Answer:
The reason it has few heavier elements is due to the fact that it's stars are widely spaced and this implies that it's stars have very low rate of formation and termination.
Explanation:
The Small Magellanic Cloud, is basically a very tiny galaxy located near the Milky Way. Although it's tiny, it's diameter is approximately 7,000 light-years while it also contains over hundred million stars which are widely spaced.
Now, the reason it has few heavier elements is due to the fact that it's stars are widely spaced and this implies that it's stars have very low rate of formation and termination.
State Newton’s second law of motion. Derive this law mathematically. A 1000 kg vehicle moving with a speed of 20m/s is brought to rest in a distance of 50 metres. Find the acceleration and calculate the force acting on the vehicle.
Answer:
-4000 N
Explanation:
newton second law F=ma
m=1000kg
vi=20m/s
vf=0
d=50 m
vf^2=vi^2+2ad
0=20^2+2a×50
100 a= -400
a= -4 m/S2
F=ma = 1000×-4 = -4000 N
A box slides down a 28.0 degree ramp with an acceleration of 1.25 m/s2. Determine the coefficient of kinetic friction between
Answer:
[tex]\mu=0.39[/tex]
Explanation:
From the question we are told that:
Angle [tex]\theta=28[/tex]
Acceleration [tex]a=1.25m/s^2[/tex]
Generally the equation for Frictional force is mathematically given by
[tex]F=\muN[/tex]
Where
[tex]N=mgcos \theta[/tex]
[tex]N=mgcos 28[/tex]
Since
Friction force is acting against move of box
Therefore
[tex]mgsin(28) - 1.25m = \mu mgcos(28)[/tex]
[tex]\mu=\frac{gsin(28) - 1.25}{gcos(28)}[/tex]
[tex]\mu=0.39[/tex]
a stone is projected horizontally with 20 m/a from top of a tall building. calculate it's position and velocity after 3s neglecting the air resistance
Answer:
1. 45 m
2. 50 m/s
Explanation:
From the question given above, the following data were obtained:
Initial velocity (u) = 20 m/s
Time (t) = 3 s
1. Determination of the position (i.e height)
Time (t) = 3 s
Acceleration due to gravity (g) = 10 m/s²
Height (h) =?
h = ½gt²
h = ½ × 10 × 3²
h = 5 × 9
h = 45 m
2. Determination of the velocity.
Initial velocity (u) = 20 m/s
Time (t) = 3 s
Acceleration due to gravity (g) = 10 m/s²
Velocity (v) =?
v = u + gt
v = 20 + (10 × 3)
v = 20 + 30
v = 50 m/s
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
please help.. i got it wrong on my last attempt
Answer:
The answer is C.
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
The core of transformer is laminated.......
a.to reduce the loss of energy in the form of heat across the transformer
b.to reduce the voltage of AC
c.to decrease the voltage of AC
d. to change the maginetic flux
Answer:
A. to reduce the loss of energy in the form of heat across the transformer
Explanation:
The core of the transformer is laminated to minimise the energy as they interfere with the efficient transfer of energy from the primary coil to the secondary one. The eddy currents cause energy to be lost from the transformer as they heat up the core - meaning that electrical energy is being wasted as heat.
The most powerful empire between the 1500s and 1600s was the __________ Empire.
A.
Ottoman
B.
Mauryan
C.
Roman
D.
Persian
Answer:
A
Explanation:
Answer:
Ottoman
Explanation:
siri told me after I asked
To overcome the problems that blur images and don't provide the best resolution from Earth, astronomers have started using flexible mirrors that change shape many times each second. This technique is called:
Answer:
adaptive optics
Explanation:
simple
Calculate the distance travelled by the car in part Q use the equation distance travelled= average speed x time
Explanation:
distance travelled = average speed x time
=30m/s*100s
=3000m
Answer:
3000m
Explanation:
30m/s*100s
3000m
How did the English bill of rights impact the colonists views of government?
The English Bill of Rights helped to shape the colonists' views of government by promoting the ideas of individual liberty, representative democracy, and limited government.
What is English bill of rights?The English Bill of Rights is a document that was passed by the Parliament of England in 1689. It was a key moment in the development of modern democratic government, as it established certain rights and protections for English citizens, such as the right to bear arms, the right to a fair trial, and the prohibition of cruel and unusual punishment.
The English Bill of Rights also limited the power of the monarchy, requiring the king or queen to obtain the consent of Parliament before levying taxes or making other important decisions. Its influence can still be seen today in many democracies around the world.
Learn more about English bill of rights, here:
https://brainly.com/question/20986147
#SPJ7
three condensers are connected in series across a 150 volt supply, the voltages across them are 40,50 and 60 volts respectively, and the charge on each condenser is 6×10^-8 c.calculate (a) the capacitance of each condenser (b)the effective capacitance of the combination
Answer:
(a) 1.5 nF, 1.2 nF, 1 nF
(b) 0.4 nF
Explanation:
V = 150 V
V' = 40 V, V'' = 50 V, V''' = 60 V, q = 6 x 10^-8 C
(a) C' = q/V' = 6 x 10^-8 / 40 = 1.5 x 10^-9 F
C'' = q/V'' = 6 x 10^-8 / 50 = 1.2 x 10^-9 F
C''' = q/V''' = 6 x 10^-8 / 60 = 1 x 10^-9 F
(b) The effective capacitance is
[tex]\frac{1}{C}=\frac{1}{C'}+\frac{1}{C''}+\frac{1}{C'''}\\\\\frac{1}{C}=\frac{10^9}{1.5}+\frac{10^9}{1.2}+\frac{10^9}{1}\\\\C = 0.4\times 10^{-9} F[/tex]
A car travelling an unbanked curve of radius 200 ft notices a truckstopped on the road ahead. The driverimmediately applies brakes causing the speed of the carto decrease at the rate of 10 ft/s2. If at that instant, the stationary truckis 100 ft ahead (the distance is measured along the path) and the car is travelling at a speed of 40ft/s, whatis the magnitude of the relative velocity ofthe truck perceived by the driver of the car (i.e. from the moving frame of referenceof the car).
Answer:
u = - 40 ft / s
Explanation:
The Galilean relation for the relative velocity is
v ’= v + u
where u is the speed between the two reference frames, v is the speed of the fixed system and v 'the speed of the mobile system.
In this case the truck has a speed with respect to the ground (fixed system) 0 m / s (it is stopped), the car has a speed with respect to the ground of v = 40 ft / s,
u = v'- v
u = 0 - 40
u = - 40 ft / s
the speed perceived by the car if the system is fixed on it is -40 ft / s
Using a scale diagram, calculate the resultant force acting on a sailing boat when an easterly wind provides 2, point, 50, k, N,2.50kN of force, the tide provides 1, point, 20, k, N,1.20kN of force from the direction 30, point, 0, degrees,30.0 ∘ more northerly than the wind. Give your answer to 2 significant figures. Remember that 'an easterly wind' means a wind coming from the East
Answer:
F = 3.6 kN, direction is 9.6º to the North - East
Explanation:
The force is a vector, so one method to find the solution is to work with the components of the vector as scalars and then construct the resulting vector.
Let's use trigonometry to find the component of the forces, let's use a reference frame where the x-axis coincides with the East and the y-axis coincides with the North.
Wind
X axis
F₁ = 2.50 kN
Tide
cos 30 = F₂ₓ / F₂
sin 30 = F_{2y} / F₂
F₂ₓ = F₂ cos 30
F_{2y} = F₂ sin 30
F₂ₓ = 1.20cos 30 = 1.039 kN
F_{2y} = 1.20 sin 30 = 0.600 kN
the resultant force is
X axis
Fₓ = F₁ₓ + F₂ₓ
Fₓ = 2.50 +1.039
Fₓ = 3,539 kN
F_y = F_{2y}
F_y = 0.600
to find the vector we use the Pythagorean theorem
F = [tex]\sqrt{F_x^2 +F_y^2}[/tex]
F = [tex]\sqrt{ 3.539^2 + 0.600^2 }[/tex]
F = 3,589 kN
the address is
tan θ = F_y / Fₓ
θ = tan⁻¹ [tex]\frac{F_y}{F_x}[/tex]
θ = tan⁻¹ [tex]\frac{0.6}{3.539}[/tex]0.6 / 3.539
θ = 9.6º
the resultant force to two significant figures is
F = 3.6 kN
the direction is 9.6º to the North - East
Two parallel conducting plates are separated by 12.0 cm, and one of them is taken to be at zero volts. (a) What is the magnitude of the electric field strength between them, if the potential 5.6 cm from the zero volt plate is 450 V
Answer:
-8.036 kV/m
Explanation:
The electric field E = -ΔV/Δx where ΔV = change in electric potential = V - V' where V = electric potential at x = 5.6 cm = 450 V and V' = electric potential at x = 0 cm, = 0 V . So, ΔV = V - V' = 450 V - 0 V = 450 V.
Δx = distance between the 0 V plate and the 450 V point = 5.6 cm = 0.056 m
So, E = -ΔV/Δx
Substituting the values of the variables into the equation, we have
E = -ΔV/Δx
E = -450 V/0.056 m
E = -8035.7 V/m
E = -8.0357 kV/m
E ≅ -8.036 kV/m
Since the electric field between two parallel conducting plates is constant, the electric field between the plates is E = -8.036 kV/m
Give reason:
Mass and volume are called physical quantities.
a.
b
Ionath is called
Mass and volume are called physical quantities because they can be measured by using physical devices i.e. mass is measured by using beam balance and volume is measured by using metre scale .
Noticing large amounts of algal growth in her small farm pond, a farmer adds about 20 grass carp to feed on the abundant algae and plants. After several years, the carp grow large, exceeding 20 pounds each in size. Late one summer, the farmer notices that the carp and most of the other fish are dead. The water also smells very bad. Which one of the following is the most likely explanation for the death of these fish?
A. the algae overgrew the pond and produced toxic levels of ozone
B. carbon dioxide released by the carp and algae eventually suffocated the fish in the pond
C. the large carp grew so large that they could not get enough oxygen
D. bacteria feeding on the large volume of carp feces depleted the oxygen
Answer:
D. bacteria feeding on the large volume of carp feces depleted the oxygen
Explanation:
In the context, it is given that in a small farm pond, the owner added 20 grass carp and feed them with the abundant plants and algae that is found on her pond. The carp grew large after many years but one summer the owner found the grass carps along with other fishes were dead.
The most possible explanation for the dead of the fishes in the pond because the bacteria feeds on the carp feces which depletes the dissolved oxygen present in the water. Thus the fishes could not breathe and were finally dead.
Therefore, the correct option is (D).
is acting on an object that is constrained to move along the x-axis. If the force moves the object 3m, from the origin, what is the potential energy difference due to this force acting on the object
Answer:
U = - ∫ F dx
U = - 3F₀
Explanation:
Force and potential energy are related
F = [tex]- \frac{dU}{dx}[/tex]
dU = - F dx
∫ dU = - ∫ F dx
U- U₀ = - ∫ F dx
to solve the problem we must know the form of the force, if we assume that F = F₀ and U₀ = 0 for x = 0
U = - F₀ 3
Quickly pls!!! A wave with a wavelength of 0.5 m moves with a speed of 1.5 m/s. What is the frequency of the wave?
A. 2.0 Hz
B. 1.0 Hz
C. 0.33 Hz
D. 3.0 Hz
Name One formula that uses joules
Answer:
[tex]{ \bf{power \: { \tt{(watts)}} = \frac{workdone \: { \tt{(joules)}}}{time \: { \tt{(seconds)}}} \: }}[/tex]
It takes 130 j of work to compress a certain spring 0.10m. (a) what is the force constant of this spring? To compress the spring an additional 0.10m, does it take 130
Answer:
Explanation:
The PE equation for a mass/spring system is
[tex]PE=\frac{1}{2}k[/tex]Δx² and filling in:
[tex]130=\frac{1}{2}k(.10)^2[/tex] and
[tex]k=\frac{2(130)}{(.10)^2}[/tex] so
k = 26000 N/m
If the displacement from equilibrium changes more, the PE needed to compress it will also change.
[tex]PE=\frac{1}{2}(26000)(.20)^2[/tex] gives us that
PE = 520J
a sprinter runs 100m in12.5seconds calculate the speed
Answer:
8m in 1 second
480m = 1 hour
11,520m in one day
Explanation:
Hope this is helpful
. A car accelerates uniformly from 0 to 72 km/h in 11.5 seconds.
a) What is the acceleration of the car in m/s2? b) What is the position of the car by the time it reaches the velocity of 72 km/h?
Answer:
Explanation:
First job is to convert 72 km/hr to m/s:
[tex]72\frac{km}{hr}[/tex] × [tex]\frac{1000m}{1km}[/tex] × [tex]\frac{1hr}{3600s}[/tex] = 2.0 × 10¹ m/s
Now to find the acceleration which is
[tex]a=\frac{v_f-v_0}{t}[/tex] and filling in:
[tex]a=\frac{2.0*10^1-0}{11.5}=1.7\frac{m}{s^2}[/tex] That's part a. Part b want to know how far the car can get in 11.5 seconds (because that's the time it takes for the car to get to 72 km/hr). Since we know that the car can get 2.0 × 10¹ meters in 1 second, that means that in 11.5 seconds, the car can get 11.5(2.0 × 10¹) which is 230 meters.
Apakah yang berlaku kepada kekuatan medan magnet jika satu lagi sel kering 1.5 v ditambahkan
Answer:
The magnetic field is doubled.
Explanation:
What happens to the strength of the magnetic field if one more 1.5 v dry cell is added?
The magnetic field due to a current carrying conductor is directly proportional to the current flowing in the wire.
If we connect one more battery of 1.5 V so the voltage is doubled and according to the Ohm's law, as the resistance of wire is constant, so the current in the wire is also doubled.
When the current doubles, the magnetic field produced by the wire is also doubled.
convert the following as instructed
67 kg into gram
explain step by step
please reply quickly its urgent
Answer:
1kg = 1000g
67kg therefore would be equal to
67 x 1000 = 67,000g = 6.7 x 10⁴g.
a stone of mass 250kg and another stone of mass 400 kg are kept at a distance of 100m what amount of gravitational force develops between them?
Explanation:
Hey there!
Given;
Mass of one object (m1) = 250kg
Mass of another object (m2) = 400 kg
Distance (d) = 100 m
Gravitational constant (g) = 6.67*10^-11
Now;
[tex]f = \frac{g.m1.m2}{ {d}^{2} } [/tex]
Keep all values;
[tex]f = \frac{6.67 \times {10}^{ - 11} \times 250 \times 400}{ {(100)}^{2} } [/tex]
Simplify
[tex]f = \frac{6.67 \times {10}^{ - 11} {10}^{5} }{10000} [/tex]
[tex]f = \frac{6.67 \times {10}^{ - 6} }{10000} [/tex]
Therefore, gravitational force is 6.67*10^-10.
Hope it helps!
A car is moving at 30km/h. What is its speed in m/s? Give your answer to 2 significant figures.
Explanation is in the attachment
hope it is helpful to you
person has a mass of 60kg. How much do they weigh on Earth, if the gravitational field strength is 10 N/kg?
Answer:
588n is the answer may be
A stream leaving a mountain range deposits a large part of its load in a __
Answer:
(n) alluvial fan sandbar
Explanation:
While using a digital radiography system, suppose a radiographer uses exposure factors of 10 mAs and 70 kVp with an 8:1 grid for an AP shoulder radiograph with acceptable anatomical part penetration and detector element (DEL) exposure. If the radiographer desires to increase scatter absorption using a 12:1 grid, what new exposure factors should be used to maintain the same DEL exposure
Answer:
b. 12.5 mAs, 70 kVp
Explanation:
The given parameter are;
The initial exposure factors := 10 mAs and 70 kVp
The initial Grid Ratio, G.R.₁ = 8:1
The Grid Ratio with which the radiographer desires to increase the scatter absorption, G.R.₂ = 12:1
Given that the lead content in the 12:1 grid, is higher than the lead content in 8:1 grid and that 12:1 grid needs more mAs to compensate, and provides a higher image contrast, the amount of extra mAs is given by the Grid Conversion Factors, GCF, as follows;
The GCF for G.R. 8:1 = 4
The GCF for G.R. 12:1 = 5
Therefore, given that the mAs used by the radiographer for 8:1 Grid Ratio is 10 mAs, the mAs required for a G.R. of 12:1 in order to maintain the same exposure is given as follows;
mAs for G.R. of 12:1 = 10 mAs × 5/4 = 12.5 mAs
Therefore the new exposure factors are;
12.5 mAs, 70 kVp
