Answer:
Explanation:
xf = xita + vxitb + ½axtc.
xf is displacement , dimensional formula L .
Xi initial displacement , dimensional formula L
t is time , dimensional formula T ,
vxi is velocity , dimensional formula LT⁻¹
ax is acceleration , dimensional formula = LT⁻²
xf = xi t a + vxi t b + ½ ax t c.
L = aLT + b LT⁻¹ T + c LT⁻² T
From the law of uniformity , dimensional formula of each term of RHS must be equal to term on LHS
aLT = L
a = T⁻¹
b LT⁻¹ T = L
b = 1 ( constant )
c LT⁻² T = L
c = T
so a = T⁻¹ , b = constant and c = T .
Four heavy elements (A, B, C, and D) will fission when bombarded by neutrons. In addition to fissioning into two smaller elements, A also gives off a beta particle, B gives off gamma rays, C gives off neutrons, and D gives off alpha particles. Which element would make a possible fuel for a nuclear reactor
Answer:
Element C will be best for a nuclear fission reaction
Explanation:
Nuclear fission is the splitting of the nucleus of a heavy atom by bombarding it with a nuclear particle. The reaction leads to the the atom splitting into two smaller elements and a huge amount of energy is liberated in the process. For the reaction to be continuous in a chain reaction, the best choice of element to use as fuel for the reaction should be the element whose nucleus also liberates a neutron particle after fission. The neutron that is given off by other atoms in the reaction will then proceed to bombard other atoms of the element in the reaction, creating a cascade of fission and bombardment within the nuclear reactor.
Two stones are thrown vertically upward from the ground, one with three times the initial speed of the other. Assume free fall. Part A If the faster stone takes 12.0 s to return to the ground, how long will it take the slower stone to return
Answer:
36s
Explanation:
Let the objects be A and B.
Let the initial velocity of A be U and the initial velocity of B be 3U
The height sustain by A will be;
The final velocity would be zero
V2 = U2-2gH
Hence
0^2= U2 -2gH
H = U^2/2g
Similarly for object B, the height sustain is;
V2 = (3U)^2-2gH
Hence
0^2= 3U^2 -2gH
U2-2gH
Hence
0^2= U2 -2gH
H = 3U^2/2g
By comparism. The object with higher velocity sustains more height and so should fall longer than object A.
Now object A would take;
From V=U+gt as the object falls freely, the initial velocity is zero hence and the final velocity of the object is;
V=10×12=120m/s let g be 10m/S2
Similarly for object B,
The final velocity for B when it's falling it should be 3×that of A
Meaning
3V= gt
t =3V/g = 3× 120/10 = 36s
A spaceship travels toward the Earth at a speed of 0.97c. The occupants of the ship are standing with their torsos parallel to the direction of travel. According to Earth observers, they are about 0.50 m tall and 0.50 m wide. Calculate what the occupants’ height and width according to the others on the spaceship?
Answer:
Explanation:
We shall apply length contraction einstein's relativistic formula to calculate the length observed by observer on the earth . For the observer , increased length will be observed for an observer on the earth
[tex]L=\frac{.5}{\sqrt{1-(\frac{.97c}{c})^2 } }[/tex]
[tex]L=\frac{.5}{.24}[/tex]
L= 2.05
The length will appear to be 2.05 m . and width will appear to be .5 m to the observer on the spaceship. . It is so because it is length which is moving parallel to the direction of travel. Width will remain unchanged.
What is a major criticism of Maslow's hierarchy of needs? Select one: a. It is subjective. b. It does not take gender differences into account. c. It is humanistic. d. It only accounts for the objective world.
The correct answer is A. It is subjective
Explanation:
In 1943, the recognized psychologist Abraham Maslow proposed a theory to understand and classify human needs. The work of Maslow included five different categories to classify all basic needs, psychological needs, and self-esteem needs; additionally, in this, Maslow proposed individuals need to satisfy the needs of previous levels to satisfy more complex needs. For example, the first level includes physiological needs such as hunger and these are necessary to get to more complex needs such as the need for safety or self-satisfaction.
This hierarchy is still used all around the world to understand human needs; however, it was been widely criticized because the classification itself is related to Maslow's perspective as this was mainly based on Maslow's ideas about needs, which makes the hierarchy subjective. Also, due to its subjectivity, the hierarchy may apply only in some individuals or societies.
Which statement best explains why the overall charge on an atom is zero?
ОООО
The positive charge of the neutrons in the nucleus equals the negative charge in the electron cloud.
The positive charge of the protons in the nucleus equals the negative charge in the electron cloud.
The negative charge of the neutrons in the nucleus equals the positive charge in the electron cloud.
The negative charge of the protons in the nucleus equals the positive charge in the electron cloud.
Answer:
B) The positive charge of the protons in the nucleus equals the negative charge in the electron cloud.
Explanation:
For every negative charge of an electron, there is an equal positively charged proton in the nucleus of the atom. This is why the overall charge on an atom is zero.
Answer:
B
Explanation:
This is described by Gauss a scientist.
The positive charge is found in the proton in the nucleus.
The neutron has no charge.
The positive charge radiates in all directions and a counter negative charge ensues.
А
mass exerts force of
5.6 X 10^-10N on
another mass
when
seperated 93cm apart. If
one mass is the square root of
the other
Find
value of
the two masses.
Answer:
i hope it will be useful for you
Explanation:
F=5.6×10^-10N
R=93cm=0.93m
let take m1 and m2 =m²
according to newton's law of universal gravitation
F=m1m2/r²
F=m²/r²
now we have to find masses
F×r²=m²
5.6×10^10N×0.93m=m²
5.208×10^-9=m²
taking square root on b.s
√5.208×10^-9=√m²
so the two masses are m1=7.2×10^-5
and m2=7.2×10^-5
Which of the following is NOT a type of electromagnetic wave?
Seismic waves
Visible light
Radio waves
Microwave
Answer:
Seismic waves
Explanation:
seismic waves are not represented by electromagnetic graphs, nor can they be reflected on an electromagnetic spectrum. Visible light, radio waves, and microwaves are all electromagnetic waves, which are represented by graphs and electromagnetic spectrums.
Answer:
Siesmic waves
Explanation:
The equation = [(8.0 )(4.0)]cos(1.0) describes a wave generated by a sonometer. State whether or not the wave is a stationary wave and find its amplitude, with given that = 2.0 cm
Answer:
Yes. the wave is a stationary wave
Amplitude=7.91 cm
Explanation:
We are given that an equation
[tex]y=[8.0sin(4.0x)]cos(1.0t)[/tex]
We have to find that the given wave is a stationary wave or not and find its amplitude at x=2.0 cm
We know that equation of stationary wave
[tex]y(x,t)=2Asin(kx)cos(\omega t)[/tex]
Where
Amplitude=[tex]2Asin(kx)[/tex]
The given equation is of the form of stationary wave equation.
By comparing we get
[tex]2Asin(kx)=8sin(4.0x)[/tex]
Substitute the value of x
Therefore, the amplitude of the wave
Amplitude=[tex]8sin(4.0\times 2.0)=7.91 cm[/tex]
f a curve with a radius of 97 m is properly banked for a car traveling 75 km/h , what must be the coefficient of static friction for a car not to skid when traveling at 100 km/h ?
Answer:
The coefficient of static friction is 0.26
Explanation:
Given;
radius of the road, R = 97 m
banking velocity, V₁ = 75 km/h = 20.83 m/s
velocity of the moving car, V₂ = 100 km/h = 27.78 m/s
Car in a banked circular turn:
[tex]\theta = tan^{-1}(\frac{V_1^2}{gR} )[/tex]
where;
θ is the angle between the horizontal ground and road in which the car move on
[tex]\theta = tan^{-1}(\frac{V_1^2}{gR} ) \\\\\theta = tan^{-1}(\frac{20.83^2}{9.8*97} ) \\\\\theta = 24.5^o[/tex]
During this type of motion, the body acquires some acceleration which tends to retain the circular motion towards its center, known as centripetal acceleration.
There are two components of this acceleration;
Parallel acceleration, [tex]a_|_|[/tex] = a*Cosθ
Perpendicular acceleration, a⊥ = a * Sinθ
Parallel acceleration, [tex]a_|_|[/tex] [tex]= \frac{V^2*Cos \theta}{R}[/tex]
Perpendicular acceleration, a⊥ [tex]= \frac{V^2*Sin \theta}{R}[/tex]
Apply Newton's second law of motion;
sum of perpendicular forces acting on the car;
ma⊥ [tex]= F_N - mg*cos \theta[/tex]
[tex]m(\frac{V^2*Sin \theta}{R} ) = F_N - mg*Cos \theta\\\\F_N = mg*Cos \theta + m(\frac{V^2*Sin \theta}{R} )[/tex] --------equation (1)
sum of parallel forces acting on the car
m[tex]a_|_|[/tex] [tex]= mg*Sin \theta - F_s[/tex]
[tex]m(\frac{V^2*Cos \theta}{R} ) = mg*Sin \theta - F_s\\\\F_s = mg*Sin \theta - m(\frac{V^2*Cos \theta}{R} )[/tex] ---------equation (2)
Coefficient of static friction is given as;
[tex]\mu = \frac{F_s}{F_N}[/tex]
Thus, divide equation (2) by equation (1)
[tex]\frac{F_s}{F_N} = \frac{mg*Sin \theta - m(\frac{V^2*Cos \theta}{R}) }{mg*Cos \theta + m(\frac{V^2*Sin \theta}{R}) } \\\\\frac{F_s}{F_N} = \frac{g*Sin \theta - (\frac{V^2*Cos \theta}{R}) }{g*Cos \theta + (\frac{V^2*Sin \theta}{R}) }[/tex]
V = V₂ = 27.78 m/s
θ = 24.5°
R = 97 m
g = 9.8 m/s²
Substitute in these values and solve for μ
[tex]\frac{F_s}{F_N} = \frac{9.8*Sin(24.5)+ (\frac{27.78^2*Cos (24.5)}{97}) }{9.8*Cos (24.5) + (\frac{27.78^2*Sin (24.5)}{97}) }\\\\\frac{F_s}{F_N} = \frac{4.0641 \ - \ 7.2391}{8.91702 \ + \ 3.299} = -0.26\\\\| \mu| = 0.26[/tex]
Therefore, the coefficient of static friction is 0.26
A thin, metallic spherical shell of radius 0.347 m0.347 m has a total charge of 7.53×10−6 C7.53×10−6 C placed on it. A point charge of 3.65×10−6 C3.65×10−6 C is placed at the center of the shell. What is the electric field magnitude EE a distance 0.795 m0.795 m from the center of the spherical shell?
Answer:
E = 12640.78 N/C
Explanation:
In order to calculate the electric field you can use the Gaussian theorem.
Thus, you have:
[tex]\Phi_E=\frac{Q}{\epsilon_o}[/tex]
ФE: electric flux trough the Gaussian surface
Q: net charge inside the Gaussian surface
εo: dielectric permittivity of vacuum = 8.85*10^-12 C^2/Nm^2
If you take the Gaussian surface as a spherical surface, with radius r, the electric field is parallel to the surface anywhere. Then, you have:
[tex]\Phi_E=EA=E(4\pi r^2)=\frac{Q}{\epsilon_o}\\\\E=\frac{Q}{4\pi \epsilon_o r^2}[/tex]
r can be taken as the distance in which you want to calculate the electric field, that is, 0.795m
Next, you replace the values of the parameters in the last expression, by taking into account that the net charge inside the Gaussian surface is:
[tex]Q=7.53*10^{-6}C+3.65*10^{-6}C=1.115*10^{-5}C[/tex]
Finally, you obtain for E:
[tex]E=\frac{1.118*10^{-5}C}{4\pi (8.85*10^{-12C^2/Nm^2})(0.795m)^2}=12640.78\frac{N}{C}[/tex]
hence, the electric field at 0.795m from the center of the spherical shell is 12640.78 N/C
(III) A baseball is seen to pass upward by a window with a vertical speed of If the ball was thrown by a person 18 m below on the street, (a) what was its initial speed, (b) what altitude does it reach, (c) when was it thrown, and (d) when does it reach the street again? Giancoli, Douglas C.. Physics (p. 45). Pearson Education. Kindle Edition.
Answer:
Assuming that the vertical speed of the ball is 14 m/s we found the given values:
a) V₀ = 23.4 m/s
b) h = 27.9 m
c) t = 0.96 s
d) t = 4.8 s
Explanation:
a) Assuming that the vertical speed is 14 m/s (founded in the book) the initial speed of the ball can be calculated as follows:
[tex] V_{f}^{2} = V_{0}^{2} - 2gh [/tex]
Where:
[tex]V_{f}[/tex]: is the final speed = 14 m/s
[tex]V_{0}[/tex]: is the initial speed =?
g: is the gravity = 9.81 m/s²
h: is the height = 18 m
[tex] V_{0} = \sqrt{V_{f}^{2} + 2gh} = \sqrt{(14 m/s)^{2} + 2*9.81 m/s^{2}*18 m} = 23.4 m/s [/tex]
b) The maximum height is:
[tex] V_{f}^{2} = V_{0}^{2} - 2gh [/tex]
[tex] h = \frac{V_{0}^{2}}{2g} = \frac{(23. 4 m/s)^{2}}{2*9.81 m/s^{2}} = 27.9 m [/tex]
c) The time can be found using the following equation:
[tex] V_{f} = V_{0} - gt [/tex]
[tex] t = \frac{V_{0} - V_{f}}{g} = \frac{23.4 m/s - 14 m/s}{9.81 m/s^{2}} = 0.96 s [/tex]
d) The flight time is given by:
[tex] t_{v} = \frac{2V_{0}}{g} = \frac{2*23.4 m/s}{9.81 m/s^{2}} = 4.8 s [/tex]
I hope it helps you!
Two forces are acting on an object as shown in Fig. on the right. What is the magnitude of the resultant force?
A) 47.5 N
B) 185 N
C) 198 N
D) 200 N
Answer:
185 N
Explanation:
Sum of forces in the x direction:
Fₓ = -(80 N cos 75°) + (120 N cos 60°)
Fₓ = 39.3 N
Sum of forces in the y direction:
Fᵧ = (80 N sin 75°) + (120 N sin 60°)
Fᵧ = 181.2 N
The magnitude of the net force is:
F = √(Fₓ² + Fᵧ²)
F = √((39.3 N)² + (181.2 N)²)
F = 185 N
We have that for the Question "Two forces are acting on an object as shown in Fig. on the right. What is the magnitude of the resultant force?" it can be said that the magnitude of the resultant force is
R=200N
From the question we are told
Two forces are acting on an object as shown in Fig. on the right. What is the magnitude of the resultant force?
A) 47.5 N
B) 185 N
C) 198 N
D) 200 N
Generally the equation for the Resultant force is mathematically given as
For x axis resolution
[tex]Fx=80cos75+120cos60\\\\Fx=80.7N[/tex]
For y axis resolution
[tex]Fx=80sin75+120sin60\\\\Fx=181.2N[/tex]
Therefore
[tex]R=\sqrt{80.7^2+181.2N^2}\\\\R=200N[/tex]
For more information on this visit
https://brainly.com/question/23379286
An aluminium bar 600mm long, with diameter 40mm, has a hole drilled in the center of the bar. The hole is 30mm in diameter and is 100mm long. If the modulus of elasticity for the aluminium is 85GN/m2, calculate the total contraction on the bar due to a compressive load of 180kN.
Answer:
[tex]\delta = 0.385\,m[/tex] (Compression)
Explanation:
The aluminium bar is experimenting a compression due to an axial force, that is, a force exerted on the bar in its axial direction. (See attachment for further details) Under the assumption of small strain, the deformation experimented by the bar is equal to:
[tex]\delta = \frac{P\cdot L}{A \cdot E}[/tex]
Where:
[tex]P[/tex] - Load experimented by the bar, measured in newtons.
[tex]L[/tex] - Length of the bar, measured in meters.
[tex]A[/tex] - Cross section area of the bar, measured in square meters.
[tex]E[/tex] - Elasticity module, also known as Young's Module, measured in pascals, that is, newtons per square meter.
The cross section area of the bar is now computed: ([tex]D_{o} = 0.04\,m[/tex], [tex]D_{i} = 0.03\,m[/tex])
[tex]A = \frac{\pi}{4}\cdot (D_{o}^{2}-D_{i}^{2})[/tex]
Where:
[tex]D_{o}[/tex] - Outer diameter, measured in meters.
[tex]D_{i}[/tex] - Inner diameter, measured in meters.
[tex]A = \frac{\pi}{4}\cdot [(0.04\,m)^{2}-(0.03\,m)^{2}][/tex]
[tex]A = 5.498 \times 10^{-4}\,m^{2}[/tex]
The total contraction of the bar due to compresive load is: ([tex]P = -180\times 10^{3}\,N[/tex], [tex]L = 0.1\,m[/tex], [tex]E = 85\times 10^{9}\,Pa[/tex], [tex]A = 5.498 \times 10^{-4}\,m^{2}[/tex]) (Note: The negative sign in the load input means the existence of compressive load)
[tex]\delta = \frac{(-180\times 10^{3}\,N)\cdot (0.1\,m)}{(5.498\times 10^{-4}\,m^{2})\cdot (85\times 10^{9}\,Pa)}[/tex]
[tex]\delta = -3.852\times 10^{-4}\,m[/tex]
[tex]\delta = -0.385\,mm[/tex]
[tex]\delta = 0.385\,m[/tex] (Compression)
Consider a situation in which you are moving two point charges such that the potential energy between them decreases. (NOTE: ignore gravity).
This means that you are moving the charges:
a) Closer to each other
b) Farther apart
c) Either A or B
Answer: Option A
Explanation:
The potential energy decreases in the case when the charges are opposite and they attract each other.
In this case there is no external energy required in order to put the charges together.
This is so because the charges are opposite and they will attract each other. Yes, the only condition should be that the charges should be alike.
Example: a negative charge and a positive charge.
What is the impulse on a car (750 kg) that accelerates from rest to 5.0 m/s in 10 seconds
Explanation:
impulse J = m × (v2-v1) =750 × ( 5 - 0 ) =3750( N×s)
Answer:
3750Ns
Explanation:
Impulse is defined as Force × time
Force = mass × acceleration,
Hence impulse is;
mass × acceleration × time.
From Newton's second law
Force × time = mass × ∆velocity
750× 5 = 3750Ns
∆velocity = Vfinal-Vinitial ; the initial velocity is zero since the body starts from rest.
A baton twirler is twirling her aluminum baton in a horizontal circle at a rate of 2.33 revolutions per second. A baton held horizontally. Calculate the induced emf between the center and one end of the baton if the magnetic field of the Earth is 0.500 gauss and is oriented at 14.42 with respect to the horizontal. Assume the baton is 60.1 cm in length.
Answer:
Explanation:
Given that;
horizontal circle at a rate of 2.33 revolutions per second
the magnetic field of the Earth is 0.500 gauss
the baton is 60.1 cm in length.
the magnetic field is oriented at 14.42°
we wil get the area due to rotation of radius of baton is
[tex]\Delta A = \frac{1}{2} \Delta \theta R^2[/tex]
The formula for the induced emf is
[tex]E = \frac{\Delta \phi}{\Delta t}[/tex]
[tex]\phi = \texttt {magnetic flux}[/tex]
[tex]E=\frac{\Delta (BA) }{\Delta t}[/tex]
[tex]=B\frac{\Delta A}{\Delta t}[/tex]
B is the magnetic field strength
substitute
[tex]\texttt {substitute}\ \frac{1}{2} \Delta \theta R^2 \ \ for \Delta A[/tex]
[tex]E=B\frac{(\Delta \theta R^3/2)}{\Delta t} \\\\=\frac{1}{2} BR^2\omega[/tex]
The magnetic field of the earth is oriented at 14.42
[tex]\omega =2.33\\\\L=60.1c,\\\\\theta=14.42\\\\B=0.5[/tex]
we plug in the values in the equation above
so, the induce EMF will be
[tex]E=\frac{1}{2} \times (B\sin \theta)R^2\omega\\\\E=\frac{1}{2} \times (B\sin \theta)(\frac{L}{2} )\omega[/tex]
[tex]=\frac{1}{2} \times0.5gauss\times\frac{0.0001T}{1gauss} \times\sin 14.42\times(\frac{60.1\times10^-^2m}{2} )^2(2.33rev/s)(\frac{2\pi rad}{1rev} )\\\\=2.5\times10^-^5\times0.2490\times0.0903\times14.63982\\\\=2.5\times10^-^5\times0.32917\\\\=8.229\times10^-^6V[/tex]
11. A vector M is 15.0 cm long and makes an angle of 20° CCW from x axis and another vector N is 8.0 cm long and makes an angle of 40° clockwise from the x axis. Find out resultant vector with its magnitude and direction using components method.
Answer:
The magnitude of the resultant vector is 22.66 cm and it has a direction of 29.33°
Explanation:
To find the resultant vector, you first calculate x and y components of the two vectors M and N. The components of the vectors are calculated by using cos and sin function.
For M vector you obtain:
[tex]M=M_x\hat{i}+M_y\hat{j}\\\\M=15.0cm\ cos(20\°)\hat{i}+15.0cm\ sin(20\°)\hat{j}\\\\M=14.09cm\ \hat{i}+5.13\ \hat{j}[/tex]
For N vector:
[tex]N=N_x\hat{i}+N_y\hat{j}\\\\N=8.0cm\ cos(40\°)\hat{i}+8.0cm\ sin(40\°)\hat{j}\\\\N=6.12cm\ \hat{i}+5.142\ \hat{j}[/tex]
The resultant vector is the sum of the components of M and N:
[tex]F=(M_x+N_x)\hat{i}+(M_y+N_y)\hat{j}\\\\F=(14.09+6.12)cm\ \hat{i}+(5.13+5.142)cm\ \hat{j}\\\\F=20.21cm\ \hat{i}+10.27cm\ \hat{j}[/tex]
The magnitude of the resultant vector is:
[tex]|F|=\sqrt{(20.21)^2+(10.27)^2}cm=22.66cm[/tex]
And the direction of the vector is:
[tex]\theta=tan^{-1}(\frac{10.27}{20.21})=29.93\°[/tex]
hence, the magnitude of the resultant vector is 22.66 cm and it has a direction of 29.33°
How many times can a three-dimensional object that has a radius of 1,000 units fit something with a radius of 10 units inside of it? How many times can something with a radius of 2,000 units fit something with a radius of 1 unit?
Answer:
# _units = 1000
Explanation:
This exercise we can use a direct proportion rule.
If a volume of radius r = 1 is one unit, how many units can fit in a volume of radius 10?
# _units = V₁₀ / V₁
The volume of a body of radius 1 is
V₁ = 4/3 π r₁³
V₁ = 4/3π
the volume of a body of radius r = 10
V₁₀ = 4/3 π r₂³
V10 = 4/3 π 10³
the number of times this content is
#_units = 4/3 π 1000 / (4/3 π 1)
# _units = 1000
Please help in the 2nd question
Answer:
[tex]q=4\times 10^{-16}\ C[/tex]
Explanation:
It is given that,
The charge on an object is 2500 e.
We need to find how many coulombs in the object. The charge remains quantized. It says that :
q = ne
[tex]q=2500\times 1.6\times 10^{-19}\ C\\\\q=4\times 10^{-16}\ C[/tex]
So, the charge on the object is [tex]4\times 10^{-16}\ C[/tex].
Two masses, each weighing 1.0 × 103 kilograms and moving with the same speed of 12.5 meters/second, are approaching each other. They have a head-on collision and bounce off away from each other. Assuming this is a perfectly elastic collision, what will be the approximate kinetic energy of the system after the collision?
A. 1.6 × 105 joules
B. 2.5 × 105 joules
C. 1.2 × 103 joules
D. 2.5 × 103 joules
Answer:
A. 1.6 × 105 joules
Explanation:
As per the question, the data given in the question is as follows
Number of masses = 2
Each weightage = [tex]1.0 \times 2.3[/tex]
Speed = 12.5 meters/ second
Based on the above information, the approximate kinetic energy after the collision is
A perfectly elastic collision is described as one where the collision does not cause any loss of kinetic energy.
So we sum the kinetic energy of each kind of system which is given below:
Kinetic energy is
[tex]= 0.5(1.0 \times 10^3) (12.5)^2 + 0.5 (1.0 \times 10^3) (12.5 )^2[/tex]
= 156250 J
[tex]= 1.6 \times 10^5 J[/tex]
Consider the binding energy of two stable nuclei, one with 60 nucleons and one with 200 nucleons. a. Is the total binding energy of the nucleus with 200 nucleons more than, less than, or equal to the total binding energy of the nucleus with 60 nucleons. Justify your reasoning.
Answer:
The total binding energy of the nucleus with 200 nucleons more than the total binding energy of the nucleus with 60 nucleons
Explanation:
Binding energy can be given by the formula:
Binding energy = Binding energy per nucleon * Number of nucleons
This means that if the binding energy per nucleon = x MeV
Where x is a positive real number
The nucleus with 60 nucleons will have Binding energy = 60x MeV
The nucleus with 200 nucleons will have binding energy = 200x MeV
For a +ve x, 200x > 60x
Binding energy is proportional (directly) to nucleon volume. A further explanation is provided below.
Binding energy
Binding energy involving 200 nucleons would've been greater than 60 nucleons because so many more nucleons result in what seems like a stronger reaction.
It makes absolutely no difference out whether nucleon seems to be a proton as well as a neutron because they just have a similar strong coupling relatively steady.
Thus the response above is correct.
Find out more information about binding energy here:
https://brainly.com/question/23942204
Two identical point charges q=71.0 pCq=71.0 pC are separated in vacuum by a distance of 2d=29.0 cm.2d=29.0 cm. Calculate the total electric flux ΦΦ through the infinite surface placed at a distance dd from each charge, perpendicular to the line on which the point charges are located.
Answer:
The electric flux at the infinite surface is ZERO
Explanation:
From the question we are told that
The point charge are identical and the value is [tex]q = 71.0 pC = 71 * 10^{-12} \ C[/tex]
The distance of separation is [tex]D = 29.0 \ cm = 0.29 \ m[/tex]
The distance of both from the infinite surface is d
Generally the electric force exerted by each of the charge on the infinite surface is
[tex]\phi = \frac{q}{\epsilon_o}[/tex]
Now given from the question that they are identical, it then means that the electric flux of the first charge on the infinite surface will be nullified by the electric flux of the second charge hence the electric flux at that infinite surface due to this two identical charges is ZERO
A particle initially located at the origin has an acceleration of = 1.00ĵ m/s2 and an initial velocity of i = 6.00î m/s. (a) Find the vector position of the particle at any time t (where t is measured in seconds). ( t î + t2 ĵ) m (b) Find the velocity of the particle at any time t. ( î + t ĵ) m/s (c) Find the coordinates of the particle at t = 4.00 s. x = m y = m (d) Find the speed of the particle at t = 4.00 s.
Answer:
a) d = (6.00 t i ^ + 0.500 t²) m , b) v = (6.00 i ^ + 1.00 t j ^) m / s
c) d = (24.00 i ^ + 8.00 j^ ) m , d) v = (6.00 i ^ + 5 j^ ) m/s
Explanation:
This exercise is about kinematics in two dimensions
a) find the position of the particle on each axis
X axis
Since there is no acceleration on this axis, we can use the relation of uniform motion
v = x / t
x = v t
we substitute
x = 6.00 t
Y Axis
on this axis there is an acceleration and there is no initial speed
y = v₀ t + ½ a t²
y = ½ at t²
we substitute
y = ½ 1.00 t²
y = 0.500 t²
in vector position is
d = x i ^ + y j ^
d = (6.00 t i ^ + 0.500 t²) m
b) x axis
as there is no relate speed is concatenating
vₓ = v₀
vₓ = 6.00 m / s
y Axis
there is an acceleration and the initial speed is zero
[tex]v_{y}[/tex] = v₀ + a t
v_{y} = a t
v_{y} = 1.00 t
the velocity vector is
v = vₓ i ^ + v_{y} j ^
v = (6.00 i ^ + 1.00 t j ^) m / s
c) the coordinates for t = 4 s
d = (6.00 4 i ^ + 0.50 4 2 j⁾
d = (24.00 i ^ + 8.00 j^ ) m
x = 24.0 m
y = 8.00 m
d) the velocity of for t = 4 s
v = (6 i ^ + 1 5 j ^)
v = (6.00 i ^ + 5 j^ ) m/s
When light with a wavelength of 225 nm is incident on a certain metal surface, electrons are ejected with a maximum kinetic energy of 2.98 × 10-19 J. Determine the wavelength (in nm) of light that should be used to double the maximum kinetic energy of the electrons ejected from this surface.
Answer:
The new wavelength is 112.5 nm.
Explanation:
It is given that,
When light with a wavelength of 225 nm is incident on a certain metal surface, electrons are ejected with a maximum kinetic energy of 2.98 × 10⁻¹⁹ J. We need to find the wavelength (in nm) of light that should be used to double the maximum kinetic energy of the electrons ejected from this surface.
The energy of incident electron is given by :
[tex]E=\dfrac{hc}{\lambda}[/tex]
New energy is 2 E and new wavelength is [tex]\lambda'[/tex]. So,
[tex]\dfrac{E}{2E}=\dfrac{\lambda'}{\lambda}\\\\\dfrac{1}{2}=\dfrac{\lambda'}{\lambda}\\\\\lambda'=\dfrac{\lambda}{2}\\\\\lambda'=\dfrac{225}{2}\\\\\lambda=112.5\ nm[/tex]
So, the new wavelength is 112.5 nm.
In Parts A, B, C consider the following situation. In a baseball game the batter swings and gets a good solid hit. His swing applies a force of 12,000 N to the ball for a time of 0.70×10−3s. Part A Assuming that this force is constant, what is the magnitude J of the impulse on the ball?
Answer:
J = 8.4 kg*m/s
Explanation:
The magnitude of the impulse, J, on the ball can be calculated using the following equation:
[tex] J = F*t [/tex] (1)
Where:
F: is the force = 12000 N
t: is the time = 0.70x10⁻³ s
So, by entering the values above into equation (1) we can find the impulse on the ball:
[tex] J = F*t = 12000 N*0.70 \cdot 10^{-3} s = 8.4 kg*m/s [/tex]
Therefore, the impulse on the ball is 8.4 kg*m/s.
I hope it helps you!
The magnitude J of the impulse on the ball is 8.4 kg m/s.
Calculation of the impulse:Since it applies a force of 12,000 N to the ball for a time of 0.70×10−3s.
So,
The magnitude is
[tex]= Force \times time\\\\= 12,000 \times 0.70\times 10^{-3}[/tex]
= 8.4 kg m/s
Here we basically multiplied the force with the time to determine the magnitude.
Learn more about impulse here: https://brainly.com/question/16324614
If the mass of the ladder is 12.0 kgkg, the mass of the painter is 55.0 kgkg, and the ladder begins to slip at its base when her feet are 70% of the way up the length of the ladder, what is the coefficient of static friction between the ladder and the floor
Answer:
μ = 0.336
Explanation:
We will work on this exercise with the expressions of transactional and rotational equilibrium.
Let's start with rotational balance, for this we set a reference system at the top of the ladder, where it touches the wall and we will assign as positive the anti-clockwise direction of rotation
fr L sin θ - W L / 2 cos θ - W_painter 0.3 L cos θ = 0
fr sin θ - cos θ (W / 2 + 0,3 W_painter) = 0
fr = cotan θ (W / 2 + 0,3 W_painter)
Now let's write the equilibrium translation equation
X axis
F1 - fr = 0
F1 = fr
the friction force has the expression
fr = μ N
Y Axis
N - W - W_painter = 0
N = W + W_painter
we substitute
fr = μ (W + W_painter)
we substitute in the endowment equilibrium equation
μ (W + W_painter) = cotan θ (W / 2 + 0,3 W_painter)
μ = cotan θ (W / 2 + 0,3 W_painter) / (W + W_painter)
we substitute the values they give
μ = cotan θ (12/2 + 0.3 55) / (12 + 55)
μ = cotan θ (22.5 / 67)
μ = cotan tea (0.336)
To finish the problem, we must indicate the angle of the staircase or catcher data to find the angle, if we assume that the angle is tea = 45
cotan 45 = 1 / tan 45 = 1
the result is
μ = 0.336
HELP ASAP!
There is a lever with 5 m long. The fulcrum is 2 m from the right end. Each end hangs a box. The whole system is in balance. If the box hung to the right end is 12 kg, then what is the mass of the box hung to the left end?
Answer:
8kg
Explanation:
For the box to be in equilibrium. The clockwise moment ensued by the box on the right should be same as that ensued by the one on the right. Hence :
M ×3 = 12 ×2
M = 24/3 = 8kg
Note mass is used because trying to compute the weight by multiplying by the acceleration of free fall due to gravity on both sides will cancel out.
You have two charges; Q1 and Q2, and you move Q1 such that the potential experienced by Q2 due to Q1 increases.
Gravity should be ignored.
Then, you must be:
a) Moving Q1 further away from Q2.
b) Moving in the opposite direction to that of the field due to Q1
c) Moving Q1 closer to Q2.
d) Moving in the same direction as the field due to Q1.
e) Any of the above
Given that,
First charge = Q₁
Second charge = Q₂
The potential experienced by Q2 due to Q1 increases
We know that,
The electrostatic force between two charges is defined as
[tex]F=\dfrac{kQ_{1}Q_{2}}{r^2}[/tex]
Where,
k = electrostatic constant
[tex]Q_{1}[/tex]= first charge
[tex]Q_{2}[/tex]= second charge
r = distance
According to given data,
The potential experienced by Q₂ due to Q₁ increases.
We know that,
The potential is defined from coulomb's law
[tex]V=\dfrac{Q_{1}}{4\pi\epsilon_{0}r}[/tex]
[tex]V\propto\dfrac{1}{r}[/tex]
If r decrease then V will be increases.
If V decrease then r will be increases.
Since, V is increases then r will decreases that is moving Q₁ closer to Q₂.
Hence, Moving Q₁ closer to Q₂.
(c) is correct option.
An uncharged, nonconducting, hollow sphere of
radius 10.0 cm surrounds a 10.0-uC charge located
at the origin of a Cartesian coordinate system. A drill
with a radius of 1.00 mm is aligned along the z axis,
and a hole is drilled in the sphere. Calculate the elec-
tric flux through the hole.
Answer:
Explanation:
The whole surface of hollow sphere = 4π r²
= 4 x 3.14 x (10 x 10⁻²)²
= 12.56 x 10⁻² m²
Area of the hole ( both side ) = 2 x π r²
= 2 x 3.14 x (10⁻³)²
= 6.28 x 10⁻⁶ m²
flux coming out of given charge at the centre as per Gauss's theorem
= q / ε₀ where q is charge at the centre and ε₀ is permittivity of the medium .
= 10 x 10⁻⁶ / 8.85 x 10⁻¹²
= 1.13 x 10⁶
This flux will pass through the surface of sphere so flux passing through per unit area
= 1.13 x 10⁶ / 12.56 x 10⁻²
= 8.99 x 10⁶ weber per m²
flux through area of hole
= 8.99 x 10⁶ x 6.28 x 10⁻⁶
= 56.45 weber .
A turtle and a rabbit are in a 150 meter race. The rabbit decides to give the turtle a 1 minute head start. The turtle moves at a constant speed of 0.500 m/s through the whole race (in fact the turtle even starts at a velocity of 0.500 m/s as while he was still approaching the starting line he was allowed to continue to keep going without stopping). The rabbit starts the race from rest and accelerates at a rate or 1.50 m/s2 until she reaches her top speed of 10 m/s. She then finishes the race running at a constant speed of 10 m/s. a) What is the turtle’s position when the rabbit starts to run (1 minute into the race)? b) How long does it take the turtle to finish the race? c) How long does it take the rabbit to reach max speed? d) What is the rabbit’s position when she reaches max speed? e) How long does it take the rabbit to finish the race? f) Who won?
Answer:
a) [tex]s_{T} = 30\,m[/tex], b) [tex]t = 5\,min[/tex], c) [tex]\Delta t = 6.667\,s[/tex], d) [tex]\Delta s_{R} = 33.333\,m[/tex], e) [tex]t' = 11.667\,s[/tex], f) The rabbit won the race.
Explanation:
a) As turtle moves at constant speed, its position is determined by the following formula:
[tex]s_{T} = v_{T}\cdot t[/tex]
Where:
[tex]t[/tex] - Time, measured in seconds.
[tex]v_{T}[/tex] - Velocity of the turtle, measured in meters per second.
[tex]s_{T}[/tex] - Position of the turtle, measured in meters.
Then, the position of the turtle when the rabbit starts to run is:
[tex]s_{T} = \left(0.5\,\frac{m}{s} \right)\cdot (60\,s)[/tex]
[tex]s_{T} = 30\,m[/tex]
The position of the turtle when the rabbit starts to run is 30 meters.
b) The time needed for the turtle to finish the race is:
[tex]t = \frac{s_{T}}{v_{T}}[/tex]
[tex]t = \frac{150\,m}{0.5\,\frac{m}{s} }[/tex]
[tex]t = 300\,s[/tex]
[tex]t = 5\,min[/tex]
The time needed for the turtle to finish the race is 5 minutes.
c) As rabbit experiments a constant acceleration until maximum velocity is reached and moves at constant speed afterwards, the time required to reach such speed is:
[tex]v_{R} = v_{o,R} + a_{R}\cdot \Delta t[/tex]
Where:
[tex]v_{R}[/tex] - Final velocity of the rabbit, measured in meters per second.
[tex]v_{o,R}[/tex] - Initial velocity of the rabbit, measured in meters per second.
[tex]a_{R}[/tex] - Acceleration of the rabbit, measured in [tex]\frac{m}{s^{2}}[/tex].
[tex]\Delta t[/tex] - Running time, measured in second.
[tex]\Delta t = \frac{v_{R}-v_{o,R}}{a_{R}}[/tex]
[tex]\Delta t = \frac{10\,\frac{m}{s}-0\,\frac{m}{s}}{1.50\,\frac{m}{s^{2}} }[/tex]
[tex]\Delta t = 6.667\,s[/tex]
The time taken by the rabbit to reach maximum speed is 6.667 s.
d) On the other hand, the position reached by the rabbit when maximum speed is reached is determined by the following equation of motion:
[tex]v_{R}^{2} = v_{o,R}^{2} + 2\cdot a_{R}\cdot \Delta s_{R}[/tex]
[tex]\Delta s_{R} = \frac{v_{R}^{2}-v_{o,R}^{2}}{2\cdot a_{R}}[/tex]
[tex]\Delta s_{R} = \frac{v_{R}^{2}-v_{o,R}^{2}}{2\cdot a_{R}}[/tex]
Where [tex]\Delta s_{R}[/tex] is the travelled distance of the rabbit from rest to maximum speed.
[tex]\Delta s_{R} = \frac{\left(10\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}}{2\cdot \left(1.50\,\frac{m}{s^{2}} \right)}[/tex]
[tex]\Delta s_{R} = 33.333\,m[/tex]
The distance travelled by the rabbit from rest to maximum speed is 33.333 meters.
e) The time required for the rabbit to finish the race can be determined by the following expression:
[tex]t' = \frac{\Delta s_{R}}{v_{R}}[/tex]
[tex]t' = \frac{150\,m-33.333\,m}{10\,\frac{m}{s} }[/tex]
[tex]t' = 11.667\,s[/tex]
The time required for the rabbit from rest to maximum speed is 11.667 seconds.
f) The animal with the lowest time wins the race. Now, each running time is determined:
Turtle:
[tex]t_{T} = 300\,s[/tex]
Rabbit:
[tex]t_{R} = 60\,s + 6.667\,s + 11.667\,s[/tex]
[tex]t_{R} = 78.334\,s[/tex]
The rabbit won the race as [tex]t_{R} < t_{T}[/tex].