Initial horizontal speed and flight time of a projectile are proportional to of the initial velocity.
What is projectile motion?A projectile is any object that is sent into space with only gravity acting on it. The primary force affecting a projectile is gravity.
This doesn't mean that other forces don't have an impact; it just means that they have a much smaller one compared to gravity. The trajectory of a projectile is the path it takes after being fired. The projectile is something that is batted or hurled, similar to a baseball.
Initial horizontal speed projectile= u
flight time of a projectile = 2usinθ/g.
So, these two quantities of projectile motion are proportional to the initial velocity.
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What is the general result of the proton-proton chain?
The genral result of the proton-proton chain is the formation of Helium Nucleus i.e., 4 hydrogen combined to form one helium nucleus.
Proton-proton chains are also called p-p chains, proton-proton cycles, or proton-proton reactions.
Roton-Proton Chain.
⁴H ---> He + energy + other
In the proton-proton chain, four hydrogen nuclei (protons) combine to form a helium nucleus. This results in a loss of the original mass. But some energy escapes in the form of neutrinos. First, two hydrogen nuclei (¹H) combine to release a positive electron (e+, positron) and a neutrino (ν) to form a hydrogen 2 nucleus (²H, deuterium).
¹H + ¹H --> ²H + e + ν
The hydrogen 2 nucleus quickly captures another proton to form a helium 3 (³He) nucleus while emitting a gamma (γ) ray.
²H + ¹H --> ³He + γ
From this point, the reaction chain can follow one of several paths, but always leads to a helium-4 nucleus emitting a total of two protons.
³He + He --> ⁴He + ¹H + ¹H
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calculate the energy in electron volts of a photon having a wavelength
a) in the microwave range, 5.00cm
b) in the visible light range, 500nm
c) in the x-ray range, 5.00nm
[tex]E=nhf[/tex]
The energies of the photon in each case are;
a) 2.5 * 10^-5 eV
b) 2.5 eV
c) 2.5 * 10^2 eV
What is the energy of the photon?We know that a photon is known to have an energy that can be measured or calculated by the use of the formula;
E = hc/λ
E = energy of the photon
h = Plank's constant
λ = Wavelength
c = Speed of light
Then;
a) E = 6.6 * 10^-34 * 3 * 10^8/5 * 10^-2 m
E = 3.96 * 10^-24 J or 2.5 * 10^-5 eV
b) E = 6.6 * 10^-34 * 3 * 10^8/500 * 10^- 9 m
E = 3.96 * 10^-19 J or 2.5 eV
c) E = 6.6 * 10^-34 * 3 * 10^8/5 * 10^- 9 m
E = 3.96 * 10^-17 or 2.5 * 10^2 eV
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A 2 kg mass is placed on an inclined plane. A 3 N, a 4 N, and a 5 N force each act
on the mass, as shown on the free body diagram below. There are no other forces
acting. What is the magnitude of acceleration of mass.
When there is no other force is acting other than the given forces the magnitude of acceleration of mass is 6 m/[tex]s^{2}[/tex].
What is Magnitude?Magnitude is defined simply as “distance or quantity.” It depicts the absolute or relative direction or size in which an object moves in the sense of motion.
What is acceleration?Acceleration is the rate of change of the velocity of an object with respect to time.
Acc. to the given data :
Object’s mass = 2kg
Forces acting = 3N, 4N, 5N
Now to find is the magnitude of acceleration of mass.
We use the following equation of Newton’s law
m = F net / a
rearranging the above equation
a = F net / m
now net force F= f1+f2+f3
net F= 3N+4N+5N=12N
so a=Fnet/m
a=12N/2kg
a=6 m/s2
Hence, the magnitude of acceleration of mass is 6 m/[tex]s^{2}[/tex].
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How much work does a 100W motor perform in 5 minutes?
The amount of work done by the 100 W motor in 5 minutes is 30000 J.
What is workdone?Work is said to be done when a force moves a body through a certain distance.
To calculate the amount of work done by the motor, we use the formula below.
Formula:
W = Pt.............. Equation 1Where:
W = Work done by the motorP = Power of the motort = TimeFrom the question,
Given:
P = 100 Wt = 5 minutes = (5×60) seconds = 300 secondsSubstitute these values into equation 1
W = 100×300W = 30000 JHence, the amount of work done by the motor is 30000 J.
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an object travels at a speed of 1 m/s. How far does it travel in 2 seconds?
Answer: 2m
Explanation:
x = v(t)
x = (1 m/s)(2s)
x = 2m
What are the conserved quantities of black holes?
Answer: Hawking radiation is a result of doing quantum mechanics outside a black hole! In short, a stationary observer outside a black hole, according to quantum mechanics, measures a precisely thermal spectrum of radiation originating from the black hole. What form this radiation takes depends on what quantum fields inhabit the universe, but this crucial result would hold in any case; photons or no photons!
Ultimately I don’t think that there’s really any way to visualise what’s going on. For me I just take it as a consequence of doing quantum mechanics near a black hole *. I suppose that I think (and believe that physicists should perhaps most correctly think) of Hawking radiation as a process, not a “noun”. Some people have cooked up explanations involving virtual particle pairs popping out of the vacuum on the horizon, where one exits and one falls in etc. and the outgoing guy is the Hawking radiation but I don’t really buy this, chiefly because I don’t think that virtual particles exist. Hawking’s original calculation doesn’t care about any of this anyway.
Black holes in nature, naively certainly do appear to violate the conservation of information . This is precisely because of the nature of Hawking radiation: it’s exactly thermal. This means that, no matter what goes in, the same spectrum of radiation emerges. Evidently, it’s then naively impossible to reconstruct the information pertaining to the in-falling stuff from what is, in this sense, just whitenoise. Some argue that if you do the calculation carefully, you can in principle recover this information but as far as I’m aware there’s no consensus as to whether or not this is possible.
As far as things falling into the horizon are concerned (such as your photons), there are two camps. One (I agree with them) maintains that due to Einstein's equivalence principle (loosely speaking, that no local patch of spacetime is in any sense “special”), an observer falling through the event horizon wouldn’t notice anything particularly special (though it’d probably be an interesting light-show!), they’d just go right on in. It’s only when they approached the singularity that things would get tense: the tidal forces near the singularity in the centre would ultimately tear any matter to shreds! The second camp support the firewall argument, which claims that the paradox is solved by an impassible wall of extremely high-energy quanta inhabiting the region just behind the horizon. As a relativist I’m less sympathetic to this view since I’d rather like to hope that solving the information paradox doesn’t require giving up Einstein’s equivalence principle (upon which Einstein’s theory of General Relativity is based). Nonetheless, if this is true, anything entering the black hole horizon would be immediately destroyed by this “firewall”.
*The reason that this happens is rather involved, and, most elegantly, requires understanding the path-integral formulation of quantum mechanical states. At the very least, what one can do is to see that the creation/annihilation operators acting on the vacuum state in Minkowski space correspond to those in a thermal state from the point of view of an observer at a fixed distance outside the black hole (and therefore a uniformally accelerating observer). This latter approach is rather ugly but tractable with some work. This is all a consequence of the fact that, in general, the ontology of quantum mechanical states are dependent on a choice of coordinates. In other words, what one observer calls a “vacuum state”, another observer (with a different coordinate system) might call an “excited state”. Indeed, the example at hand is precisely an instance of this.
Explanation:
nshbwlhcj.s,ncdj dhd
Answer: yes
Explanation:
Identify what is happening at each location of the water cycle. location 3
The water cycle moves from one pond to another i.e. from the river to the ocean, or from the ocean to the atmosphere.
How does location affect the water cycle?The water cycle is the path that all water accompanies as it moves around Earth in other states. The warm dry air at this distance then increases evaporation, which leaves the land beneath even drier. There are three basic locations of water storage that occur in the erratic water cycle. Water is stored in the atmosphere; water is deposited on the surface of the earth, and water is stored in the ground. There is no start or end to the water cycle, but for classification purposes, we will start at the sun.
So we can conclude that the water cycle relates to water being exchanged through Earth's land, ocean, and atmosphere.
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4. A cylindrical gas tank 1 m long, inside diameter of 20 cm, is evacuated and then filled with carbon dioxide gas at 25°C. To what pressure should it be charged if there should be 1.2 kg of carbon dioxide?
The pressure of the carbon dioxide is 21.2 kPa.
What is the pressure of the carbon dioxide?We know that the pressure of the gas is the force with which the gas does hit the walls of the container. In this case, we are told that a cylindrical gas tank 1 m long, inside diameter of 20 cm, is evacuated and then filled with carbon dioxide gas at 25°C.
Thus;
Volume of the cylinder = πd^2/4
Volume = 3.142 * (20 * 10^-2)^2/4
= 0.03142 m^3
From the ideal gas law;
PV = nRT
P = pressure
V = volume
n = Number of moles
R = gas constant
T = temperature
Then;
n = 1.2 * 10^3 g/44 g/mol = 27.3 moles
P = nRT/V
P = 27.3 * 0.082 * 298/0.03142
P = 21.2 kPa
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.Given the displacement vectors A = 6f-4j+8k B = 2î+4ƒ^ − 14k
Find the magnitude and the unit vector of the vector 0.5A + 0.5B
Answer:
Explanation:
A = 6i - 4j + 8k
B = 2i + 4j - 14k
A + B = (6+2)i + (-4+4)j + (8-14)k
A + B = 8i +0j -6k
A + B = 8i - 6k
(1/2)(A + B) = 4i - 3k
d = √ (4² + 3²) = 5
e = (4/5)i -(3/4)j
Consider the Atwood Machine, assuming a massless pulley. Assume m1 > m2. Using N2, we found the acceleration of the system is: a = g(m1 − m2)/(m1 + m2) and the tension was equal. Now take into account the mass of the pulley:
a) Suppose m1 = 0.7 kg and m2 = 0.4 kg. What would the system acceleration be for the pulley?
b) What is the net torque acting on the pulley in terms of the tensions T1, T2, and the pulley radius R? What is the net force acting on each mass in terms of T and weight?
c) Use Newton's 2nd, Rotational N2, and assume the string doesn’t “slip” (at = αR) to find the acceleration of the system if the mass of the pulley is 5 kg.
The system acceleration for the pulley having masses m1 = 0.7 kg and m2 = 0.4 kg Is 35.93 [tex]m/s^2[/tex].
What is Newton's Second Law?Newton's second law provides a detailed explanation of how a force can alter a body's motion. It is believed that a body's momentum varies over time at a rate equal to the force acting on it in both direction and amplitude. The combined effects of a body's mass and velocity determine its momentum.
Given:
The mass, m₁ = 0.7 kg,
m₂ = 0.4 kg,
Calculate the system acceleration by the following formula,
a = g(m1 − m2)/(m1 + m2)
Here a is the acceleration.
Substitute the values,
a = 9.8 * (0.7 - 0.4 ) / (0.7 + 0.4)
a = 35.93 [tex]m/s^2[/tex]
Therefore, the system acceleration for the pulley is 35.93 [tex]m/s^2[/tex].
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Part 1/2 : A neutron in a reactor makes an elastic head- on collision with the nucleus of an atom ini- tially at rest.
Assume: The mass of the atomic nucleus is about 12.4 the mass of the neutron. What fraction of the neutron's kinetic energy is transferred to the atomic nucleus?
Part 2/2 : If the initial kinetic energy of the neutron is 6.97 x 10 ^-13 J, find its final kinetic energy. Answer in units of J.
1. The fraction of the neutron's kinetic energy transferred to the atomic nucleus is 8% or 0.08.
2. The final kinetic energy of the neutron KE₂ (neutron), is 6.42 x 10⁻¹³ J.
What is kinetic energy?Kinetic energy is the energy possessed by a body by virtue of its motion.
The fraction of the neutron's kinetic energy that is transferred to the atomic nucleus is calculated as follows:
The final velocity of the atom is found using the principle of conservation of linear momentum.
initial momentum of the neutron = final momentum of the atom
m₁u₁ = m₂u₂
where;
m₁ = mass of the neutron
u₁ = initial velocity of the neutron
m₂ = mass of the atomic nucleus
u₂ = final velocity of the atomic nucleus
m₂ = 112.4 m₁
u₂ = m₁u₁ / m₂
u₂ = m₁u₁ / (12.4 m₁)
u₂ = 0.08 u₁
The initial kinetic energy of the neutron is determined as follows:
KE₁ = ¹/₂m₁u₁²
The final kinetic energy of the atomic nucleus is determined as follows:
KE₂ = ¹/₂m₂u₂²
KE₂ = ¹/₂(12.4 m₁)(0.08u₁)²
KE₂ = 0.08 (¹/₂m₁u₁²)
KE₂ = 0.08 (KE₁)
The fraction of the neutron's kinetic energy transferred to the atomic nucleus is determined as follows:
Fraction = 0.08 (KE₂) / KE₁
Fraction= 0.08
Fraction = 8 %
The final kinetic energy of the neutron is calculated as follows;
KE₂ (neutron) = (1 - 0.08) x (6.97 x 10⁻¹³ J)
KE₂ (neutron) = 6.42 x 10⁻¹³ J
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25pts
When Joel is riding his bicycle, he and the bicycle have a combined mass of 91.4 kg. The coefficient of kinetic friction between the bicycle's tires and the road is 0.61. If the bicycle was to skid to a stop, what would the force of friction be while the bicycle was sliding?
Ff=897 N
Ff=443 N
Ff=546 N
Ff=55.8 N
The Force of friction on the bicycle of mass 91.4 kg, while moving is 546 N.
What is force?Force is the product of mass and acceleration.
To calculate the force of friction while the bicycle moves, we use the formula below.
Formula:
F = mgμ..................... Equation 1Where:
F = Force of frictionm = Massg = Acceleration due to gravityμ = Cofficient of kinetic frictionFrom the question,
Given:
m = 91.4 kgg = 9.8 m/s²μ = 0.61Substitute these values into equation 1
F = 91.4×9.8×09.61F = 546.39F ≈ 546 NHence, the force of friction on the bicycle is 546 N.
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when does Neuroplasticity occur in the brain????
Answer:
Neuroscientists believed that neuroplasticity only manifested itself in childhood, but late 20th-century research suggests that many aspects of the brain can change (or become "plasticized") even in adulthood.
Explanation:
You plan to take a spaceship to the photon sphere
and hover above the black hole to observe the back
of your head. What sort of acceleration will you
experience as you hover at this point? (Answer
qualitatively, e.g., small, comparable to one g, several times g, much bigger than g, incredibly huge.)
The photon rapidly goes towards the 'singularity' at the center of the black hole, and acceleration will increase tremendously towards center of black hole.
The rate at which an item changes its velocity is known as acceleration, a vector variable. If an object's velocity is changing, it is accelerating. A moving object can occasionally alter its velocity by the same amount every second. a moving object that changes its speed by 10 m/s per second. Since the velocity is changing by a fixed amount every second, this is known as a constant acceleration. It is important to distinguish between an item with a constant acceleration and one with a constant velocity. Be not deceived! An object is accelerating if its velocity is changing, whether by a fixed amount or a variable quantity. Additionally, a moving item with a constant speed is not accelerating.
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A force F~ = Fx ˆı + Fy ˆ acts on a particle that
undergoes a displacement of ~s = sx ˆı + sy ˆ
where Fx = 3 N, Fy = −2 N, sx = 5 m, and
sy = 2 m.
Find the work done by the force on the
particle.
Answer in units of J.
Find the angle between F~ and ~s.
Answer in units of ◦
Answer:
Explanation:
Given:
F = Fₓ·i + Fy·j
S = Sₓ·i + Sy·j
Fₓ = 3 N
Fy = - 2 N
Sₓ = 5 m
Sy = 2 m
_________
A - ? - Work
α - ? - Angle
F = 3·i - 2·j
S = 5·i + 2·j
The work is numerically equal to the scalar product of the displacement force
A = (F·S) = 3·5 + (-2)·2 = 15 - 4 = 11 J
Modules:
| F | = √ (3² + (-2)² ) = √ (9 + 4) = √ 13
| S | = √ (5² + 2² ) = √ (25 + 4) = √ 29
Angle:
cos α = (F·S) / ( |F| · |S| ) = 11 / ( √13 · √29) ≈ 0,5665
α ≈ 55.5°
An adventurous ant finds herself at the end of a fan blade when it is switched on. It is a high speed fan with blades measuring 0.30 m long. If she has a mass of 0.22 g and can hold on to the fan blade with a maximum force of 0.0137 N, what is the maximum number of revolutions per minute the fan can run at before she will be flung off?
The maximum number of revolutions per minute the fan can run at before she will be flung off is 1,982.24 rev/min.
What is the maximum angular speed of the fan?
The maximum angular speed of the fan is calculated by applying the following kinematic equation.
From Newton's second law of motion,
F = ma
F = mv/t
F = m(ωr/t)
where;
m is the mass ω is the angular speed in rad/sr is the radius of the bladet is the time of the bladeω/t = F/mr
ω/t = (0.0137 N) / (0.00022 kg x 0.3 m)
ω/t = 207.58 rad/s
ω/t = (207.58 rad/s) x (1 rev / 2π rad) x (60 s / min)
ω/t = 1,982.24 rev/min
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What is the best definition of a
wave
A. A wave is a disturbance that transfers matter.
B. A wave is electric charges that flow through
wires really fast.
C. A wave is a disturbance that transfers energy.
Answer: C
Explanation:
Which of the following statements about energy is true?
I. Kinetic energy cannot be transformed into another type of energy.
II. Kinetic energy can be transformed into potential energy.
III. Potential energy can be transformed into kinetic energy.
IV. Potential energy cannot be transformed into another type of energy.
the RIGHT answer is II and III only - kinetic energy can be transformed into potential, potential energy can be transformed into kinetic energy
The statement that is true about energy is as follows;
Kinetic energy can be transformed into potential energy (option II)Potential energy can be transformed into kinetic energy (option III)What is kinetic energy?Kinetic energy is the energy possessed by an object because of its motion, equal (nonrelativistically) to one half the mass of the body times the square of its speed.
Potential energy is the energy possessed by an object because of its position (in a gravitational or electric field), or its condition (as a stretched or compressed spring, as a chemical reactant, or by having rest mass).
The law of conservation of energy is a principle stating that energy may not be created or destroyed but instead can be transformed from one form to another.
This suggests that kinetic energy as a type of energy can be changed into the resting form called potential energy and vice versa.
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A sprinter with a mass of 68 kg reaches a speed of 8 m/s during a race. Find the sprinter's linear momentum (in kg · m/s). (Enter the magnitude.)
The momentum of the sprinter of mass and velocity of 68 kg and 8 m/s respectively is 544 kgm/s
What is momentum?Momentum can be defined as the product of mass and velocity of a body.
To calculate the linear momentum of the sprinter, we use the formula below.
Formula:
M = mv........... Equation 1Where:
M = Linear momentum of the sprinterm = Mass of the sprinterv = Velocity of the sprinterFrom the question,
Given:
m = 68 kgv = 8 m/sSubstitite the values into equation 1
M = 68×8M = 544 kgm/sHence, the momentum of the sprinter is 544 kgm/s.
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The linear momentum of the sprinter with a mass of 68 kg that reaches a speed of 8 m/s during a race is 544kgm/s.
How to calculate momentum?The momentum of a body in motion is the tendency of a body to maintain its inertial motion. It is the product of its mass and velocity, or the vector sum of the products of its masses and velocities.
The linear momentum of a body can be calculated as follows:
p = m × v
According to this question, a sprinter has a mass of 68kg and reaches a speed of 8 m/s during a race. The momentum can be calculated as follows:
p = 68kg × 8m/s
p = 544kgm/s
Therefore, 544kgm/s is the linear momentum of the sprinter.
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Your partner in caring for the four-year-olds at a day care center is an older male who has been too friendly since you began work there three weeks ago. He usually behaves in a professional manner when you are both at the job site; but on sunny afternoons when you take the children to a nearby park, he talks about personal topics and has lately been insisting that you go out with him. Another co-worker, who said she transferred to caring for the two-year-olds because he was so obnoxious, has warned you about him. You've tried being assertive, but he persists. During the past few days, he has been so physically bold that even the kids are beginning to notice. You are fed up trying to deal with it alone.
To whom should you speak about the problem?
What could be the outcome?
Describe an environment which nurtures and promotes empathy in the early childhood setting.
How can you ensure an inclusive setting in your classroom?
The answers include the following:
We should speak to the director about the problem.The outcome could be punishment, warning or termination of the job.An environment which nurtures and promotes empathy in the early childhood setting should be safe and be able to serve each other.You can ensure an inclusive setting in your classroom by creating a supportive and respectful environment.What is a Daycare?This is an institution which provides supervision and care of infants and young children especially so that parents can focus on their jobs.
The director is the head of the daycare and issues such as this should be reported to him/her so as to enable the best possible resolution.
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A 12-kg sled is lying on a hill with an incline of 21 degrees.
If the sled it not moving what must the coefficient of
static friction be (at least)?
The coefficeint of static friction, given that the 12 Kg sled is lying on the hill with an incline of 21 degrees is 0.38
How do I determine the coefficient of static friction?We know that the coefficient of static friction is related to frictional force according to the following formula:
Frictional force (N) = coefficient of friction (μ) × normal reaction (N)
F = μN
μ = F / N
For inclined plane, we have:
F = mgSineθ
N = mgCosθ
Thus,
μ = mgSineθ / mgCosθ
Recall
Sineθ / Cosθ = Tanθ
μ = Tanθ
Where
m is the mass of objectg is the acceleration due to gravityNow, we shall determine the coefficient of static friction as follow:
Mass of sled(m) = 12 KgAngle of inclination (θ) = 21 degreesCoefficient of static friction (μ) =?μ = Tanθ
μ = Tan21
μ = 0.38
Thus, the coefficient of static friction is 0.38
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Very Important, I need the answer
Answer:
A
Explanation:
A constant velocity means the position graph has a constant slope. It's a straight line sloping up.
A 5 kg block is moved up a 30 degree incline by a force of 50 N, parallel to the incline. The coefficient of kinetic friction between the block and the incline is 0.25. What is the net work done on the block over this distance?
PLEASE HELP ME WITH THIS QUESTION
The net work done on the block over the given distance is 39.4d (joules)
What is the net work done on the block over this distance?The net work done on the block over the given distance is calculated by applying the following equation as shown below;
W(net) = F(net) x d
where;
F(net) is the net force on the blockd is the distance moved by the blockF(net) = F - μmgcosθ
where;
μ is the coefficient of kinetic frictionm is the mass of the blockg is acceleration due to gravityθ is the angle of inclination of the planeF(net) = 50 N - (0.25 x 5 x 9.8 x cos30)N
F(net) = 50 N - 10.6 N
F(net) = 39.4 N
The net work done on the block over the given distance is calculated as;
W = 39.4 N x d
where;
d is the distance moved by the block = length of the inclineW = 39.4d (joules)
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A mobile starts from rest after 20 s reaches a speed of 90 km / h determine its speed and distance traveled
We first apply the data to the problem.
Data:
[tex] \bold{V = 90km/h}[/tex]
[tex] \bold{T = 20s}[/tex]
[tex] \bold{D = ?}[/tex]
Now, we convert km/h to m/s.
Conversion:
[tex] \bold{90km/h * (1000m/1km) * (1h/3600s)}[/tex]
[tex] \boxed{ \boxed{ \bold{V = 25m/s}}}[/tex]
Then, we apply the formula that is.
Formula:
[tex] \bold{D = V * T}[/tex]
To determine the distance traveled we develop the problem.
Developing:
[tex] \bold{D = (25m/s) * (20s)}[/tex]
[tex] \boxed{\boxed{ \bold{D = 500m}}}[/tex]
Its speed is 25 meters per second and the distance traveled is 500 meters.
The rock to the right is sitting at the top of a ramp.
I wonder how much work it required to get that rock up
there.
[ ]
Can you figure it out? (This is not a yes or no question:
solve!)
The amount of work required to move the rock of mass 95 kg up a ramp of 100 m is 93100 J.
What is work?Work can be fined as the product of force and distance.
To calculate the amount of work required to get the rock up, we use the formula below.
Formula:
W = mgh........... Equation 1Where:
W = Amount of workm = Mass of the rockg = Acceleration due to gravityh = Height of the rampFrom the question,
Given:
m = 95 kgh = 100 mg = 9.8 m/s²Substitute these values into equation 1
W = 95×100×9.8W = 93100 JHence, the amount of work required is 93100 J.
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A 76 kg bike racer climbs a 1500-m-long section of road that has a slope of 4.3 ∘ . you may want to review ( pages 296 - 298) . part a by how much does his gravitational potential energy change during this climb
The gravitational potential energy change during this climb is 83,790 J.
What is the change in the gravitational potential energy?
The change in the gravitational potential energy of the bike racer is calculated as follows;
ΔP.E = mgh
where;
m is the mass of the bike racerh is the vertical height travelledg is acceleration due to gravityThe vertical height of the road is calculated as follows;
h = L sinθ
where;
θ is the slope of the roadL is the length of the roadh = 1500 x sin(4.3)
h = 112.5 m
ΔP.E = 76 x 9.8 x 112.5
ΔP.E = 83,790 J
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How would you describe the mass and size the players need to have to play the game?
Answer:
mass and size are the main important to any playing person for more the two extremely don't Matcha the person doesn't play any exercise
A student has a power output of 250 Watts of poweHow much does the student weigh if she travels 15 meters in 30 seconds?
Answer:
Thank you for helping use
8. An ice skater at rest on ice catches a dance partner moving 1.5 m/s during a performance.
The ice skater has a mass of 75 kg and the dance partner has a mass of 50 kg. What is the
speed of the ice skater and dance partner after the collision? (Show your work)
Answer:
[tex]0.60\; {\rm m \cdot s^{-1}}[/tex], assuming that the friction between the skates and the ice is negligible.
Explanation:
Under the assumptions, the total momentum of the two skaters will be conserved. In other words, the sum of the momentum of the two skaters will be the same before and after the collision.
When an object of mass [tex]m[/tex] moves at velocity [tex]v[/tex], the momentum [tex]p[/tex] of that object will be [tex]p = m\, v[/tex].
The [tex]m_{b} = 50\; {\rm kg}[/tex] skater was initially moving with a velocity of [tex]v_{b} = 1.5\; {\rm m\cdot s^{-1}}[/tex]. The momentum of this skater will be:
[tex]m_{b}\, v_{b} = 50\; {\rm kg }\times 1.5\; {\rm m\cdot s^{-1}} = 75\; {\rm kg \cdot m \cdot s^{-1}}[/tex].
Since the [tex]m_{a} = 75\; {\rm kg}[/tex] skater was initially not moving (velocity is [tex]v_{b} = 0\: {\rm m\cdot s^{-1}}[/tex],) the momentum of that skater will be:
[tex]m_{a}\, v_{a} = 75\; {\rm kg }\times 0\; {\rm m\cdot s^{-1}} = 0\; {\rm kg \cdot m \cdot s^{-1}}[/tex].
Thus, the total momentum of the two skaters was [tex]75\; {\rm kg \cdot m \cdot s^{-1}}[/tex] before the collision.
Since the two skaters held on to each other, the two will travel at the same velocity after the collision. Let [tex]v[/tex] denote this velocity. The total momentum of the two skaters after the collision will be [tex]m_{a}\, v + m_{b}\, v = (m_{a} + m_{b})\, v[/tex].
Under the assumptions, momentum will be conserved in the collision. Hence, the total momentum after collision [tex](m_{a} + m_{b})\, v[/tex] should be equal to the total momentum before the collision, [tex]75\; {\rm kg \cdot m \cdot s^{-1}}[/tex]. In other words:
[tex](m_{a} + m_{b})\, v = 75\; {\rm kg \cdot m\cdot s^{-1}}[/tex].
Rearrange this equation and solve for the velocity [tex]v[/tex] of the two skaters after the collision:
[tex]\begin{aligned}v &= \frac{75\; {\rm kg \cdot m\cdot s^{-1}}}{m_{a} + m_{b}} \\ &= \frac{75\; {\rm kg \cdot m\cdot s^{-1}}}{75\; {\rm kg} + 50\; {\rm kg}} \\ &= 0.60\; {\rm m\cdot s^{-1}}\end{aligned}[/tex].
In other words, the two skaters will travel at approximately [tex]0.60\; {\rm m\cdot s^{-1}}[/tex] after the collision.