Answer:
d=1/2 (a)(t^2)
D = distance
A = acceleration
T = time
acceleration due to gravity is 32 ft/second
so, d=1/2 (32)(5^2)
d=16(25)
d=400
Explanation:
Answer:
400 ft.
Explanation:
D= 1/2 gt^2
=1/2(-32 ft/sec^2)(5 sec^2)
= -(1/2)(32)(25) ft
D= -400 ft, down
Example Problem
The potential energy of an object is given by U(x) = 8x2 - x4, where U is in joules and x is in
(a) Determine the force acting on this object.
(b) At what positions is this object in equilibrium?
(c) Which of these equilibrium positions are stable and which are unstable?
metres.
111 Unit 2 Concepts and Definitions Prof Mark Lester
Exam Part B Example
A neutron of mass m moving with velocity v collides head-on and elastically with a stationary nucleus of mass M.
(a) Show that the velocity of the nucleus after the collision, U, is given by
U= 2m v (m+M)
(b) Hence show that the neutron loses a fraction f of its energy where
f= 4mM (m+M)
10marks 5 marks
(c) A fast neutron enters a target of carbon nuclei which may be assumed to have masses 12 times that of the neutron. How many head-on collisions will it take
before the neutron loses 95% of its energy?
4 marks
(d) Suggest one reason why in a real reactor a neutron is likely to make more
collisions with the moderator nuclei before losing this much energy
2
1 mark
Answer:
Part A
a) F = -16x + 4, b) x = 0.25 m, c) STABLE
Explanation:
Part A
a) Potential energy and force are related
F = [tex]- \frac{dU}{dx}[/tex]- dU / dx
F = - (8 2x -4)
F = -16x + 4
b) The object is in equilibrium when the forces are zero
0 = -16x + 4
x = 4/16
x = 0.25 m
c) An equilibrium position is called stable if with a small change in position, the forces make it return to the initial position, in case the forces make it move away it is called unstable.
In this case there is only one equilibrium point
by changing the position a bit
x ’= x + Δx
we substitute
F ’= - 16 x’ + 4
F ’= - 16 (x + Δx) + 4
F ’= (-16x +4) - 16 Δx
at equilibrium position F = 0
F ’= 0 - 16 Δx
we can see that the body returns to the equilibrium position, therefore it is STABLE
PART B
This is an exercise in body collisions, let's define the system formed by the two bodies in such a way that the forces during the collisions are internal and the moment is conserved
initial instant. Before the shock
p₀ = m v
final instant. After the crash
p_f = (m + M) v_f
We have two possibilities: an elastic collision in which the bodies separate, each one maintaining its plus, and an INELASTIC collision where the neutron is absorbed by the nucleus and the final mass is M '= m + M, in this case they indicate that the collision is elastic
p₀ = pf
mv = mv ’+ M v_f
in the case of the elastic collision, the kinetic energy is conserved
K₀ = K_f
½ m v² = ½ m v’² + ½ M v_f²
we write the system of equations
mv = mv ’+ M v_f (1)
m (v² -v'²) = M v_f ²
m (v - v ’) = M v_f
m (v-v ’) (v + v’) = M v_f
v + v ’= v_f
we substitute in equation 1 and solve
v ’=[tex]\frac{m -M }{m+M } \ vo[/tex]
v_f = [tex]\frac{2m}{m+M} \ v_o[/tex]
the mechanical energy of the neutron is
initial
Em₀ = K = ½ m v²
final moment
Em_f = K + U = ½ m v_f ² + U
U is the energy lost in the collision
total energy is conserved
Em₀ = Em_f
½ m v² = ½ m v_f ² + U
U = ½ m (v² -v_f ²)
U = ½ m [v² - ( [tex]\frac{m-M}{m+M}[/tex] v)² ]
U = ½ m v² [1- ( [tex]\frac{m-M}{m+M}[/tex] )² ]
U = ½ m v2 [ [tex]\frac{2M}{m+M}[/tex]]
U = [tex]\frac{2 mM}{m +M } \ v^2[/tex]
Let's do the same calculations for the nucleus
initial Em₀ = 0
final Em_f = K + U = ½ M v_f ² + U
Em₀ = Em_f
0 = K + U
U = -K
U = - ½ M v_f ²
U = - ½ M [ [tex]\frac{2m}{m+M} \ v[/tex] ]²
U = [tex]\frac{2 m M }{m+M} \ v^2[/tex]
We can see that we obtain the same result, that is, the potential energy lost by the neutron is equal to the potential energy gained by the nucleus.
b) the fraction of energy lost
f = U / Em₀
f = 4 m M / m + M
c) let's calculate the fraction of energy lost in a collision
m = 1.67 10⁻²⁷ kg
M = 12 1.67 10⁻²⁷= 20 10⁻²⁷ kg
f = 4 1.6 20 / (1.6+ 20) 10⁻²⁷
f = 5.92 10⁻²⁷ J
the energy of a fast neutron is greater than 1 eV
Eo = 1 eV (1.67 10⁻¹⁹ J / 1eV) = 1.67 10⁻¹⁹ J
Let's use a direct portion rule if in a collision f loses in how many collisions it loses 0.95Eo
#_collisions = 0.95 Eo / f
#_collisions = 0.95 1.67 10⁻¹⁹ / 5.92 10⁻²⁷
#_collisions = 2.7 10⁷ collisions
How long will it take a car to acceleration from 15.2ms to 23.Ms if the car has an average acceleration of 3.2m\s
Answer: 2.43 s
Explanation:
Initial velocity is [tex]u=15.2\ m/s[/tex]
Final velocity [tex]v=23\ m/s[/tex]
Average acceleration is [tex]a_{avg}=3.2\ m/s[/tex]
Average acceleration is change in velocity in the given amount of time
[tex]\therefore a_{avg}=\dfrac{v-u}{t}\\\\\Rightarrow 3.2=\dfrac{23-15.2}{t}\\\\\Rightarrow t=\dfrac{7.8}{3.2}\\\\\Rightarrow t=2.43\ s[/tex]
Thus, 2.43 s is required to acquire that average acceleration with 23 m/s velocity .
The idea that the universe began from a single point and expanded to its current size explains a large number of observations, including those in the table below.
Observations Explained
The universe consists mostly of low-mass elements.
Cosmic microwave background is nearly the same in all directions.
Light from other galaxies shows that these galaxies are moving away from Earth.
In addition, many predictions based on the idea have led to additional observations that support it. Which best describes this idea of the origin of the universe?
Answer:
theory
Explanation:
took the quiz
4. Which of the following statement describes the particle model.
A. The average kinetic energy of the particles changes easily
B. All matter is made up of large triangular particles
C. There are no spaces between the particles
D. The average kinetic energy of all the particles remains constant
E.Other:
A 0.15-mm-wide slit is illuminated by light of wavelength 462 nm. Consider a point P on a viewing screen on which the diffraction pattern of the slit is viewed; the point is at 26.9° from the central axis of the slit. What is the phase difference between the Huygens' wavelets arriving at P from the top and midpoint of the slit?
Answer:
[tex]\triangle \phi=461.5rad[/tex]
Explanation:
From the question we are told that:
Silt width [tex]w=0.15=>0.1510^{-3}[/tex]
Wavelength [tex]\lambda=462nm=462*10^{-9}[/tex]
Angle [tex]\theta=26.9[/tex]
Generally the equation for Phase difference is mathematically given by
[tex]\triangle \phi=\frac{2 \pi}{\lambda}(\frac{wsin\theta }{2})[/tex]
[tex]\triangle \phi=\frac{2 \pi}{462*10^{-9}}(\frac{0.1510^{-3}*sin 26.9 }{2})[/tex]
[tex]\triangle \phi=461.5rad[/tex]
[tex]\triangle \phi=146.89\pi[/tex]
A Bullet Off mass 100 gm is fired From A Gun Off mass 5 Kg. If the backward velocity of the gun's 5 m / s, what is forward velocity of the bullet?
Answer:
250 m/s
Explanation:
The mass of the bullet, m₁ = 100 g = 0.1 kg
The mass of the gun, m₂ = 5 kg
The backward velocity of the gun, v₂ = -5 m/s
Given that the momentum is conserved, we have;
The total initial momentum = The total final momentum
The gun and the bullet are at rest, therefore, we have;
The initial momentum = 0
The total final momentum = m₁·v₁ + m₂·v₂
Where;
v₁ = The forward velocity of the bullet
Therefore, we get;
m₁·v₁ + m₂·v₂ = 0
0.1 kg × v₁ + 5 kg × (-5 m/s) = 0
0.1 kg × v₁ = 5 kg × 5 m/s
v₁ = (5 kg × 5 m/s)/(0.1 kg) = 250 m/s
The forward velocity of the bullet, v₁ = 250 m/s
A ball is at the top of the hill. As the ball rolls down the hill, its total mechanical energy will:
Answer:
To explain what happens with the ball we must remember the Law of Conservation of Energy.
This law states that the energy can be neither created nor destroyed only converted from one form of energy to another.
Then,
At the top of the hill, the potential energy is maximum and the kinetic energy equals to zero.
When the ball starts to roll down the potential energy will be lower and the kinetic energy will have a low value.
At the middle of the hill, both energies have the same values.
At the end of the hill, the potential energy will be equal to zero and the kinetic energy will be maximum.
Correct me if I'm wrong...
hope it helps...
Name One formula that uses joules
Answer:
[tex]{ \bf{power \: { \tt{(watts)}} = \frac{workdone \: { \tt{(joules)}}}{time \: { \tt{(seconds)}}} \: }}[/tex]
definition of matter . A object which cover the place and have mass is called matter
Answer:
you have written the definition so what are you asking
help helphelp it is 90km per hr
Answer:
Explanation:
The equation for acceleration is
[tex]a=\frac{v_f-v_0}{t}[/tex] which is the final velocity minus the initial velocity divided by the time. I first need to put the units all in the same terms. We have the velocity given as km/hr, but the time is given in seconds and that's not gonna work. I will change the velocity to km/sec:
[tex]90\frac{km}{hr}*\frac{1hr}{3600s}=.025\frac{km}{s}[/tex] That's the value we will use for the final velocity of this car in the equation for acceleration:
[tex]a=\frac{.025-0}{10}=.0025\frac{km}{s^2}[/tex]
The second part of this problem asks how far the car travels in this 10 seconds. We just determined that the car can travel .025 km in 1 second, so in 10 seconds the car travels 10(.025) = .25 km
what are MA and VR of a lever?
Explanation:
Mechanical advantage (MA) = Load/Effort. Velocity ratio (VR) = distance effort moves/ distance load moves in the same time
Please help it's for a test that is due right now.
A car of mass 1000kg is traveling 30m/s
a) What is the kinetic energy?
b) How high will it have to travel up a hill to have the same potential as kinetic energy as this speed? Remember Ep-Ek
Answer:
a. 15,000J
b. .76m
Explanation:
KE = (1/2)m*v²
KE = .5*1000kg*30m/s
KE = 15000J
PE = m*g*h
7500J = 1000kg*9.81m/s²*h
7500J = 9810*h
h = .76m
train starting from a railway station and moving with a uniform accleration attains a speed of 90km/hr in 10s .Find the accleration
Options
90m/s2
1m/s2
10m/s2
0.1m/s2
Answer:
a= 2.5m/s²
Explanation:
U=0
V=90km/hr
T= 10s
Convert 90km/hr to m/s
1km= 1000m
1hr= 3600s
(60×60)
therefore, 90km/hr = 90000/3600
90km/hr= 25m/s
From Newton First Equation,
V=U + AT
25=0+ A(10)
25= 10A
25/10 =10A/10
A= 2.5m/s²
Why would an airplane flying at 10,000 meters above the ground have more gravitational potential
energy than the same airplane flying at 1,000 meters above the ground?
Answer:
The gravitational potential energy (gpe) possessed by an object or body is directly proportional to the height of the object or body.
Explanation:
Gravitational potential energy (GPE) is an energy possessed by an object or body due to its position above the earth.
Mathematically, gravitational potential energy is given by the formula;
G.P.E = mgh
Where;
G.P.E represents potential energy measured in Joules.
m represents the mass of an object.
g represents acceleration due to gravity measured in meters per seconds square.
h represents the height measured in meters.
Generally, the gravitational potential energy (gpe) of an object or body is directly proportional to the height of the object or body. Thus, the gravitational potential energy of a body increases as the height of the body increases.
In conclusion, an object with a higher height would have a higher gravitational potential energy.
A child is playing on a swing. As long as he does not swing too high the time it takes him to complete one full oscillation will be independent of
Answer:
We know that for a pendulum of length L, the period (time for a complete swing) is defined as:
T = 2*pi*√(L/g)
where:
pi = 3.14
L = length of the pendulum
g = gravitational acceleration = 9.8 m/s^2
Now, we can think on the swing as a pendulum, where the child is the mass of the pendulum.
Then the period is independent of:
The mass of the child
The initial angle
Where the restriction of not swing to high is because this model works for small angles, and when the swing is to high the problem becomes more complex.
1. A rocket is forced forward by the ______ force of its engines, expelling gases out the rear of the rocket.
There are two forces acting on a rocket at the moment of lift off: Thrust pushes the rocket upwards by pushing gases downwards in the opposite direction.Weight is the force due to gravity pulling the rocket downwards towards the centre of the earth.So I'm thinking the answer is THRUST.
A 1000 kg dragon is at rest sleeping in outer space. A 50 kg unicorn runs into the dragon with a velocity of 600 ms . The final velocity of the dragon is 40 ms . What is the final velocity of the unicorn?
Answer:
Explanation:
This is a Law of Momentum Conservation problem where, in particular, our problem looks like this:
[tex][m_dv_d+m_uv_u]_b=[m_dv_d+m_uv_u]_a[/tex] in other words, the momentum before they collide has to be equal to the momentum after they collide. Knowing that the dragon is initially at rest:
[1000(0) + 50(600)] = [1000(40)m + 50v] and
0 + 30,000 = 40,000 + 50v and
-10,000 = 50v so
v = -200 m/s or
200 m/s in the direction opposite to its initial direction
A car moving east at a velocity of 16.0 m/s collides with a stationary truck with exactly twice the mass. If the two vehicles lock together, calculate the velocity of their combined mass immediately after collision
Answer:
5.33ms-¹
Explanation:
that is the procedure above
what force to be required to accelerate a car of mass 120 kg from 5 m/s to 25m/s in 2s
Answer:1200
Explanation:
F=ma =m(Vf-Vi)/t
F=120(25-5)/2 =1200N
Answer:
[tex]f = m \frac{v1 - v2}{t} \\ = 120 \times \frac{25 - 5}{2} \\ = 120 \times 10 \\ = 1200N \\ thank \: you[/tex]
4. An object is thrown from from the ground upward with an initial speed of 3.75 m/s. How long will the object be in the air before it lands on the ground?
Answer:
Explanation:
There's an easy way to answer this and then an easier way. I'll do both since I'm not sure what you're doing this for: physics or calculus. Calculus is the easier way, btw.
Going with the physics version first, here's what we know:
a = -9.8 m/s/s
v₀ = 3.75 m/s
t = ??
That's not a whole lot...at least not enough to directly solve the problem. What we have to remember here is that at the max height of a parabolic path, the final velocity is 0. So we can add that to our info:
v = 0 m/s. Use the one-dimensional equation that utilizes all that info and allows us to solve for time:
v = v₀ +at and filling in:
0 = 3.75 + (-9.8)t and
-3.75 = -9.8t so
t = .38 seconds. This is how long it takes to get to its max height. Another thing we need to remember (which is why calculus is so much easier!) is that at the halfway point of a parabolic path (the max height), the object has traveled half the time it takes to make the whole trip. In other words, if .38 is how long it takes to go halfway, then 2(.38) is how long the whole trip takes:
2(.38) = .76 seconds. Now onto the calculus way:
The position function is
[tex]s(t)=-4.9t^2+3.75t[/tex] The first derivative of this is the velocity function and, knowing that when the velocity is 0, the time is halfway gone, we will find the velocity function and then set it equal to 0 and solve for t:
v(t) = -9.8t + 3.75 and
0 = -9.8t + 3.75 and
-3.75 = -9.8t so
t = ,38 and multiply that by 2 to find the time the whole trip took:
2(.38) = .76 seconds.
a ball is thrown straight up into the air while the ball is traveleling upwards what are the magnitue and direction
Answer: hi your question is incomplete attached below is the complete question
answer :
magnitude of acceleration : | a | = g = 9.81 m/s^2
direction : a = - g j
Explanation:
Neglecting Air resistance
magnitude of acceleration :
| a | = g = 9.81 m/s^2
Direction of acceleration
a = - g j ( given that the direction of acceleration is against the acceleration due to gravity i.e. in the opposite direction )
What is potential energy? What are some of its examples.
Answer:
the energy stored in an object because of its specific state or position is called its potential energyexamples:-a compressed springWater that is behind a dam.A car that is parked at the top of a hill. a moving car.Etc....
Explanation:
❣️jess bragoli❣️#keep learning!!
Explanation:
POTENTIAL energy is the energy that is stored in an object due to its position relative to some zero position.
A cup of mass 0.4 kg with its base radius of 4 cm is kept on a table. Calculate the pressure exerted by the cup on the table.
Answer:
779.87 N/m²
Explanation:
Pressure = Force (F) / Area (A)
Recall :
Force = ma ; mass * acceleration due to gravity
a = 9.8 m/s² ; mass = 0.4kg
Force, F = 0.4 * 9.8 = 3.92 N
Area = πr² ; r = Radius = 4cm = 4 / 100 = 0.04 m
Area = π0.04² = 0.0050265 m²
Hence,
Pressure = 3.92 / 0.0050265
Pressure = 779.86670 N/m²
Pressure = 779.87 N/m²
A boy im50kg at rest on a skateboard is pushed by another boy who exerts a force of 200 N on him. If the first boy's
final velocity is 8 m/s, what was the contact time?
seconds
Answer:
Time, t = 2 seconds
Explanation:
Given the following data;
Mass, m = 50 kg
Initial velocity, u = 0 m/s (since it's starting from rest).
Final velocity, v = 8 m/s
Force, F = 200 N
To find the time, we would use the following formula;
[tex] F = \frac {m(v - u)}{t} [/tex]
Making time, t the subject of formula, we have;
[tex] t = \frac {m(v - u)}{F} [/tex]
Substituting into the formula, we have;
[tex] t = \frac {50(8 - 0)}{200} [/tex]
[tex] t = \frac {50*(8)}{200} [/tex]
[tex] t = \frac {400}{200} [/tex]
Time, t = 2 seconds
The area of a position-time graph is the
Answer:
mmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmm
Explanation:
..........
A distant planet with a mass of (7.2000x10^26) has a moon with a mass of (5.0000x10^23). The distance between the planet and the moon is (6.10x10^11). What is the gravitational force between the two objects?
A distant planet with a mass of (7.2000x10^26) has a moon with a mass of (5.0000x10^23). The distance between the planet and the moon is (6.10x10^11). What is the gravitational force between the two objects?
Answer:
Explanation:
This is a simple gravitational force problem using the equation:
[tex]F_g=\frac{Gm_1m_2}{r^2}[/tex] where F is the gravitational force, G is the universal gravitational constant, the m's are the masses of the2 objects, and r is the distance between the centers of the masses. I am going to state G to 3 sig fig's so that is the number of sig fig's we will have in our answer. If we are solving for the gravitational force, we can fill in everything else where it goes. Keep in mind that I am NOT rounding until the very end, even when I show some simplification before the final answer.
Filling in:
[tex]F_g=\frac{(6.67*19^{-11})(7.2000*10^{26})(5.0000*10^{23})}{(6.10*10^{11})^2}[/tex] I'm going to do the math on the top and then on the bottom and divide at the end.
[tex]F_g=\frac{2.4012*10^{40}}{3.721*10^{23}}[/tex] and now when I divide I will express my answer to the correct number of sig dig's:
[tex]Fg=[/tex] 6.45 × 10¹⁶ N
How many more neutrons are in a I SOTOPE of copper-14 than in standard carbon atom
Answer:
2 more neutrons
Explanation:
To obtain the answer to the question, let us calculate the number of neutrons in carbon–14 and standard carbon (i.e carbon–12). This can be obtained as follow:
For carbon–14:
Mass number = 14
Proton number = 6
Neutron number =?
Mass number = Proton + Neutron
14 = 6 + Neutron
Collect like terms
14 – 6 = Neutron
8 = Neutron
Neutron number = 8
For carbon–12:
Mass number = 12
Proton number = 6
Neutron number =?
Mass number = Proton + Neutron
12 = 6 + Neutron
Collect like terms
12 – 6 = Neutron
6 = Neutron
Neutron number = 6
SUMMARY:
Neutron number of carbon–14 = 8
Neutron number of carbon–12 = 6
Finally, we shall determine the difference in the neutron number. This can be obtained as follow:
Neutron number of carbon–14 = 8
Neutron number of carbon–12 = 6
Difference =?
Difference = (Neutron number of carbon–14) – (Neutron number of carbon–12)
Difference = 8 – 6
Difference = 2
Therefore, carbon–14 has 2 more neutrons than standard carbon (i.e carbon–12)
Students are completing a lab in which they let a lab cart roll down a ramp. The students record the mass of the cart, the height of the ramp, and the velocity at the bottom of the ramp. The students then calculate the momentum of the cart at the bottom of the ramp.
A 4 column table with 3 rows. The first column is labeled Trial with entries 1, 2, 3, 4. The second column is labeled Mass of Cart in kilograms with entries 200, 220, 240, 260. The third column is labeled Height of ramp in meters with entries 2.0, 2.1, 1.5, 1.2. The fourth column is labeled Velocity at Bottom in meters per second with entries 6.5, 5.0, 6.4, 4.8.
Which trial’s cart has the greatest momentum at the bottom of the ramp?
Answer:
second column
Explanation:
Answer:
Trial 3 is the answer.
Explanation:
In Boolean Algebra zero represent
1) Zero potential
2) Ground potential
3) low potential
4) Both 1 &2
Answer:
Option 1 & 2
Explanation:
The area of mathematics known as Boolean algebra concerns with operations on logical quantities with binary variables. To express truths, Boolean features are transformed as binary numbers: 1 = true and 0 = false. Boolean algebra concerns with logistical processes, whereas fundamental algebra deals with machine based.
the unit of energy is same as that of work i.e joule give reason
"Energy" is the ability to do work.
"Work" is the process of using energy.