Answer:
the answer is 0
Explanation:
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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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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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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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.
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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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°
You are walking toward the back of a bus that is moving forward with a constant velocity. describe your motion relative to the bus and a point on the ground
You are travelling towards the back of the bus. If your walking speed is slower than the speed of the moving bus (which it generally is), your motion relative to a point on the ground will be travelling in the same direction as the moving bus, but at a slower rate.
What is speed?
Speed can be defined as the distance travelled by an object in relation to the time it takes to travel that distance. In other words, it measures how rapidly an object moves but does not provide direction. When we get direction with speed, we term it "velocity."
The SI unit system is most typically used to express speed. Speed is measured in metres per second, or m/s, because distance is measured in metres and time is recorded in seconds.
Speed is measured in metres per second (ms-1) and is symbolised by the letter "s."
The distance travelled is denoted by d and measured in meters (m).
The distance travelled per unit of time is referred to as speed. It is the rate at which an object moves. The magnitude of the velocity vector is represented by the scalar quantity speed. It lacks a sense of direction. A higher speed indicates that an object is travelling faster. Lower speed indicates that it is travelling more slowly. It has zero speed if it is not moving at all.
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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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A car rounding a corner has an acceleration
of 5m/s². If it rounds a corner with twice the
radius at the same speed, what would be its new
acceleration?
5/4 m/s^{2} would be its new acceleration .
What is centrifugal acceleration ?
The simplest case of circular motion is uniform circular motion, where an object moves in a circular trajectory with constant velocity. Note that the linear velocity of an object in circular motion is constantly changing because, unlike velocity, it is constantly changing direction. From kinematics we know that acceleration is a change in velocity, either in magnitude or direction, or both. Therefore, an object in uniform circular motion will always be accelerated even if the magnitude of its velocity is constant. You can feel this acceleration every time you get in the car while cornering. Keeping the handle steady while turning and moving at a constant speed creates a smooth circular motion. What you will notice is a feeling of skidding (or skidding depending on speed) out of the center of the turn.
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what happens to the weight and mass of an object transported from earth to the moon
When an object transported from earth to the moon, its mass remains same but weight becomes 1/6th that on earth..
What is mass?In physics, mass is a quantitative measurement of inertia, a basic characteristic of all matter.
It essentially refers to a body of matter's resistance to changing its speed or location in response to the application of a force. The change caused by an applied force is smaller the more mass a body has.
What is weight?The amount of a body's weight indicates how much gravity is pulling on it. Weight is calculated using the method w = mg. Since weight is a force, it has the same SI unit as a force, which is the Newton (N).
So, mass of an object is intrinsic property of the object - it remains same in earth and in moon. But weight is the amount of gravitational attraction force. Moon has nearly 1/6th gravitational attraction field. So, the weight of the object on moon will be 1/6th that on earth.
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what is the angle between a = 3.0i - 4.0j and b=2.0i +3.0k?
The answer is 70.6°
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:
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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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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Superman must stop a 120−km/h train in 150 m to keep it from hitting a stalled car on the tracks. If the train’s mass is 3.6×105 kg, how much force must he exert? Compare to the weight of the train (give as %). How much force does the train exert on Superman?
The force that Superman must exert is -1.33 * 10⁶ N
The force that must be exerted by Superman is 37.6 % of the weight of the train.
The force the train exerts on superman is 1.33 * 10⁶ N
What is the average acceleration of the train?The average acceleration of the train is determined from the equation of motion as follows:
v² = u² + 2as
where
v is final velocity = 0 m/s
u is initial velocity = 120 km/h
To convert km/h to m/s, we multiply km/h by 1000/3600
u = 120 * 1000/3600
u = 33.3 m/s
a = ?
s = 150 m
a = v² - u² / 2s
a = 0 - 33.3² / 2 * 150
a = 3.696 m/s²
Force that must be exerted will be:
Force = mass * accelerationForce = 3.6 × 10⁵ * 3.696 m/s²
Force = -1.33 * 10⁶ N
b. Weight of train = 3.6 × 10⁵ * 9.81
Weight of train = 3.53 * 10⁶ N
The ratio of the force and the weight of the train = (-1.33 * 10⁶ / 3.53 * 10⁶) * 100%
The ratio of the force and the weight of the train = 37.6%
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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.
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
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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The absolute Temprature of an ideal diatomic gas is quadrupied. What happens to the average sp eed of molecules?
If the absolute temperature of an ideal diatomic gas is quadrupled, then, the average speed of molecules is doubled.
[tex][V_avg=\sqrt(8k *4T)/\pi m],[V_avg=2\sqrt(8kT) \pi m],[V_avg=2v_avg][/tex]
Speed (often abbreviated as "s") is a scalar variable that describes how much an object's location changes over time or how much it changes per unit of time. The average speed of an item in a period of time is the distance traveled by the object divided by the duration of the period.
Time divided by distance are the speed-related metrics.. The meter per second (m/s) is the SI unit of speed, while the kilometer per hour (kph) is the most often used unit of speed in daily life.
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An object of mass 3.0 kg starts from rest and moves along the x-axis. A net horizontal force is applied to the object in the +x direction. The Force-time graph is shown below. What is the net impulse delivered by the applied force?
Answer:
120
Explanation:
because it is horizontal
An object of mass 3.0 kg starts from rest and moves along the x-axis. A net horizontal force is applied to the object in the +x direction. The Force-time graph is shown below. The net impulse delivered by the applied force is 102 joule.
What is force ?A force is an influence that has the power to alter an object's motion. An object with mass can change its velocity, or accelerate, as a result of a force. An obvious way to describe force is as a push or a pull. A force is a vector quantity since it has both magnitude and direction.
Force is used to describe a body's tendency to modify or change its state as a result of an external cause. When force is applied, the body can also alter its size, shape, and direction. kicking a ball, pushing and pulling on the door, or kneading dough are a few examples.
We know the Impulse = Area of the Graph
= ( 6 × 4) + 0.5 × 2 × 6
= 30 sec
For First 4 sec
Acceleration = 6 ÷ 3
= 2 m/sec²
S = 0.5 × 2 × 4²
= 16 m
Therefore, work Done = 6 × 16
= 96 J
Then, Average force
= ( 6 - 0 ) ÷ 2
= 3 N
Acceleration = 1 m/sec²
S = 0.5 × 1 × 2²
= 2 m
Work = 3 × 2
= 6 J
Total Work = 96 + 6
= 102 J
Thus, The net impulse delivered by the applied force is 102 joule.
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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
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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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 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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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.
A rubber band (in the form of a loop) is to be launched horizontally from a height of 1.0m. During the first launch, the rubber band is stretched 2.0 cm and the rubber band lands at a horizontal distance of 3.5 m from the point of launch. For the second launch the rubber band is stretched 4.0 cm. How far does it travel horizontally this time? If the same rubber band were to be stretched 4.0 cm, but aimed vertically, how high does it go?
The rubber band travels horizontally 7.0 cm in second time.
Given parameters:
the rubber band is stretched = 2.0 cm.
And, the rubber band travels = 3.5 cm.
Secondly, the rubber band is stretched = 4.0 cm.
As force is directly proportional with stretching and landing at a horizontal distance, it travels horizontally in second time
= 4.0 cm x 3.5 cm/ 2.0 cm.
= 7.0 cm.
The vertical movement can't be measured as a new gravitational force comes in play in this case.
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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:
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:
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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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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