Answer:
A) Bohr’s work with atomic spectra led him to say that the electrons were limited to existing in certain energy levels, like standing on the rungs of a ladder.
Explanation:
The change to the atomic model that helped solve the problem seen in Rutherford's model was the discovery of the strong nuclear force.
Explanation:Rutherford's model required that the electrons be in motion. Positive and negative charges attract each other, so stationary electrons would fall into the positive nucleus. However, Rutherford's model failed to explain why electrons were not pulled into the atomic nucleus by this attraction. The change to the atomic model that helped solve this problem was the discovery of the strong nuclear force, which is much stronger than electrostatic interactions and holds the protons and neutrons together in the nucleus.
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The cheetah is one of the fastest-accelerating animals, because it can go from rest to 19.6 m/s (about 44 mi/h) in 2.9 s. If its mass is 108 kg, determine the average power developed by the cheetah during the acceleration phase of its motion. Express your answer in the following units.
(a) watts(b) horsepower.
Answer:
a)P =14288.4 W
b)P = 19.16 horsepower
Explanation:
Given that
m= 108 kg
Initial velocity ,u= 0 m/s
Final velocity ,v= 19.6 m/s
t= 2.9 s
Lets take acceleration of Cheetah is a m/s²
We know that
v= u + a t
19.6 = 0 + a x 2.9
a= 6.75 m/s²
Now force F
F= m a
F= 108 x 6.75 N
F= 729 N
Now the power P
P = F.v
P = 729 x 19.6 W
P =14288.4 W
We know that
1 W= 0.0013 horsepower
P = 19.16 horsepower
P =14288.4 W
The purpose of studying a system is to identify the strengths and weaknesses of the existing system and examine current inputs, outputs, processes, security and controls, and system performance.
The purpose of system is to make system more reliable and efficient for desired work completion
Explanation:
In most of the times, the system analysts operate in a dynamic environment where change is the necessary process . A business application, a business firm or a computer system may be the required environment. In order To rebuild a system, the key elements must be considered which we need to examine are as follows:
1. Outputs and inputs.
2. Security and Control.
3. processor
4. Environment.
5. Feedback.
6. Boundaries and interface.
A "seconds pendulum" is one that moves through its equilibrium position once each second. (The period of the pendulum is precisely 2.000 s.) The length of a seconds pendulum is 0.9923 m at Tokyo and 0.9941 m at Cambridge, England. What is the ratio of the free-fall accelerations at these two locations?
Explanation:
We know the equation
[tex]T=2\pi \sqrt{\frac{l}{g}}[/tex]
where l is length of pendulum, g is acceleration due to gravity and T is period.
Rearranging
[tex]g= \frac{4\pi^2l}{T^2}[/tex]
Length of pendulum in Tokyo = 0.9923 m
Length of pendulum in Cambridge = 0.9941 m
Period of pendulum in Tokyo = Period of pendulum in Cambridge = 2s
We have
[tex]\frac{ g_{\texttt{Tokyo}}}{ g_{\texttt{Cambridge}}}= \frac{\frac{4\pi^2 l_{\texttt{Tokyo}}}{ T_{\texttt{Tokyo}}^2}}{\frac{4\pi^2 l_{\texttt{Cambridge}}}{ T_{\texttt{Cambridge}}^2}}\\\\\frac{ g_{\texttt{Tokyo}}}{ g_{\texttt{Cambridge}}}=\frac{\frac{0.9923}{2^2}}{\frac{0.9941}{2^2}}=0.998[/tex]
Ratio of free fall acceleration of Tokyo to Cambridge = 0.998
In a coffee-cup calorimeter experiment, 10.00 g of a soluble ionic compound was added to the calorimeter contained 75.0 g H2O initially at 23.2°C. The final temperature of the solution was 31.8°C. What was the change in enthalpy for the dissolution of this compound?
Answer:
The enthalpy for dissolution is - 305.558 J/g
Solution:
Mass of the ionic compound, m = 10.00 g
Mass of water, m' = 75.0 g
Initial temperature, T = [tex]23.2^{\circ}C[/tex]
Final Temperature, T' = [tex]31.8^{\circ}C[/tex]
Now,
To calculate the change in enthalpy:
We know that the specific heat of water is 4.18 [tex]J/g^{\circ}C[/tex]
Total mass of the solution, M = m + m' = 10.00 + 75.0 = 85.0 g
Temperature, difference, [tex]\Delta T = T' - T = 31.8 - 23.2 = 8.6^{\circ}C[/tex]
Thus
The heat absorbed by the solution is given by:
[tex]Q = MC_{w}\Delta T = 85.0\times 4.18\times 8.6 = 3055.58\ J[/tex]
Enthalpy, [tex]\Delta H = -\frac{Q}{m} = - \frac{3055.58}{10} = - 305.558\ J/g[/tex]
Two identical strings, of identical lengths of 2.00 m and linear mass density of μ=0.0065kg/m, are fixed on both ends. String A is under a tension of 120.00 N. String B is under a tension of 130.00 N. They are each plucked and produce sound at the n=10 mode. What is the beat frequency?
Answer:
beat frequency = 13.87 Hz
Explanation:
given data
lengths l = 2.00 m
linear mass density μ = 0.0065 kg/m
String A is under a tension T1 = 120.00 N
String B is under a tension T2 = 130.00 N
n = 10 mode
to find out
beat frequency
solution
we know here that length L is
L = n × [tex]\frac{ \lambda }{2}[/tex] ........1
so λ = [tex]\frac{2L}{10}[/tex]
and velocity is express as
V = [tex]\sqrt{\frac{T}{\mu } }[/tex] .................2
so
frequency for string A = f1 = [tex]\frac{V1}{\lambda}[/tex]
f1 = [tex]\frac{\sqrt{\frac{T}{\mu } }}{\frac{2L}{10}}[/tex]
f1 = [tex]\frac{10}{2L} \sqrt{\frac{T1}{\mu } }[/tex]
and
f2 = [tex]\frac{10}{2L} \sqrt{\frac{T2}{\mu } }[/tex]
so
beat frequency is = f2 - f1
put here value
beat frequency = [tex]\frac{10}{2*2} \sqrt{\frac{130}{0.0065}}[/tex] - [tex]\frac{10}{2*2} \sqrt{\frac{120}{0.0065} }[/tex]
beat frequency = 13.87 Hz
A molecule moves down its concentration gradient using a transport protein in the plasma membrane. This is an example of
Final answer:
Facilitated transport, also known as facilitated diffusion, is the process by which a molecule moves down its concentration gradient using transport proteins in the plasma membrane.
Explanation:
Facilitated transport, also known as facilitated diffusion, is the process by which a molecule moves down its concentration gradient using transport proteins in the plasma membrane. This process does not require the input of energy and allows substances to diffuse across the membrane more easily. For example, glucose is transported into cells using glucose transporters that utilize facilitated transport. This process is important for the movement of larger or charged molecules that cannot freely diffuse across the cell membrane.
Determine whether the following actions cause the fission reaction in the reactor to speed up or slow down.
a. speeds up fission: Adding the moderator to the reactor
b. speeds up fission: Removing the control rods from the reactor
c. slows down fission: Inserting the control rods into the reactor
d. slows down fission: Removing the moderator from the reactor
e. slows down fission: A sudden loss of primary coolant water in a pressurized water reactor
Answer:
option B and C
Explanation:
Control rod are used to regulate the nuclear reactor.
When you insert control rod in the reactor it slows down the nuclear fission inside the reactor and the energy produced in the reactor will be less.
When you remove control road from the reactor the nuclear fission increase inside the reactor and the energy production is high.
Control rod consist of boron, boron absorb the neutron which help to control the nuclear fission.
Hence, the correct answer is option B and C
A Biologists have studied the running ability of the northern quoll, a marsupial indigenous to Australia In one set of experiments, they studied the maximum speed that quolls could run around a curved path without slipping. One quoll was running at 2.4 m/s around a curve with a radius of 1.6 m when it started to slip.
What was the coefficient of static friction between the quoll's feet and the ground in this trial?
Answer:
Coefficient of static friction = 0.37
Explanation:
At the point the the quoll slides, quoll attains its maximum velocity.
So Ne = (mv^2)/r ....equa 1
And N =mg....equ 2
Where N vertical force of qoull acting on the surface, e = coefficient of friction, m=mass, g=9.8m/s^2, r =radius =1.6m, v= max velocity of quill = 2.4m/s
Sub equ 2 into equ 1
Mge= (mv^2)/r ...equa3
Simplfy equ3
e = v^2/(gr)...equ 4
Sub figures above
e = 5.76/(9.8*1.6)
e = 0.37
A car traveling 6.0 m/s is uninformly accelerating at a rate of 3.0 m/s^2 for 15 seconds. What is it’s final velocity?
Answer:
The answer to your question is 11.2 m/s
Explanation:
Data
Initial speed (vo) = 6.0 m/s
Acceleration (a) = 3.0 m/s²
time = 15 s
Final speed = ?
Formula
d = vot + [tex]\frac{1}{2} at^{2}[/tex]
vf² = vo² + 2ad
Process
d = (6)(15) + [tex]\frac{1}{2} (3)(15)^{2}[/tex]
d = 90 + 337.5
d = 427.5 m
vf² = (6)² + 2(3)(15)
vf² = 36 + 90
vf² = 126
vf = 11.2 m/s
A ledge on a building is 23 m above the ground. A taut rope attached to a 4.0-kg can of paint sitting on the ledge passes up over a pulley and straight down to a 3.0-kg can of nails on the ground. If the can of paint is accidentally knocked off the ledge, what time interval does a carpenter have to catch the can of paint before it smashes on the ground?
Answer:
The time can catch before it smashes on the ground is [tex]t=5.73 s[/tex]
Explanation:
Using the force equation
[tex]F=m*a[/tex]
[tex]F_{net}=m*a[/tex]
So replacing and solving to find the acceleration
[tex]a = (m_1*g-m_2*g) / m_1+m_2[/tex]
Finding the factor
[tex]a = g *( m_1-m_2)/m_1+m_2[/tex]
[tex]a=9.8m/s^2 *( 4.0 kg- 3.0 kg) / (4.0 + 3.0) kg[/tex]
[tex]a=1.4 m/s^2[/tex]
Now replacing in Newtons law to find the time before can catch so:
[tex]d= \frac{1}{2}*a*t^2[/tex]
[tex]t=\sqrt{\frac{2*d}{a}}=\sqrt{\frac{2* 23m}{1.4 m/s^2}}[/tex]
[tex]t=5.73 s[/tex]
The end point of a spring vibrates with a period of 2.1 seconds when a mass m is attached to it. When this mass is increased by 6.810×101 kg, the period is found to be 3.4 seconds. Find the value of m.
Answer:
Mass attached to the spring is 41.95 kg
Explanation:
We have given time period of the spring T = 2.1 sec
Let the mass attached is m
And spring constant is k
We know that time period is given by
[tex]T=2\pi \sqrt{\frac{m}{k}}[/tex]
[tex]2.1=2\pi \sqrt{\frac{m}{k}}[/tex]---------eqn 1
Now if the mass is increased by 68.10 kg then time period become 3.4 sec
So [tex]3.4=2\pi \sqrt{\frac{m+68.10}{k}}[/tex]------eqn 2
Now dividing eqn 1 by eqn 2
[tex]\frac{2.1}{3.4}=\sqrt{\frac{m}{m+68.10}}[/tex]
[tex]0.381=\frac{m}{m+68.10}[/tex]
[tex]m=41.95 kg[/tex]
So mass attached to the spring is 41.95 kg
Final answer:
To find the value of mass m, use the formula for the period of a mass-spring system.
Explanation:
In order to find the value of mass m, we can use the formula for the period of a mass-spring system:
T = 2π√(m/k)
Where T is the period, m is the mass, and k is the spring constant.
For the initial system with period 2.1 seconds, we have:
2.1 = 2π√(m/k)
For the system with mass increased by 6.810×10^1 kg and period 3.4 seconds, we have:
3.4 = 2π√((m + 6.810×10^1)/k)
Using these two equations, we can solve for the value of m.
A tennis ball is thrown upward from the top of a 680 foot high building at a speed of 56 feet per second. The tennis ball's height above ground can be modeled by the equation . When does the tennis ball hit the ground?
Answer:
t = 8.5 s
Explanation:
Kinematic equation of the movement of the tennis ball that is thrown upwards :
y = y₀ + v₀*t -½ g*t² Equation (1)
Where :
y : position of the ball as a function of time
y₀ : Initial position of the ball
t: time
g: acceleration due to gravity in m/s²
Known data
g = 32 ft/s²
y₀ = 680 ft
v₀ = 56 ft/s
Calculation of the time it takes for the ball to thit the ground
We replace data en the equation (1)
y = y₀ + v₀*t -½ g*t²
0 = 680+(56)*t -½( 32) *t²
16*t²-(56)*t- 680 = 0 equation (2)
solving equation (2) quadratic:
t₁ = 8.5 s
t₁ = -5 s
Time cannot be negative so the time it takes for the ball to hit the ground is t = 8.5 s
You observe a spiral galaxy with a large central bulge and tightly wrapped arms. It would be classified a
Answer:
Sa
Explanation:
Spiral Galaxies -
It is a disk shaped galaxies which have spiral structure , is refereed to as spiral galaxies .
According to Hubble , these galaxies are classified as Sa , Sb , Sc .
Where ,
Sa - have the structure , which is bulged from the central portion , along with a tightly wrapped spiral structure .
Sb - have a lesser bulge and the spiral is looser .
Sc - It has very weak bulge with the open spiral structure .
Hence , from the question ,
The given information is about the Sa .
Beth exerts 14 Newton’s of force to propel a 4.5 kilogram bowling ball down the lane. Describe how the ball will travel.
The ball will accelerate at a rate of [tex]3.11 m/s^2[/tex]
Explanation:
We can describe the motion of the ball by using Newton's second law, which states that the net force exerted on an object is equal to the product between the mass of the object and its acceleration:
[tex]F=ma[/tex]
where
F is the net force
m is the mass
a is the acceleration
In this problem,
F = 14 N is the force exerted on the ball
m = 4.5 kg is the mass of the ball
Solving the equation, we find its acceleration:
[tex]a=\frac{F}{m}=\frac{14}{4.5}=3.11 m/s^2[/tex]
So, the ball will accelerate at a rate of [tex]3.11 m/s^2[/tex].
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Six baseball throws are shown below. In each case the baseball is thrown at the same initial speed and from the same height h above the ground. Assume that the effects of air resistance are negligible. Rank these throws according to the speed of the baseball the instant before it hits the ground.
Answer:
The final velocities of all the six balls will be same.
Explanation:
According to law of conservation of energy:
Gain in K.E = Loss in potential energy
½ mv^2 = mgh
Where “m” and “g” are constant. The interchange in energies will occur only with the change in velocity and height. Since, balls are thrown from the same hight with the same initial velocity so, their final velocities will also be same just before striking the ground.
The six balls will reach the ground at the same time, hence the final velocity of the balls will be the same.
During a downward motion of an object, the speed of the object increases as the object moves downwards and becomes maximum before the object hits the ground.
The equation for estimating the final velocity of the six balls is given as;
[tex]v_f = v_i + gt[/tex]
If air resistance is negligible, the six balls will reach the ground at the same time, hence the final velocity of the balls will be the same.
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Describe what happens, at a microscopic level, when an object is charged by rubbing. For instance, what happens when a plastic pipe is rubbed with a cloth? Describe the specific case where the rod becomes negatively charged
Answer:
Explanation:
The static charges are generated due to excess or deficiency of electrons, because these are the smallest quanta of charge available at the molecular level which can get transferred with minimal energy requirement.
These charges are usually generated by friction between the two surfaces leading to the transfer of electron from one to another.
When a plastic pipe is rubbed with a cloth then due to friction the surface of the cloth loses electron which gets stuck at the surface of the pipe making it negatively charged.
Assume the speed of light to be 299 792 458 m/s. If the frequency of an electromagnetic wave is 80,000 GHz (GHz = gigahertz = 109 Hz), what is the wavelength of that radiation? Express your answer in micrometres (μm)
Answer:
3.747 μm
Explanation:
To answer this question, the fundamental wave equation will be used. Light is an electromagnetic wave so we will use the speed of light for this electromagnetic wave.
v = fλ
299 792 458 m/s = 80,000 *10^9 * λ
λ = 3.747 *10^-6 = 3.747 μm
A father racing his son has half the kinetic energy of the son, whohas three-fifths the mass of the father. The father speeds up by2.5 m/s and then has the same kinetic energy as the son.a) What is the original speed of the father?b) What is the original speed of the son?
Answer:
a) 6.04 m/s
b) 11.02 m/s
Explanation:
a) Let the father mass be M, and his speed be V. His son mass is m = 3M/5. Since his kinetic energy initially is half of after he increases his speed by 2.5m/s
[tex]E_2 = 2E_1[/tex]
[tex]\frac{M(V+2.5)^2}{2} = 2\frac{MV^2}{2}[/tex]
[tex]V^2 + 5V + 6.25 = 2V^2[/tex]
[tex]V^2 - 5V - 6.25 = 0[/tex]
[tex]V \approx 6.04m/s[/tex]
b) The son kinetic energy initially is:
[tex]E_s = 2E_1 = 2\frac{MV^2}{2} = MV^2 = M*6.04^2 = 36.43M J[/tex]
We can solve for the son speed by the following formula
[tex]E_s = \frac{mv^2}{2}[/tex]
[tex]v^2 = \frac{2E_s}{m} = \frac{2*36.43M}{3M/5} = \frac{10*36.43}{3} = 121.4m/s[/tex]
[tex]v = \sqrt{121.4} = 11.02 m/s[/tex]
Consider the following electron configurations to answer the question:
(i) 1s2 2s2 2p6 3s1
(ii) 1s2 2s2 2p6 3s2
(iii) 1s2 2s2 2p6 3s2 3p1
(iv) 1s2 2s2 2p6 3s2 3p4
(v) 1s2 2s2 2p6 3s2 3p5
The electron configuration of the atom that is expected to have a positive electron affinity is ________.
Answer:
(ii) 1s2 2s2 2p6 3s2
Explanation:
Electron Affinity is the energy change that occur when an atom gains an electron.
X₍₉₎ + e⁻ → X⁻ ΔE = Eea
ΔE is change in energy
Eea is electron affinity
Often, electron affinity has negative energy values. The more negative the electron affinity, the easier it is to add an electron to a particular atom. Electron affinity increases across the period in the periodic table. However, there are few exceptions:
1. The electron affinities of group 18 (8A) elements are greater than zero. This is because the atom has a filled valence shell, an addition of electron causes the electron to move to a higher energy shell.
2. The electron affinities of group 2 (2A) elements are more positive because addition of an electron requires it to reside in the previously unoccupied p sub-shell.
3. The electron affinities of group 15 (5A) elements are more positive because addition of an electron requires it to be put in an already occupied orbital.
Applying these consideration to the elements given in the question:
(i) The sum of the electron in 1s²2s² 2p⁶ 3s¹ = 2+2+6+1 = 11 Sodium (Na)
(ii) The sum of the electron in 1s² 2s² 2p⁶ 3s² = 2+2+6+2 = 12 Magnesium (Mg)
(iii) The sum of the electron in 1s² 2s² 2p⁶ 3s² 3p¹ = 2+2+6+2+1 = 13 Aluminium (Al)
(iv) The sum of the electron in 1s² 2s² 2p⁶ 3s² 3p⁴ = 2+2+6+2+4 = 16 Sulfur (S)
(v) The sum of the electron in 1s² 2s² 2p⁶ 3s² 3p⁵ = 2+2+6+2+5 = 17 Chlorine (Cl)
The atom that is expected to have a positive electron affinity is Magnesium which is a group 2A element with electron configuration of 1s² 2s² 2p⁶ 3s².
An atom with a positive electron affinity attracts electrons more than one with negative electron affinity. Typically, these atoms tend to have half-filled or filled subshell configurations. In the given examples, the atom with electron configuration 1s2 2s2 2p6 3s2 3p1 is likely to have a positive electron affinity.
Explanation:The electron affinity of an atom is a measure of the energy change when an electron is added to a neutral atom to form a negative ion. An atom with a positive electron affinity attracts electrons more than one with negative electron affinity. Usually, the atom would tend to have a half-filled (e.g. 3p3) or filled (e.g. 3p6) subshell configuration.
If we consider the given electron configurations: (i) 1s2 2s2 2p6 3s1 (ii) 1s2 2s2 2p6 3s2 (iii) 1s2 2s2 2p6 3s2 3p1 (iv) 1s2 2s2 2p6 3s2 3p4 (v) 1s2 2s2 2p6 3s2 3p5, it appears that (iii) 1s2 2s2 2p6 3s2 3p1 should have a positive electron affinity. This atom is a Potassium atom (K). It has a positive electron affinity because it is just one electron away from obtaining a full orbit shell (in the 3s and 3p orbitals) which would provide a more stable electron configuration.
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Sound wave A delivers 2J of energy in 2s. Sound wave length B delivers 10J of energy in 5s. Sound wave C delivers 2mJ of energy in 1ms. Rank in order, from largest to smallest, the sound powers of Pa, Pb, Pc of these three waves.Explain. What equation would you use to determine this?
Answer:
[tex]P_c=P_b>P_a[/tex]
Explanation:
E = Energy
T = Time
Power is given by the equation
[tex]P=\frac{E}{T}[/tex]
For first case
[tex]P_a=\frac{2}{2}\\\Rightarrow P_a=1\ W[/tex]
For second case
[tex]P_b=\frac{10}{5}\\\Rightarrow P_b=2\ J[/tex]
For third case
[tex]P_c=\frac{2\times 10^{-3}}{1\times 10^{-3}}\\\Rightarrow P_b=2\ J[/tex]
The rank of power would be [tex]P_c=P_b>P_a[/tex]
An 19-cm-long bicycle crank arm, with a pedal at one end, is attached to a 23-cm-diameter sprocket, the toothed disk around which the chain moves. A cyclist riding this bike increases her pedaling rate from 65 rpm to 90 rpm in 10 s .
Answer:
The tangential acceleration of the pedal is 0.0301 m/s².
Explanation:
Given that,
Length = 19 cm
Diameter = 23 cm
Time = 10 sec
Initial angular velocity = 65 rpm
Final velocity = 90 rpm
Suppose we need to find the tangential acceleration of the pedal
We need to calculate the tangential acceleration of the pedal
Using formula of tangential acceleration
[tex]a_{t}=r\alpha[/tex]
[tex]a_{t}=\dfrac{23\times10^{-2}}{2}\times\dfrac{\omega_{2}-\omega_{1}}{t}[/tex]
[tex]a_{t}=\dfrac{23\times10^{-2}}{2}\times\dfrac{90\times\dfrac{2]pi}{60}-65\times\dfrac{2\pi}{60}}{10}[/tex]
[tex]a_{t}=0.0301\ m/s^2[/tex]
Hence, The tangential acceleration of the pedal is 0.0301 m/s².
Listed following are four models for the long-term expansion (and possible contraction) of the universe. Rank the models from left to right based on their predictions for the average distance between galaxies five billion years from now, from smallest to largest.
a. recollapsing universe
b. accelerating universe
c. coasting universe
d. critical universe
Answer:
gdsz
Explanation:
dsgzz cxvzdgctfgdsvftgdsftrdsfdtsardtgasfd5t6sgftsfdrstfdtgsv6cr5vsd5rw5
As an intern with an engineering firm, you are asked to measure the moment of inertia of a large wheel, for rotation about an axis through its center. Since you were a good physics student, you know what to do. You measure the diameter of the wheel to be 0.88 m and find that it weighs 280 N . You mount the wheel, using frictionless bearings, on a horizontal axis through the wheel's center. You wrap a light rope around the wheel and hang a 6.32 kg mass from the free end of the rope. You release the mass from rest; the mass descends and the wheel turns as the rope unwinds. You find that the mass has speed 4.0 m/s after it has descended 2.5 m .(a) What is the moment of inertia of the wheel for an axis perpendicular to the wheel at its center?
Final answer:
The moment of inertia of the wheel for an axis perpendicular to the wheel at its center is 0.964 kg * m^2.
Explanation:
To calculate the moment of inertia of the wheel, we can use the principle of conservation of energy. The initial gravitational potential energy of the mass is equal to the final rotational kinetic energy of the wheel. This can be represented by the equation:
mg * h = 1/2 * Iω^2
Where m is the mass, g is the acceleration due to gravity, h is the distance the mass has descended, I is the moment of inertia of the wheel, and ω is the angular velocity of the wheel. Rearranging the equation:
I = 2mg * h / ω^2
Substituting the given values:
I = 2 * 6.32 kg * 9.8 m/s^2 * 2.5 m / (4.0 m/s)^2
I = 0.964 kg * m^2
Therefore, the moment of inertia of the wheel for an axis perpendicular to the wheel at its center is 0.964 kg * m^2.
You throw a 50.0g blob of clay directly at the wall with an initial velocity of -5.00 m/s i. The clay sticks to the wall, and the collision takes about 20.0 ms (2.00 x 10^-2 s). a) What is the change in momentum for the blob of clay?
Answer:0.25 kg-m/s
Explanation:
Given
mass of blob [tex]m=50 gm [/tex]
initial velocity [tex]u=-5 m/s\ \hat{i}[/tex]
time of collision [tex]t=20 ms[/tex]
we know Impulse is equal to change in momentum
initial momentum [tex]P_i=mu[/tex]
[tex]P_i=50\times 10^{-3}\times (-5)=-0.25 kg-m/s[/tex]
Final momentum [tex]P_f=50\times 10^{-3}v[/tex]
[tex]P_f=0[/tex] as final velocity is zero
Impulse [tex]J=P_f-P_i[/tex]
[tex]J=0-(-0.25)[/tex]
[tex]J=0.25 kg-m/s[/tex]
Suppose we consider the system of the three capacitors as a single "equivalent" capacitor. Given the charges of the three individual capacitors calculated in the previous part, find the total charge Qtot for this equivalent capacitor.
Answer:
Qtot = 6C * deltaV
Explanation:
you can find the total capacitance from adding 1C+2C+3C=6C. and the total voltage is 1V. Capacitance = charge/voltage--> C = Q / V--> 6C = Q / deltaV. this makes Qtot = 6C* deltaV
The Total charge for the equivalent circuit is = [tex]Q_{tot}[/tex] = 6c * ΔV
Although your question is incomplete I found the missing part online and used it to resolve the question
Given data :
Total capacitance ( C ) = 6C ( 1 + 2 + 3 )
voltage = 1 V
Three capacitors having values of ; 1 C, 2 C, 3 C
Determine the total charge ( Qtot )
Applying the formula ; Q = CV ---- ( 1 )
where; Q = charge
C = capacitance
change in V = ΔV
∴ [tex]Q_{tot}[/tex] = 6c * ΔV
Hence the total charge Qtot for the equivalent capacitor = 6c * ΔV
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An astronaut drops a hammer on the moon . It takes 1 second to hit the ground after being dropped, and it is going 1.6m/s when it lands. What is the acceleration due to gravity on thr moon?
Answer:
the value of acceleration due to gravity in moon is 1.6m/[tex]s^{2}[/tex] along downward direction
Explanation:
Here, the acceleration is constant and it is equal to acceleration due to gravity in moon. Therefore the question depicts a situation of uniformly accelerated motion in a straight line. So, let us refresh the three equations of uniformly accelerated straight line motion.
v = u + at
[tex]s = ut + \frac{1}{2}at^{2}[/tex]
[tex]v^{2} = u^{2} +2as[/tex]
where,
u = initial velocity
v = final velocity
s = displacement
a = acceleration
t = time
Since we are dealing with vectors (velocity, acceleration and displacement), we have to take their directions in to account. So we must adopt a coordinate system according to our convenience. Here, we are taking point of throwing as origin, vertically upward direction as positive y axis and vertically downward direction as negative y axis.
t = 1s
u = 0 (since the hammer is dropped)
v = -1.6m/s (since its direction is downward)
a = ?
The only equation that connects all the above quantities is
v = u + at
therefore,
a = [tex]\frac{v - u}{t}[/tex]
substituting the values
a = [tex]\frac{-1.6 - 0}{1}[/tex]
a = -1.6m/[tex]s^{2}[/tex]
Thus, the value of acceleration due to gravity in moon is 1.6m/[tex]s^{2}[/tex]. The negative sign indicates that it is along downward direction.
A car slams on its brakes creating an acceleration of -3.2 m/s2 it comes to a rest after traveling a distance of 210 m what was it's velocity before it began to accelerate
The initial velocity of the car is 36.6 m/s
Explanation:
The motion of the car is a uniformly accelerated motion (=constant acceleration), therefore we can apply suvat equations:
[tex]v^2-u^2=2as[/tex]
where
v is the final velocity
u is the initial velocity
a is the acceleration
s is the displacement
For the car in this problem, we have:
v = 0 is the final velocity (the car comes to a stop)
[tex]a=-3.2 m/s^2[/tex] is the acceleration
s = 210 m is the displacement of the car
Solving for u, we find the initial velocity:
[tex]u=\sqrt{v^2-2as}=\sqrt{0-(2)(-3.2)(210)}=36.6 m/s[/tex]
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Calculate the work required to move a planet’s satellite of mass 571 kg from a circular orbit of radius 2R to one of radius 3R, where 8.8 × 106 m is the radius of the planet. The mass of the planet is 7.76 × 1024 kg. Answer in units of J]
Final answer:
The work required to move a satellite from an orbit of radius 2R to 3R around a planet is calculated using the gravitational potential energy formula and is found to be 3.897×1010 J.
Explanation:
To calculate the work required to move a satellite from one circular orbit to another around a planet, we must consider the gravitational potential energy differences in the two orbits.
The gravitational potential energy (U) of an object of mass m in orbit around a planet of mass M at a distance r is given by U = -GmM/r, where G is the gravitational constant (6.67×10-11 N m2/kg2).
For the initial orbit at radius 2R, the potential energy is U1 = -GmM/(2R), and for the final orbit at radius 3R, the potential energy is U2 = -GmM/(3R). The work done (W) in moving the satellite is the difference in gravitational potential energy, W = U2 - U1. Substituting the values, we get:
W = (-GmM/3R) - (-GmM/2R) = (GmM/6R)
Let's calculate the work required using the given values: G = 6.67×10-11 N m2/kg2, m = 571 kg, M = 7.76×1024 kg, R = 8.8×106 m.
W = (6.67×10-11 N m2/kg2 × 571 kg × 7.76×1024 kg) / (6 × 8.8×106 m)
W = 3.897×1010 J
Therefore, the work required to move the satellite from a circular orbit of radius 2R to one of radius 3R is 3.897×1010 J.
Electromagnetic radiation of 5.16Ă—1016 Hz frequency is applied on a metal surface and caused electron emission. Determine the work function of the metal if the maximum kinetic energy (Ek) of the emitted electron is 4.04Ă—10-19 J.
Answer:
Work function of the metal, [tex]W_o=3.38\times 10^{-17}\ J[/tex]
Explanation:
We are given that
Frequency of the electromagnetic radiation, [tex]f=5.16\times 10^{16}[/tex] Hz
The maximum kinetic energy of the emitted electron, [tex]K=4.04\times 10^{-19}\ J[/tex]
We need to find the work function of the metal.
We know that the maximum kinetic energy of ejected electron
[tex]K=h\nu-w_o[/tex]
Where h=Plank's constant=[tex]6.63\times 10^{-34} J.s[/tex]
[tex]\nu[/tex] =Frequency of light source
[tex]w_o[/tex]=Work function
Substitute the values in the given formula
Then, the work function of the metal is given by :
[tex]W_o=h\nu -K[/tex]
[tex]W_o=6.63\times 10^{-34}\times 5.16\times 10^{16}-4.04\times 10^{-19}[/tex]
[tex]W_o=3.38\times 10^{-17}\ J[/tex]
So, the work function of the metal is [tex]3.38\times 10^{-17}\ J[/tex]. Hence, this is the required solution.
A 4.9 kg block slides down an inclined plane that makes an angle of 27◦ with the horizontal. Starting from rest, the block slides a distance of 2.7 m in 5.4 s. The acceleration of gravity is 9.81 m/s 2 . Find the coefficient of kinetic friction between the block and plane.
Answer:
μk = 0.488
Explanation:
Newton's second law:
∑F = m*a Formula (1)
∑F : algebraic sum of the forces in Newton (N)
m : mass s (kg)
a : acceleration (m/s²)
We define the x-axis in the direction parallel to the movement of the block on the inclined plane and the y-axis in the direction perpendicular to it.
Forces acting on the block
W: Weight of the block : In vertical direction
FN : Normal force : perpendicular to the inclined plane
fk : kinetic Friction force: parallel to the inclined plane
Calculated of the W
W= m*g
W= 4.9 kg* 9.8 m/s² = 48.02 N
x-y weight components
Wx = Wsin θ = 48.02*sin27° = 21.8 N
Wy = Wcos θ = 48.02*cos27° = 42.786 N
Calculated of the FN
We apply the formula (1)
∑Fy = m*ay ay = 0
FN - Wy = 0
FN = Wy
FN = 42.786 N
Calculated of the fk
fk = μk* FN= μk*42.786 Equation (1)
Kinematics of the block
Because the block moves with uniformly accelerated movement we apply the following formula to calculate the acceleration of the block :
d = v₀*t+(1/2)*a*t² Formula (2)
Where:
d:displacement (m)
v₀: initial speed (m/s)
t: time interval (m/s)
a: acceleration ( m/s²)
Data:
d= 2.7 m
v₀ = 0
t= 5.4 s
We replace data in the formula (2)
d = v₀*t+(1/2)*a*t²
2.7 = 0+(1/2)*a*( 5.4)²
2.7 = (14.58)*a
a = 2.7 / (14.58)
a= 0.185 m/s²
We apply the formula (1) to calculated μk:
∑Fx = m*ax , ax= a : acceleration of the block
Wx-fk= m*a , fk=μk*42.786 of the Equation (1)
21.8 - (42.786)*μk = (4.9)*(0.185)
21.8 -0.907= (42.786)*μk
20.89 = (42.786)*μk
μk = (20.89) / (42.786)
μk = 0.488