Answer:
Explained
Explanation:
Dynamically continuous innovation:
- Falls in between continuous and discontinuous innovation.
-Changes in customer habits are not as large as in discontinuous innovation and not as negligeble as in continuous innovation.
best example can as simple as transformation in Television. New HD TVs have flat panels, wide screens and improved performance The Added features are considered dynamically improved.
Discontinuous innovation:
- discontinuous innovation comprise of new to world product only so they are discontinuous to every customer segment.
- these product are so fundamentally different from the the product that already exist that they reshape market and competition.
For example- the mobile and the internet technology are reshaping the market through regular innovation and change.
Dynamically continuous innovation involves moderate changes to an existent product, seen, for example, in the evolution from traditional cell phones to smartphones. Discontinuous innovation involves the creation of entirely new products, directly impacting consumer behavior, such as the invention of the internet.
Explanation:Dynamically continuous innovation and discontinuous innovation are two concepts centered around improvements to existing products or introduction of radically new products. A dynamically continuous innovation signifies moderate changes to existing products. An example could be the transition from a traditional cell phone to a smartphone. The base product- phone remained the same, only the features were enhanced. This innovation still requires consumers to adapt to new functionalities, but to a lesser degree than discontinuous innovations.
On the other hand, discontinuous innovation involves creating products that did not exist before, fundamentally changing consumer behavior. An example would be the creation of the internet. Before its invention, nothing similar existed and it greatly impacted how people live and work.
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A 34 kg block slides with an initial speed of 9 m/s up a
rampinclined at an angle of 10o with the horizontal.
Thecoefficient of kinetic friction between the block and the ramp
is0.6. Use energy conservation to find the distance the block
slidesbefore coming to rest.
Answer:
5.4 m
Explanation:
mass, m = 34 kg
initial velocity, u = 9 m/s
final velocity, v = 0 m/s
coefficient of friction, μ = 0.6
Angle of inclination, θ = 10°
Let teh distance traveled before to come into rest is d.
According to the work energy theorem, the work done by all the forces is equal to the change in kinetic energy of the block.
Work done by the gravitational force = W1 = - mg Sinθ x d
Work done by the frictional force = W2 = - μ N = - μ mg Cosθ x d
negative sign shows that the direction of force and the direction of displacement is opposite to each other.
Total work done W = W1 + w2
W = - 34 x 9.8 x Sin 10 x d - 0.6 x 34 x 9.8 x cos 10 x d
W = - 254.86 d
Change in kinetic energy = 0.5 x m (v^2 - u^2)
= 0.5 x 34 (0 - 81) = - 1377
So, W = change in KE
- 254.86 d = - 1377
d = 5.4 m
The acceleration of automobiles is often given in terms of the time it takes to go from 0 mi/h to 60 mi/h. One of the fastest street legal cars, the Bugatti Veyron Super Sport, goes from 0 to 60 in 2.70 s. To the nearest integer, what is its acceleration in m/s^2? Please show work.
Answer:
9.934 m/s²
Explanation:
Given:
Initial speed of the Bugatti Veyron Super Sport = 0 mi/h
Final speed of the Bugatti Veyron Super Sport = 60 mi/h
Now,
1 mi/h = 0.44704 m / s
thus,
60 mi/h = 0.44704 × 60 = 26.8224 m/s
Time = 2.70 m/s
Now,
The acceleration (a) is given as:
[tex]a=\frac{\textup{Change in speed}}{\textup{Time}}[/tex]
thus,
[tex]a=\frac{26.8224 - 0 }{2.70}[/tex]
or
a = 9.934 m/s²
If there is no air resistance, a quarter dropped from the top of New York’s Empire State Building would reach the ground 9.6 s later. (a) What would its speed be just before it hits the ground? (b) How high is the building?
Answer:
a) 94.176 m/s
b) 452.04 m
Explanation:
t = Time taken by the quarter to reach the ground = 9.6 s
u = Initial velocity
v = Final velocity
s = Displacement
a)
[tex]v=u+at\\\Rightarrow v=0+9.81\times 9.6\\\Rightarrow v=94.176 m/s[/tex]
Speed of the quarter just before it hits the ground is 94.176 m/s
b)
[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow s=0\times t+\frac{1}{2}\times 9.81\times 9.6^2\\\Rightarrow s=452.04\ m[/tex]
The building is 452.04 m high
Two objects are dropped at the same time from heights of 0.49 m and 1.27 m. How long after the first object hits the ground does the second object hit the ground?
Answer:
After 0.19288 sec second object hit the ground after hitting first object
Explanation:
We have given two objects are dropped from at the same time from height 0.49 m and 1.27 m
As the object is dropped so its initial velocity will be zero u = 0 m/sec
For object 1 height h = 0.49 m
From second equation of motion
[tex]h=ut+\frac{1}{2}gt^2[/tex]
[tex]0.49=0\times t+\frac{1}{2}\times 9.8\times t^2[/tex]
[tex]t^2=0.1[/tex]
t = 0.31622 sec
For second object
[tex]h=ut+\frac{1}{2}gt^2[/tex]
[tex]1.27=0\times t+\frac{1}{2}\times 9.8\times t^2[/tex]
[tex]t^2=0.1[/tex]
t = 0.5091 sec
So difference in time = 0.5091-0.31622 = 0.19288 sec
So after 0.19288 sec second object hit the ground after hitting first object
A 0.676 m long section of cable carrying current to a car starter motor makes an angle of 57.7º with the Earth’s 5.29 x 10^−5T field. What is the current when the wire experiences a force of 6.53 x 10^−3 N?
Answer:
The current in the wire under the influence of the force is 216.033 A
Solution:
According to the question:
Length of the wire, l = 0.676 m
[tex]\theta = 57.7^{\circ}[/tex]
Magnetic field of the Earth, [tex]B_{E} = 5.29\times 10^{- 5} T[/tex]
Forces experienced by the wire, [tex]F_{m} = 6.53\times 10^{-3} N[/tex]
Also, we know that the force in a magnetic field is given by:
[tex]F_{m} = IB_{E}lsin\theta[/tex]
[tex]I = \frac{F_{m}}{B_{E}lsin\theta}[/tex]
[tex]I = \frac{6.53\times 10^{-3}}{5.29\times 10^{- 5}\times 0.676sin57.7^{\circ}[/tex]
I = 216.033 A
HOW DOES LIGHT BEHAVE MOVING FROM A MORE DENSE MEDIUM TO ALESS
DENSE MEDIUM?
Answer:
Light travels faster in a less dense medium
Explanation:
In a less dense medium, light will travel faster than it did in the more dense medium (an example of this transition can be light going from water to air).
This is because in a less dense medium there ares less particles, so light finds less resistence in its path.
The opposite happens in a more dense medium, there are more particles so light will travel slower because more particles will get in the way.
In summary, the more dense the medium, the more particles it has, and the slower that light will travel in it.
Two very small 8.55-g spheres, 15.0 cm apart from center to center, are charged by adding equal numbers of electrons to each of them.
(a)Disregarding all other forces, how many electrons would you have to add to each sphere so that the two spheres will accelerate at 25.0g when released?
(b)Which way will they accelerate?
Answer:
1.43 x 10¹⁷.
They will accelerate away from each other.
Explanation:
Force on each charged sphere F = mass x acceleration
= 8.55 x 10⁻³ x 25 x 9.8
= 2.095 N
Let Q be the charge on each sphere
F = [tex]\frac{9\times10^9\times Q^2}{(15\times10^{-2})^2}[/tex]
2.095 =[tex]\frac{9\times10^9\times Q^2}{(15\times10^{-2})^2}[/tex]
Q² =[tex]\frac{2.095\times(15)^2\times10^{-4}}{9\times10^9}[/tex]
Q = 2.289 X 10⁻⁶
No of electrons = Charge / charge on a single electron
= [tex]\frac{2.289\times10^{-6}}{1.6\times10^{-19}}[/tex]
=1.43 x 10¹³.
They will accelerate away from each other.
The number of electrons would have to add to each sphere to accelerate at 25.0g is 1.43×10¹³ and accelerate away from each other.
What is electric force?Electric force is the force of attraction of repulsion between two bodies.
According to the Coulombs law, the force of attraction of repulsion between charged two bodies is directly proportional to the product of charges of them and inversely proportional to the square of distance between them. It can be given as,
[tex]F=\dfrac{KQ_1Q_2}{r^2}[/tex]
Here, (k) is the coulombs constant, (q1 and q2) is the charges of two bodies and (r) is the distance between the two charges.
Two very small 8.55-g spheres, 15.0 cm apart from center to center, are charged by adding equal numbers of electrons to each of them.
(a) The Number of electrons would have to add to each sphere-Two spheres accelerate at 25.0g when released, and the mass of each sphere is 8.55 g. Thus, the force on the sphere can be given as,
[tex]F=8.55\times{10^{-3}}\times25\times9.8\\F=2.095\rm \;N[/tex]
Two very small spheres are 15.0 cm apart from center to center, are charged by adding equal numbers of electrons to each of them.As the charge on both the sphere is same (say Q) Put these values in the above formula as,
[tex]2.095=\dfrac{(9\times10^{9})QQ}{(15\times10^{-2})^2}\\Q=2.289\times10^{-6}\rm \;C[/tex]
It is known that the charge on one electron is 1.6×10⁻¹⁹ C. Thus the number of electron in the above charge is,
[tex]n=\dfrac{2.289\times10^{-6}}{1.6\times10^{-19}}\\n=1.43\times10^{13}[/tex]
(b)Direction of acceleration-The direction of acceleration of both the sphere will be opposite to each other. Thus, they accelerate away from each other.
Hence, the number of electrons would have to add to each sphere to accelerate at 25.0g is 1.43×10¹³ and accelerate away from each other.
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A football is kicked from ground level at an angle of 38 degrees. It reaches a maximum height of 9.7 meters before returning to the ground. How long will the football spend in the air, in seconds?
Answer:
Football will be in air for 2.8139 sec
Explanation:
We have given maximum height h = 9.7 meters
Angle of projection [tex]\Theta =38^{\circ}[/tex]
We know that maximum height is given by [tex]h=\frac{u^2sin^2\Theta }{2g}[/tex]
So [tex]9.7=\frac{u^2sin^{2}38^{\circ} }{2\times 9.8}[/tex]
[tex]u^2=501.5842[/tex]
u = 22.396 m/sec
Time of flight is given by
[tex]T=\frac{2usin\Theta }{g}=\frac{2\times 22.396\times \times sin38^{\circ}}{9.8}=2.8139sec[/tex]
Considere un campo vectorial U que apunta radialmente hacia afuera de un punto P. La magnitud del campo en un punto Q una distancia r del punto P es a K donde K es una constante positiva. Calcule el flujo de este campo vectorial a través de una superficie esférica de radio R centrada en el punto P
Answer:
El flujo del campo U a través de dicha superficie es [tex]\Phi_U=4\pi r^2K[/tex]
Explanation:
El flujo vectorial de un campo U a través de una superficie S está dado por
[tex]\Phi_U=\int_S \vec{U} \cdot \hat{n} \ dS[/tex]
donde [tex]\hat{n}[/tex] representa el vector unitario normal a la superficie. Teniendo en cuenta que el campo se encuentra en dirección radial, se tiene
[tex]\Phi_U=\int_S K\hat{r} \cdot \hat{r} \ dS = K\int_S dS = KS=K*4\pi r^2[/tex]
What is the velocity of a 30-kg box with a kinetic energy of 6,000 J? 64 m/s
6.0 m/s
18 m/s
20 m/s
36 m/s
Answer: 20 m/s
Explanation: To solve this problem we have to consider the expression of the kinetic energy given by:
Ekinetic=(1/2)*(m*v^2)
then E=0.5*30Kg*(20 m/s)^2=15*400=6000J
Riding in a car, you suddenly put on the brakes. As you experience it inside the car, do Newton's law apply? Do they apply as seen by someone outside the car? Please explain reasoning, thank you!
Answer with Explanation:
Newton's laws are applicable for inertial frames of reference which is a frame which is not accelerating when seen from the observer standing on earth.
For the person as he presses the brakes his frame is a decelerating frame of reference hence he cannot apply the newtons laws of motion as they are in their original form but if he analyses the motion he has to apply a correction known as pseudo-force on the object he is analyzing. Pseudo Force has no basis in newton's laws but are a correction that needs to be applied if he wishes to analyse the motion from non inertial frame of reference
While as a person standing on earth outside the car since his frame is an inertial frame of reference he can apply newton's laws of motion without any correction.
A motorist drives north for 35.0 min at 85.0 km/h and then stops for 15.0 min. He then continues north, traveling 130 km in 2.00 h. (a) what is his total displacement? b) what is his average velocity?
Final answer:
The motorist's total displacement is 179.58 km north, and his average velocity is approximately 63.38 km/h north for the entire trip.
Explanation:
Displacement and Average Velocity Calculation
For the given problem, we can find the motorist's total displacement by adding the distances traveled in each segment of the journey, all in the same direction (north). First, the motorist drives north for 35.0 minutes at 85.0 km/h. The distance for this part of the trip is:
Distance = Speed × Time = 85.0 km/h × (35.0 min / 60 min/h) = 49.58 km (rounded to 2 decimal places)
The next segment involves a stop, so the displacement remains unchanged.
Finally, the motorist drives an additional 130 km north.
Therefore, the total displacement is the sum of these distances: Total Displacement = 49.58 km + 130 km = 179.58 km north.
To find the average velocity, we need to consider the total displacement and the total time, including the stop:
Total time = Driving time + Stopping time = (35.0 min + 2.00 h + 15.0 min) = 2.833 hours (2 hours and 50 minutes).
Average Velocity = Total Displacement / Total Time = 179.58 km / 2.833 h ≈ 63.38 km/h north.
The motorist's total displacement is 179.58 km north, and his average velocity over the entire trip is approximately 63.38 km/h north.
A 2 gram marble is placed in the bottom of a frictionless, hemispherical bowl with a diameter of 4.9 m and it is set in circular motion around the lowest point in the bowl such that its radius of "orbit" is 16 cm. What is its speed? (If you find this problem confusing, consider a simple pendulum with a string length half the diameter given here and suppose its amplitude is 16 cm.)
Answer:
6.93 m/s
Explanation:
mass of marble, m = 2 g
Diameter of the bowl = 4.9 m
radius of bowl, r = half of diameter of the bowl = 2.45 m
The length of the string is 2.45 m
Now it is executing simple harmonic motion, thus the potential energy at the bottom is equal to the kinetic energy at the highest point.
[tex]mgr=\frac{1}{2}mv^{2}[/tex]
where, m be the mass of marble, v be the velocity of marble at heighest point, g be the acceleration due to gravity and r be the radius of the path which is equal to the length of the string
By substituting the values
9.8 x 2.45 = 0.5 x v^2
v = 6.93 m/s
Thus, the speed is 6.93 m/s.
The pilot of an airplane reads the altitude 6400 m and the absolute pressure 46 kPa when flying over a city. Calculate the local atmospheric pressure in that city in kPa and in mmHg. Take the densities of air and mercury to be 0.828 kg/m3 and 13,600 kg/m3, respectively.
The local atmospheric pressure in the city in kPa is ?
The local atmospheric pressure in the city in mmHg is ?
Answer:
1. 6.672 kPa
2. 49.05 mm of mercury
Explanation:
h = 6400 m
Absolute pressure, p = 46 kPa = 46000 Pa
density of air, d = 0.823 kg/m^3
density of mercury, D = 13600 kg/m^3
(a) Absolute pressure = Atmospheric pressure + pressure due to height
46000 = Atmospheric pressure + h x d x g
Atmospheric pressure = 46000 - 6400 x 0.823 x 10 = 6672 Pa = 6.672 kPa
(b) To convert the pressure into mercury pressure
Atmospheric pressure = H x D x g
Where, H is the height of mercury, D be the density of mercury, g be the acceleration due to gravity
6672 = H x 13600 x 10
H = 0.04905 m
H = 49.05 mm of mercury
To calculate the local atmospheric pressure in the city, use the formula P = P0 * e^(-Mgh/RT), where P is the pressure at the current location, P0 is the pressure at sea level, M is the molar mass of air, g is the acceleration due to gravity, h is the altitude, R is the ideal gas constant, and T is the temperature. Plug in the given values into the formula and solve for P0 to find the local atmospheric pressure in kPa. To convert this pressure to mmHg, use the conversion factor that 101.325 kPa is equal to 760 mmHg.
Explanation:To calculate the local atmospheric pressure in the city, we can use the relationship between altitude, pressure, and density of air. Altitude affects atmospheric pressure because as we go higher in the atmosphere, the air becomes less dense and therefore exerts less pressure. The formula to calculate pressure is given by:
P = P0 * e^(-Mgh/RT)
Where P is the local atmospheric pressure, P0 is the pressure at sea level, M is the molar mass of air, g is the acceleration due to gravity, h is the altitude, R is the ideal gas constant, and T is the temperature. Since we know the altitude and pressure at the current location, we can rearrange the formula to solve for P0:
P0 = P / e^(-Mgh/RT)
Using the given values, we can substitute them into the formula. The density of air is 0.828 kg/m³, the altitude is 6400 m, the absolute pressure is 46 kPa, the molar mass of air is the average molar mass of air (28.97 g/mol), the acceleration due to gravity is 9.8 m/s², the ideal gas constant is 8.314 J/(mol·K), and the temperature can be assumed to be the average temperature at that altitude (around 0°C or 273 K).
Plugging in these values into the formula, we get:
Calculating this expression will give us the local atmospheric pressure in kPa.
To convert this pressure to mmHg, we can use the conversion factor that 101.325 kPa is equal to 760 mmHg. So, to find the pressure in mmHg, we can multiply the pressure in kPa by the conversion factor.
Air around the center of surface low pressure systems in the Northern Hemisphere is spinning ______ and _______ the system's center. O counter-clockwise; converging towards O counter-clockwise; diverging away from O clockwise; diverging away from O clockwise; converging towards
Answer:
Clockwise, converging towards
Explanation:
At the center of surface , system with low pressure in Northern hemisphere have air around the center that rotates in anti clockwise direction.
As a result of low pressure, the air is directed slightly inwards thus converges towards the center of the system.
In the high pressure systems, air rotates in clockwise direction and diverges away from the center of the system.
A package is dropped from a helicopter moving upward at 15m/s. If it takes 10 s before the package strikes the ground, how high above the ground was the ground was the package when it was released if air resistance is negligible?
(A) 408 m
(B) 272 m
(C) 204 m
(D) 340 m
Answer:
The right answer is (D) 340m
Explanation:
If the package is dropped from a helicopter moving upward and the air resistance is negligible, the only acting force in the package is the gravity force. Therefore you can consider this as a Vertical Projectile Motion problem.
The initial speed of the package V₀ is 15m/s.
The acceleration of the package G is 9.8m/s².
The initial hight is Y₀. Therefore:
Y(t)=Y₀+V₀·t-0.5·G·t²
But we know than T(10s)=0m
0m=Y₀+150m-490m
Y₀=340m
A proton is accelerated down a uniform electric field of 450 N/C. Calculate the acceleration of this proton.
Answer:
The acceleration of proton will be [tex]431.137\times 10^8m/sec^2[/tex]
Explanation:
We have given electric field E = 450 N/C
Charge on proton [tex]=1.6\times 10^{-19}C[/tex]
Force on electron due to electric field is given by [tex]F=qE=1.6\times 10^{-19}\times 450=720\times 10^{-19}N[/tex]
Mass of electron [tex]m=1.67\times 10^{-27}kg[/tex]
Now according to second law of motion [tex]F=ma[/tex]
So [tex]720\times10^{-19}=1.67\times 10^{-27}a[/tex]
[tex]a=431.137\times 10^8m/sec^2[/tex]
So the acceleration of proton will be [tex]431.137\times 10^8m/sec^2[/tex]
A mechanics shop has a 16,500 N car sitting on a large car lift of 125 cm^2 piston. How much force is needed to be applied to the small piston of area 4.51 cm^2 to lift it?
Answer:595.32 N
Explanation:
Given
Mechanic shop has a car of weight([tex]F_1[/tex]) 16,500 N
Area of lift([tex]A_1[/tex])=[tex]125 cm^2[/tex]
Area of small piston([tex]A_2[/tex])=[tex]4.51 cm^2[/tex]
According to pascal's law pressure transmit will be same in all direction in a closed container
[tex]P_1=P_2[/tex]
[tex]\frac{F_1}{A_1}=\frac{F_2}{A_2}[/tex]
[tex]F_2=F_1\frac{A_2}{A_1}[/tex]
[tex]F_2=16,500\times \frac{4.51}{125}=595.32 N[/tex]
A block is placed at the top of a frictionless inclined plane with angle 30 degrees and released. The incline has a height of 15 m and the block has mass 2 kg. In the above problem the speed of the block when it reaches the bottom of the incline is 8.3 m/s 17 m/s 25 m/s 30 m/s
Answer:
The speed of the block when it reaches the bottom is 17 m/s.
(b) is correct option.
Explanation:
Given that,
Height = 15 m
Mass = 2 kg
We need to calculate the speed of the block when it reaches the bottom
Using conservation of energy
[tex]mgh=\dfrac{1}{2}mv^2[/tex]
[tex]v^2=2gh[/tex]
[tex]v=\sqrt{2gh}[/tex]
Where, g = acceleration due to gravity
h = height
Put the value into the formula
[tex]v=\sqrt{2\times9.8\times15}[/tex]
[tex]v=17\ m/s[/tex]
Hence, The speed of the block when it reaches the bottom is 17 m/s.
The neutron mass is 1.675 x 10-27 kg. Express the proton mass in eV/c, keeping 4 significant figures B) The mass of the electron is 9.109 x 10-31 kg. Express this in eV/c . C) What is the mass of the hydrogen atom in eV/c??
Answer:
Explanation:
given,
mass of neutron = 1.675 × 10⁻²⁷ kg
1 kg = 5.58 × 10³⁵ eV/c²
a) mass of proton = 1.675 × 10⁻²⁷ × 5.58 × 10³⁵ = 9.34 × 10⁸ eV/c²
b) mass of electron = 9.109 x 10⁻³¹ × 5.58 × 10³⁵ = 5.08 × 10⁵eV/c²
c) mass of hydrogen = 1.675 × 10⁻²⁷ × 5.58 × 10³⁵ = 9.34 × 10⁸ eV/c²
For a satellite already in perfect orbit around the Earth, what would happen if: - the satellite's speed is reduced? - the satellite's mass is reduced?
Answer:
Answer
Explanation:
If the satellite is in perfect orbit around the earth then gravitational pull of the earth on the satellite and moving forward satellite is in perfect balance i.e gravitational pull does not effect the orbit as the velocity of the satellite is good enough to counter the gravitational pull of the earth and inertia makes the satellite move forward with greater velocity, but when its velocity is suddenly reduced, the gravitational pull will counter on the orbiting satellite and it will crash onto the earth.
This is same as in well of death when motorist moves in circular motion inside the well with centripetal and centrifugal forces balanced, suddenly crashes on the ground when velocity of the motorist is not enough to balance forces.
the satellite's mass has no affect on its orbit around the earth rather the mass of the earth is only affecting factor .
Ropes 3 m and 5 m in length are fastened to a holiday decoration that is suspended over a town square. The decoration has a mass of 3 kg. The ropes, fastened at different heights, make angles of 52° and 40° with the horizontal. Find the tension in each wire and the magnitude of each tension. (Use g
Answer:
T₁ = 19.39N with an angle of 52° with the horizontal.
T₂ = 15.64N with an angle of 40° with the horizontal.
Explanation:
1)We calculate the weight of holiday decoration
W = m×g
W: holiday decoration weight in Newtons (N)
m: holiday decoration weight = 3 kg
g: acceleration due to gravity = 9.8 m/s²
W = m×g = 3 kg × 9.8 m/s² = 29.4 N
2)We apply Newton's first law to the system in equilibrium:
ΣFx = 0
T₁Cos52° - T₂Cos40°=0
T₁Cos52° = T₂Cos40°
T₁ = T₂(Cos40° / T₁Cos52°)
T₁ = 1,24×T₂ Equation (1)
ΣFy = 0
T₁Sin52° + T₂Sin40° - 29.4 = 0 Equation (2)
We replace T₁ of Equation (1) in the Equation (2)
1.24×T₂×Sin52° + T₂Sin40° - 29.4 = 0
1.88*T₂ = 29.4
T₂ = 29.4/1.88
T₂ = 15.64N
We replace T₂ = 15.64N in the Equation (1) to calculate T₁:
T₁ = 1.24×15.64N
T₁ = 19.39N
How would you find the total energy stored in the
capacitorsif.....
A 2.0 microF capacitor and a 4.0 microF capacitor are connected
inPARALLEL across a 300V potential difference.
Answer:
The energy of the capacitors connected in parallel is 0.27 J
Given:
C = [tex]2.0\micro F = 2.0\times 10^{- 6} F[/tex]
C' = [tex]4.0\micro F = 4.0\times 10^{- 6} F[/tex]
Potential difference, V = 300 V
Solution:
Now, we know that the equivalent capacitance of the two parallel connected capacitors is given by:
[tex]C_{eq} = C + C' = 2.0 + 4.0 = 6.0\micro F = 6.0\times 10^{- 6} F[/tex]
The energy of the capacitor, E is given by;
[tex]E = \frac{1}{2}C_{eq}V^{2}[/tex]
[tex]E = \frac{1}{2}\times 6.0\times 10^{- 6}\times 300^{2} = 0.27 J[/tex]
A standard 1 kilogram weight is a cylinder 48.0 mm in height and 55.0 mm in diameter. What is the density of the material?
Answer:8769 Kg/m^3
Explanation: First of all we have to calculate the volume of the cylinder so it is equal to:
Vcylinder= π*r^2*h where r and h are the radius and heigth, respectively.
Vcylinder= π*0.0275^2*0.048=1.14*10^-4 m^3
The density is defined by ρ=mass/volume
the we have
ρ=1Kg/1.14*10^-4 m^3=8769 Kg/m^3
What fraction of all the electrons in a 25 mg water
dropletmust be removed in order to obtain a net charge of 40
nC?
Answer:
[tex]9.11\times 10^{-15}.[/tex]
Explanation:
The water droplet is initially neutral, it will obtain a 40 nC of charge when a charge of -40 nC is removed from the water droplet.
The charge on one electron, [tex]\rm e=-1.6\times 10^{-19}\ C.[/tex]
Let the N number of electrons have charge -40 nC, such that,
[tex]\rm Ne=-40\ nC\\\Rightarrow N=\dfrac{-40\ nC}{e}=\dfrac{-40\times 10^{-9}\ C}{-1.6\times 10^{-19}\ C}=2.5\times 10^{11}.[/tex]
Now, mass of one electron = [tex]\rm 9.11\times 10^{-31}\ kg.[/tex]
Therefore, mass of N electrons = [tex]\rm N\times 9.11\times 10^{-31}=2.5\times 10^{11}\times 9.11\times 10^{-31}=2.2775\times 10^{-19}\ kg.[/tex]
It is the mass of the of the water droplet that must be removed in order to obtain a charge of 40 nC.
Let it is m times the total mass of the droplet which is [tex]25\ \rm mg = 25\times 10^{-6}\ kg.[/tex]
Then,
[tex]\rm m\times (25\times 10^{-3}\ kg) = 2.2775\times 10^{-19}\ kg.\\m=\dfrac{2.2775\times 10^{-19}\ kg}{25\times 10^{-3}\ kg}=9.11\times 10^{-15}.[/tex]
It is the required fraction of mass of the droplet.
Bragg reflection results in a first-order maximum at 15.0°. In this case, at what angle would the second-order maximum occur?
Answer:
31.174°
Explanation:
Bragg's condition is occur when the wavelength of radiation is comparable with the atomic spacing.
So, Bragg's reflection condition for n order is,
[tex]2dsin\theta=n\lambda[/tex]
Here, n is the order of maxima, [tex]\lambda[/tex] is the wavelength of incident radiation, d is the inter planar spacing.
Now according to the question, first order maxima occur at angle of 15°.
Therefore
[tex]2dsin(15^{\circ})=\lambda\\sin(15^{\circ})=\frac{\lambda}{2d}[/tex]
Now for second order maxima, n=2.
[tex]2dsin\theta=2\times \lambda\\sin\theta=\frac{2\lambda}{2d}[/tex]
Put the values from above conditions
[tex]sin\theta=2\times sin(15^{\circ})\\sin\theta=2\times 0.258819045103\\sin\theta=0.517638090206\\\theta=sin^{-1}0.517638090206\\ \theta=31.174^{\circ}[/tex]
Therefore, the second order maxima occurs at 31.174° angle.
An aircraft carrier is sailing at a constant speed of 15.0m/s. it is suddenly spotted by an enemy bomber aircraft flying in the same direction as the ship is moving. the bomber is flying at a constant speed of 100 m/s at an altitude of 1.00 km above the carrier when it decides to attack the carrier. if the bomber drops its bomb at a horizontal distance pf 1.3 km from the aircraft carrier, will it score a direct hit?
Answer:
Explanation:
Relative velocity = 100 - 15 = 85 m /s
Distance to be covered up with this relative speed = 1.3km
= 1300m
If t be the time taken
85 x t = 1300
t = 1300/ 85
= 15.29
Time of fall of the bomb t₁ by 1 km
1000 = 1/2 g t₁²
t₁² = 2000/9.8
t₁ = 14.28 s
Since t₁≠ t₂
It will not score a direct hit.
A uniformly accelerated car passes three equally spaced traffic signs. The signs are separated by a distance d = 25 m. The car passes the first sign at t = 1.3 s, the second sign at t = 3.9 s, and the third sign at t = 5.5 s.
(a) What is the magnitude of the average velocity of the car during the time that it is moving between the first two signs?
(b) What is the magnitude of the average velocity of the car during the time that it is moving between the second and third signs?
(c) What is the magnitude of the acceleration of the car?
The average velocity between the first two signs is 9.1 m/s, between the second and third signs is 10.4 m/s, and the car's acceleration is approximately 0.26 m/s².
To solve for the average velocity and acceleration of the car passing three equally spaced traffic signs, we can use the kinematic equations for uniformly accelerated motion.
Part a: average velocity between the first two signs
The average velocity is calculated by dividing the displacement (distance between the signs) by the time interval: vavg1,2 = d /
[tex](t_2 - t_1)[/tex]. Substituting the given values, we have vavg1,2 = 25 m / (3.9 s - 1.3 s), resulting in an average velocity of 9.1 m/s.
Part b: average velocity between the second and third signs
Similarly, for the average velocity between the second and third signs: vavg2,3 = d / ([tex]t_3 - t_2[/tex]). Using the given times, we calculate vavg2,3 = 25 m / (5.5 s - 3.9 s), giving an average velocity of 10.4 m/s.
Part c: magnitude of the acceleration
To find the acceleration, we can use the velocities we just calculated, and the time between the signs. We can set up an equation for motion for the time intervals. As initial velocity v0 for the second interval is the final velocity of the first interval, we get v0 = vavg1,2 and final velocity v = vavg2,3. Then we can use the equation v = v0 + a[tex](t_3 - t_2)[/tex] to find the acceleration. Rearranging for a, we have a = [tex](v - v_0) / (t_3 - t_2)[/tex]. After calculating, we find that the acceleration is approximately 0.26 m/s².
38.4 mol of krypton is in a rigid box of volume 64 cm^3 and is initially at temperature 512.88°C. The gas then undergoes isobaric heating to a temperature of 935.9°C. (a) What is the final volume of the gas?(in cm^3 ) (b) It is then isothermally compressed to a volume 29.3cm^3; what is its final pressure?(in Pa )
Answer:
Final volumen first process [tex]V_{2} = 98,44 cm^{3}[/tex]
Final Pressure second process [tex]P_{3} = 1,317 * 10^{10} Pa[/tex]
Explanation:
Using the Ideal Gases Law yoy have for pressure:
[tex]P_{1} = \frac{n_{1} R T_{1} }{V_{1} }[/tex]
where:
P is the pressure, in Pa
n is the nuber of moles of gas
R is the universal gas constant: 8,314 J/mol K
T is the temperature in Kelvin
V is the volumen in cubic meters
Given that the amount of material is constant in the process:
[tex]n_{1} = n_{2} = n [/tex]
In an isobaric process the pressure is constant so:
[tex]P_{1} = P_{2} [/tex]
[tex]\frac{n R T_{1} }{V_{1} } = \frac{n R T_{2} }{V_{2} }[/tex]
[tex]\frac{T_{1} }{V_{1} } = \frac{T_{2} }{V_{2} }[/tex]
[tex]V_{2} = \frac{T_{2} V_{1} }{T_{1} }[/tex]
Replacing : [tex]T_{1} =786 K, T_{2} =1209 K, V_{1} = 64 cm^{3}[/tex]
[tex]V_{2} = 98,44 cm^{3}[/tex]
Replacing on the ideal gases formula the pressure at this piont is:
[tex]P_{2} = 3,92 * 10^{9} Pa[/tex]
For Temperature the ideal gases formula is:
[tex]T = \frac{P V }{n R }[/tex]
For the second process you have that [tex]T_{2} = T_{3} [/tex] So:
[tex]\frac{P_{2} V_{2} }{n R } = \frac{P_{3} V_{3} }{n R }[/tex]
[tex]P_{2} V_{2} = P_{3} V_{3} [/tex]
[tex]P_{3} = \frac{P_{2} V_{2}}{V_{3}} [/tex]
[tex]P_{3} = 1,317 * 10^{10} Pa[/tex]
The brakes on your automobile are capable of creating a deceleration of 4.6 m/s^2. If you are going 114 km/h and suddenly see a state trooper, what is the minimum time in which you can get your car under the 79 km/h speed limit? (The answer reveals the futility of braking to keep your high speed from being detected with a radar or laser gun.)
Answer:
The minimum time to get the car under max. speed limit of 79 km/h is 2.11 seconds.
Explanation:
[tex]a=\frac{V_f-V_0}{t}[/tex]
isolating "t" from this equation:
[tex]t=\frac{V_f-V_0}{a}[/tex]
Where:
a=[tex]-4.6m/s^2[/tex] (negative because is decelerating)
[tex]V_f= 79 km/h[/tex]
[tex]V_0= 114 km/h[/tex]
First we must convert velocity from km/h to m/s to be consistent with units.
[tex]79\frac{km}{h}*\frac{1000m}{1 km}*\frac{1h}{3600s}=\frac{79*1000}{3600}=21.94 m/s[/tex]
[tex]114\frac{km}{h}*\frac{1000m}{1 km}*\frac{1h}{3600s}=\frac{114*1000}{3600}=31.67 m/s[/tex]
So;
[tex]t=\frac{V_f-V_0}{a}=\frac{21.94 m/s-31.66m/s}{-4.6 m/s^2}=2.11 s[/tex]