:
[tex]\begin{aligned} \text{KE} &= \frac{1}{2}\, m\, v^{2} \\ &= \frac{1}{2} \times 2.0\; {\rm kg} \times (30\; {\rm m\cdot s^{-1}})^{2} \\ &= 900\; {\rm J} \end{aligned}[/tex].
Apply the Pythagorean Theorem to find the initial speed of this projectile:
[tex]\begin{aligned}v &= \sqrt{(v_{x})^{2} + (v_{y})^{2}} \\ &= \left(\sqrt{900 + 1600}\right)\; {\rm m\cdot s^{-1}} \\ &= 50\; {\rm m\cdot s^{-1}}\end{aligned}[/tex].
Hence, the initial kinetic energy [tex]\text{KE}[/tex] of this projectile would be:
[tex]\begin{aligned} \text{KE} &= \frac{1}{2}\, m\, v^{2} \\ &= \frac{1}{2} \times 2.0\; {\rm kg} \times (50\; {\rm m\cdot s^{-1}})^{2} \\ &=2500\; {\rm J} \end{aligned}[/tex].
All that energy was from the work done in launching this projectile. Hence, the (useful) work done in launching this projectile would be [tex]2500\; {\rm J}[/tex].
*example of previous incorrect/correct answer on similar question) Inside a vacuum tube, an electron is in the presence of a uniform electric field with a magnitude of 330 N/C.
(a) What is the magnitude of the acceleration of the electron (in m/s²)?
__________m/s²
(b) The electron is initially at rest. What is its speed (in m/s) after 8.00 ✕ 10−⁹ s?
______m/s
The final speed of the electron is 4.64 * 10^5 m/s.
What is the speed of the electron?Given that the mass of the electron is obtained as 9.11 * 10^-31 Kg, we have that the charge of the electron is 1.6 * 10^-19 C.
E= F/q
F = Eq
F = 330 N/C * 1.6 * 10^-19 C
F = 5.28 * 10^-17 N
F = ma
a = F/m = 5.28 * 10^-17 N/ 9.11 * 10^-31 Kg
a = 5.8 * 10^13 m/s^2
Using
v = u + at
u = 0 m/s because the electron was initially at rest
v = at
v = 5.8 * 10^13 * 8.00 ✕ 10−⁹ s
v = 4.64 * 10^5 m/s
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A mass M is suspended from a spring and oscillates with a period of 0.840 s. Each complete oscillation results in an amplitude reduction of a factor of 0.96 due to a small velocity dependent frictional effect. Calculate the time it takes for the total energy of the oscillator to decrease to 0.50 of its initial value.
The energy becomes 0.50 times in 6.72 s.
Let E represent the oscillator's initial energy, Et be the energy's final value at time t, where A is its beginning amplitude, At amplitude at time t, be. as the oscillator's energy increases to 0.50 times its initial value. We can replace the oscillator's total energy for the energy at time t to obtain the amplitude as shown below.
Et=0.50E
1
k(4₂)² = (0.5) - kA²
(4₂)² = (0.5) A²
At = 0.71A
So, the amplitude of the oscillator becomes 0.71 times its initial ar
0.71A = = A(0.96)¹2
log(0.71)
log(0.96)
8.4
n=
So, the time taken for n oscillation is obtained as,
t = n (0.800 s)
= (8.4) (0.800)
= 6.72 s
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Can we take water instead of clock oil in Milikan oil drop experiment. Explain.
Answer:
No you can't cuz,if you put water instead of clock oil in Millikan oil drop your experiment will fail and it won't turn out the way you wanted it to be
If the gravitational potential energy of an object 10 m above the ground is 50 J, what is its Ep, if it moves to 30 m above the ground?
Answer:
150 J
Explanation:
Moving 3 times higher will increase the P E x 3 = 150 J
Why do batteries discharge more quickly in cold weather?
give me answer
who will give me answer in easy language i will give him 5 star ⭐⭐⭐
Answer:
lowering the ambient temperature causes chemical reaction to proceed more slowly,so a battery used in a low temperature produces less current than that at high temperature . As cold batteries run down they quickly reach the point where they cannot deliver enough current to keep up the demand.. hence discharge more quickly in cold weather..
A uniform plank of length 2.00 m and mass 30.23 kg is supported by three ropes. A 700 N person is a distance, of 0.55 m from the left end. A) find the magnitude of the tension in the vertical rope on the left end. Give your answers in newtons. B) find the magnitude of the tension in the rope in the right end. Give your answers in newtons C) find the magnitude of the tension in the horizontal rope on the left end. Give your answers in newtons
a. The magnitude of the tension, T₂ in the vertical rope on the left end is T₂ = 655.62 N
b. The magnitude of the tension in the rope in the right end is T₁ = 530.4 N
c. The magnitude of the tension in the horizontal rope on the left end, T₃ is T₃ = 406.3 N
What is tension force?Tension force refers to a pulling force that is exerted by a string or cable about an axis.
a. The magnitude of the tension, T₂ in the vertical rope on the left end is given as follows:
Taking moment about the vertical axis
T₂ = 30.23 * 9.81 + 700 - T₁ * Sin40°
Solving for T₁ by taking the left end as the pivot;
T₁ Sin 40° * 2.00 = 700 * 0.55 + (30.23 * 9.81) * 1.0
T₁ * 1.285 = 681.5563
T₁ = 530.4 N
Therefore;
T₂ = 30.23 * 9.81 + 700 - 530.4 * Sin 40°
T₂ = 655.62 N
b. From calculation, the magnitude of the tension in the rope in the right end is T₁.
T₁ = 530.4 N
c. The magnitude of the tension in the horizontal rope on the left end, T₃ is determined thus:
Taking moments about the left end in the horizontal direction;
T₃ = T₁ * cos 40°
T₃ = 530.4 N * cos 40°
T₃ = 406.3 N
In conclusion, the tension at the rope in the various ends is determined by taking moments about the left end.
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Write a properly formatted hypothesis statement to answer this question: How does the amount of salt added to ice affect the rate at which the ice will melt?
Specify how you plan to change the independent variable by using terms such as increase or decrease. Also, specify how the dependent variable will change in response by using terms such as increase, decrease, or stays the same.
Criteria pts
Correct placement of IV 5
Correct placement of DV 5
If, then format 5
IV indicates either "increases" or "decreases" 5
DV indicates either "increases", "decreases", or "stays the same" 5
The hypothesis will be:
H₀ = The amount of salt added to ice will not affect the rate at which the ice will melt.
H₁ = The amount of salt added to ice will affect the rate at which the ice will melt.
The independent variable which is can be changed by increasing the rate of salt added to the equation.
The dependent variable which is ice will change or melt in response as it will decrease if the rate of the salt added increases.
What is the effect of salt on the melting temperature of ice?Salt does not really lower the temperature of an ice cubes, it is known to just lowers their freezing point, that is lowers their melting point.
Note that if salt is around, ice cubes are known to be colder to be solid, and they tend to melt at a temperature that is said to be lower than the freezing point of pure water.
If the ionic compound salt is known to be added, it tends to lowers the freezing point of the water, which implies that the ice on the ground is not able to freeze that layer of water at all.
Therefore, The hypothesis will be:
H₀ = The amount of salt added to ice will not affect the rate at which the ice will melt.
H₁ = The amount of salt added to ice will affect the rate at which the ice will melt.
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quations (3.4) and (3.5) are equivalent expressions for Lagrange's equations.
Exercise 3.1 Using the Nielsen form, determine the equation of motion
for a mass m connected to a spring of constant k.
Exercise 3.2 Using the Nielsen form, determine the equations of motion
for a planet in orbit around the Sun. (Answer: mr - mro² = - GMm and
mrö +2mr00.)
3.2 Hamilton's principle
echanical system composed of N particles can be described by n = 3N
inande
Answer:Using the Nielsen form, determine the equation of motion for a mass m connected to a spring of constant k. Exercise 3.2 Using the
Explanation:
Your car rides on springs, so it will have a natural frequency of oscillation. (Figure 1) shows data for the amplitude of motion of a car driven at different frequencies. The car is driven at 29 mph over a washboard road with bumps spaced 12 feet apart; the resulting ride is quite bouncy.
1)Determine the frequency of the oscillation, caused by the bumps. 1 mile is 5280 feet.
2)Should the driver speed up or slow down for a smoother ride?
Answer:
See below
Explanation:
29 mile/hr * 5280 f/mile / 3600 s/hr = 42.53 ft/ sec
42.53 ft / sec / 12 feet = 3.54 cycles / sec = 3.54 Hz
for frequency I suppose....where is the referred to figure?
Suppose you need your silicon circuit element to run continuously for 3 minutes before it shuts off long enough to cool back down to its initial temperature. If the circuit element can withstand a temperature change of 5.1 ∘C without being damaged, what is the maximum rate at which energy can be added to the circuit element?
The maximum rate at which energy can be added to the circuit element mathematically given as
[tex]MER=5.044 \times 10^{-4} \mathrm{~J} / \mathrm{sec}[/tex]
What is the maximum rate at which energy can be added to the circuit element?Generally, the equation for P is mathematically given as
[tex]P=\ln s \frac{\Delta T}{\Delta t}[/tex]
Therefore
[tex]Rate\ of\ Change\ of\ Temp =\frac{p}{lnS}[/tex]
[tex]\frac{p}{lnS}=\frac{7.4 \times 10^{-3}}{23 \times 10^{-6} \times 705}[/tex]
[tex]\frac{p}{lnS}=0.456^{\circ \mathrm{c}} / \mathrm{sec}[/tex]
Max temp Change
[tex]MaxT=5.6^{\circ} \mathrm{C}[/tex]
[tex]\text { time }=3 \times 60[/tex]
t=180s
In conclusion, Max Energy Rate
[tex]MER =23 \times 10^{-6} \times \frac{301 \times 5.6}{180}[/tex]
[tex]MER=5.044 \times 10^{-4} \mathrm{~J} / \mathrm{sec}[/tex]
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A thin spherical shell of mass M and radius r is allowed to roll from the edge of a hemispherical bowl of radius R = 80.0 cm. It rolls down with no slipping.
1) Find the speed of the center of mass of the spherical shell when it is at the bottom of the bowl, if r is very small.
2) Repeat part 1) if r = 10.0 cm. Moment of inertia of a thing spherical shell if 2/3Mr^2.
(a) The speed of the center of mass of the spherical shell when it is at the bottom of the bowl is 3.1 m/s when the radius is 80 cm.
(b) The speed of the center of mass of the spherical shell when it is at the bottom of the bowl is 1.1 m/s when the radius is 10 cm.
Speed of the shell at the bottom of the bowl
The speed of the shell at the bottom of the bowl is calculated by applying the principle of conservation of energy.
K.E(rot) + K.E(trans) = P.E
where;
P.E is the potential energy of the ball at the initial positionK.E(rot) is rotational kinetic energyK.E(trans) is translation kinetic energy¹/₂mv² + ¹/₂Iω² = mgh
where;
I is moment of inertia of the spherical shellh is the height of fallv is the speed at the bottomω is angular speed¹/₂mv² + ¹/₂(²/₃Mr²)(v/r)² = mgh
¹/₂v² + ¹/₂(²/₃r²)(v²/r²) = gh
¹/₂v² + ¹/₂(²/₃)(v²) = gh
¹/₂v² + ¹/₃v² = gh
⁵/₆v² = gh
v² = 6gh/5
v = √(6gh/5)
Let the vertical height from the edge of bowl to the bottom , h = R = 80 cm
v = √(6 x 9.8 x 0.8 /5)
v = 3.1 m/s
When the radius = 10 cmv = √(6gh/5)
v = √(6 x 9.8 x 0.1 /5)
v = 1.1 m/s
Thus, the speed of the center of mass of the spherical shell when it is at the bottom of the bowl is 3.1 m/s when the radius is 80 cm.
The speed of the center of mass of the spherical shell when it is at the bottom of the bowl is 1.1 m/s when the radius is 10 cm.
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Suppose that the math man, a super here that fights crime with math and physics, can decelerate the rate of gravity. During a recent fight with a diamond thief, Math man fell from the top of a 500 meter building. The equation D=t^2*.25. How long us he falling from the top of the building to the ground?
Answer:
44.7 seconds
Explanation:
D = 500 m
500 = .25 t^2
500/.25 = t^2
2000 = t^2
t = 44.7 seconds
A 0.350-kg ice puck, moving east with a speed of 5.22 m/s , has a head-on collision with a 0.950-kg puck initially at rest. Assume that the collision is perfectly elastic.
a)What is the speed of the 0.350- kg puck after the collision?
Express your answer to three significant figures and include the appropriate units.
b)What is the direction of the velocity of the 0.350- kg puck after the collision?
c)What is the speed of the 0.950- kg puck after the collision?
Express your answer to three significant figures and include the appropriate units.
d)What is the direction of the velocity of the 0.950- kg puck after the collision?
A 0.350-kg ice puck, moving east with a speed of 5.22 m/s , has a head-on collision with a 0.950-kg puck initially at rest then
(a) The speed of the 0.350- kg puck after the collision - 2.40 m/s
(b) The direction of the velocity of the 0.350- kg puck after the collision is towards West.
c) The speed of the 0.950- kg puck after the collision is 2.82 m/s
d) The direction of the velocity of the 0.950- kg puck after the collision is towards East.
Given:Mass of ice puck, m₁ = 0.350 kg
Mass of another puck, m₂ = 0.950 kg
Velocity of ice puck, v₁ = 5.22 m/s
Velocity of another puck, v₂ = 0 m/s
[tex]v^{'}_1= ?[/tex]
[tex]v^{'}_2= ?[/tex]
[tex]m_{1} v_{1} + m_{2} v_{2 }= m_{1} v^{'} _{1} + m_{2} v^{'}_{2 }[/tex]
[tex]v_{1} - v_{2} = - ( v^{'}_1 - v^{'}_2 )\\v_{1} - v_{2} = - v^{'}_1 + v^{'}_2\\v^{'}_2 = v^{'}_1 + v_{1}\\\\\\m_{1} v_{1} + m_{2} v_{2 }= m_{1} v^{'} _{1} + m_{2} v^{'}_{1 } + m_{2} v_{1 }[/tex]
[tex](0.35)(5.22) + 0 = 0.350 v^{'} _{1} + 0.950 v^{'}_{1 }+ (0.950)(5.22)\\1.827 = 0.350v^{'} _{1} + 0.950v^{'}_{1 } + 4.959\\1.827 - 4.959 = 1.3 v^{'} _{1}\\- 3.132 = 1.3 v^{'} _{1}\\v^{'} _{1} = -2.40 m/s\\\\\\[/tex]
Therefore, the speed of the 0.350- kg puck after the collision is - 2.40 m/s.
[tex]v^{'}_2 = v^{'} + v_1[/tex]
[tex]= -2.40+5.22[/tex]
[tex]= 2.82 m/s[/tex]
Therefore, the speed of the 0.950- kg puck after the collision is 2.82 m/s.
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It's your first day on the job without your mentor as a member of mission control.
Your shift begins and your counterpart, who happens to be in their 10th year in this
position, speaks to you for about 45 minutes as they end their shift and you begin
yours. What were they talking to you about?
getting you up to date with recent mission developments
sanitation protocol that all employees must memorize
connecting you with each of the astronauts in space at the time
how to change settings for each system that's in use
Explanation:
getting me up to date with recent mission development
Consider the f(x) = Acos(x) function shown in the figure in blue color. What is the value of amplitude A for this function?
As a reference the g(x) = cos(x) function is shown in red color, and green tick marks are drawn at integer multiples of π.
The amplitude of the red colored wave is 1 unit and the amplitude of the red colored wave is 2.1 unit.
What is amplitude of a wave?
The amplitude of a wave is the maximum displacement of the wave. It can also be described at the maximum upward displacement of a wave curve.
Amplitude of the red colored waveFrom the graph, the amplitude of the red colored wave is 1 unit.
Amplitude of the blue colored waveFrom the graph, the amplitude of the red colored wave is 2.1 unit.
Thus, the amplitude of the red colored wave is 1 unit and the amplitude of the red colored wave is 2.1 unit.
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Consider the f(x) = cos(x-C) function shown in the figure in blue color, where 0 ≤ C ≤ 2π. What is the value of parameter C for the function in the figure?
As a reference the g(x) = cos(x) function is shown in red color, and green tick marks are drawn at integer multiples of π.
The value of parameter C for the function in the figure is 2.
What is amplitude of a wave?The amplitude of a wave is the maximum displacement of the wave. It can also be described at the maximum upward displacement of a wave curve.
f(x) = Acos(x - C)
where;
A is amplitude of the waveC is phase difference of the waveWhat is angular frequency of a wave?Angular frequency is the angular displacement of any element of the wave per unit time.
From the blue colored graph; at y = 1, x = -2 cm
1 = cos(2 - C)
(2 - C) = cos^(1)
(2 - C) = 0
C = 2
Thus, the value of parameter C for the function in the figure is 2.
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