1. The inductive time constant RL has units of seconds.
2. Doubling the inductance in an LR circuit does not affect the half-life.
3. Doubling the resistance in an LR circuit increases the half-life.
4. Doubling the charging voltage in an LR circuit does not affect the half-life.
5. To plot the equation V(1) = Vmax × [tex]e^{(tR/L)[/tex] as a straight line, plot ln(V(1)) against time and the slope is (R/L).
6. The expected self-inductance of the solenoid is calculated using the formula L = (4π × [tex]10^{-7[/tex] Tm/A) × (1600²) × (π × (0.02)²) / 0.12.
1. To show that the inductive time constant RL has units of seconds, we need to consider the units of the inductance (L) and resistance (R) individually.
The unit of inductance, L, is Henries (H).
The unit of resistance, R, is ohms (Ω).
The time constant (τ) of an RL circuit is given by the formula τ = L/R.
Substituting the units, we have:
τ = (H)/(Ω)
By rearranging the units, we can express henries (H) in terms of seconds (s):
1 H = 1 (Ω)(s)
Therefore, the units of RL, which is the time constant of an RL circuit, are seconds (s).
2. If the inductance in the LR circuit is doubled, the half-life is not affected. The half-life is a measure of the time it takes for the current (or voltage) to decrease to half of its initial value in an exponential decay process. The half-life is independent of inductance (L) and is primarily determined by the resistance (R) in the circuit.
3. If the resistance in the LR circuit is doubled, the half-life is increased. The half-life is directly proportional to the resistance (R) in the circuit. Doubling the resistance will result in a longer time for the current (or voltage) to decrease to half its initial value.
4. If the charging voltage in the circuit is doubled, the half-life is not affected. The half-life of an LR circuit depends on the resistance (R) and inductance (L) but is independent of the charging voltage. Increasing the charging voltage will result in a higher initial current (or voltage), but it will not affect the time it takes for the current (or voltage) to decrease to half its initial value.
5. To plot the equation V(1) = Vmax × [tex]e^{(tR/L)[/tex] in a way that results in a straight line, you need to plot the natural logarithm of the voltage (ln(V(1))) against time (t). The equation then becomes ln(V(1)) = (R/L) × t + ln(Vmax), which is in the form of a linear equation (y = mx + c), where m is the slope and c is the y-intercept.
The expression for the slope of this straight line is (R/L), which represents the ratio of resistance (R) to inductance (L) in the LR circuit.
6. To determine the expected self-inductance of a solenoid with the given specifications, we can use the formula for the self-inductance of a solenoid:
L = (μ₀ × N² × A) / l
Where:
L is the self-inductance
μ₀ is the permeability of free space (4π × [tex]10^{-7[/tex] Tm/A)
N is the number of windings (1600 windings)
A is the cross-sectional area of the solenoid (π × r², where r is the radius of the solenoid)
l is the length of the solenoid (12 cm)
Let's calculate the self-inductance using the given values:
N = 1600
r = 2.0 cm = 0.02 m
A = π × (0.02)²
l = 12 cm = 0.12 m
Substituting these values into the formula, we have:
L = (4π × [tex]10^{-7[/tex] Tm/A) × (1600²) × (π × (0.02)²) / 0.12
Simplifying the expression, we can calculate the expected self-inductance.
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Which one of the following items contains matter in the plasma state?
Stars
liquid water
ice cubes
Answer:
Stars i think
Explanation:
(a) An insulating sphere with radiusa has a uniform charge density rho. The sphere isnot centered at the origin but at. Show that the electric field inside thesphere is given by.
(b) An insulating sphere of radius R has a spherical holeof radius a located within its volume and centered adistance b from the center of the sphere. The solid partof the sphere has a uniform volume charge density rho. Findthe magnitude and direction of the electric fieldinside the hole, and show thatis uniform over the entire hole.[Hint: Use the principle of superposition and the resultof part (a).]
a. The electric field is E = (kρa^3/r^2) r
b. The magnitude of the electric field inside the hole of the sphere is (kρa^3/b^2 - kρ(4/3)πa^3/(b^2 - a^2)) j
a)
The electric field inside an insulating sphere with radius a and uniform charge density rho is,
E = (kq/r^2) r
Where
q = volume charge density * volume of the sphere
= rho*(4/3)πa^3
q = (4/3)πa^3ρ
The electric field is:,
E = (kq/r^2) r
= (kρa^3/r^2) r
b)
E = E_solid_sphere - E_hollow_sphere
The electric field due to the charge density inside the hole of the sphere is zero because the sphere is insulating. Therefore, the electric field inside the hole due to the charge density is zero and is uniform over the entire hole.The electric field due to the solid sphere can be found using the result from part (a),E_solid_sphere = (kρa^3/b^2) j
The electric field inside the hole due to the solid sphere is,E_hollow_sphere = -(kρ(4/3)πa^3/(b^2 - a^2)) j
Therefore, the magnitude and direction of the electric field inside the hole of the sphere areE = E_solid_sphere - E_hollow_sphere
= (kρa^3/b^2 - kρ(4/3)πa^3/(b^2 - a^2)) j
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a thin-walled, hollow sphere of mass m and radius r is free to rotate around a vertical shaft that passes through the center of the sphere. initially, the sphere is at rest. a small ball of clay of the same mass m moving horizontally at speed v grazes the surface of the sphere at its equator. after grazing the surface, the ball of clay is moving at speed v/2 . what is the angular momentum of the ball of clay about the shaft before it grazes the surface? express your answer in terms of the variables
The angular momentum of the ball of clay before grazing the surface is (2/3)mrv.
What is the angular momentum of the sphere?The angular momentum of the ball of clay about the shaft before it grazes the surface is calculated as follows;
Moment of inertia of the ball;
I = (2/3) mr²
where;
m is the mass of the spherer is its radius.Angular velocity is given as;
ω = v / r
where;
v is linear velocityr is the radiusThe angular momentum of the ball of clay before grazing the surface is calculated as;
L = Iω
L = (2/3)mr² x (v / r)
L = (2/3)mrv
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A carbon resistor is 8 mm long and has a constant cross section of 0.3 mm2. The conductivity of carbon at room temperature is σ = 3 ✕ 104 per ohm·m. In a circuit its potential at one end of the resistor is 15 volts relative to ground, and at the other end the potential is 21 volts. A thin copper wire in the same circuit is 8 mm long and has a constant cross section of 0.3 mm2. The conductivity of copper at room temperature is σ = 6 ✕ 107 ohm-1m-1. The copper wire is in series with the carbon resistor in the same circuit mentioned above, with one end connected to the 21 volt end of the carbon resistor. Calculate the resistance R of the copper wire and the potential Vat end at the other end of the wire.
R =___ ohms
V at end = ____V
You can see that for most purposes a thick copper wire in a circuit would have practically a uniform potential. This is because the small drift speed in a thick, high-conductivity copper wire requires only a very small electric field, and the integral of this very small field creates a very small potential difference along the wire.
The resistance (R) of the copper wire in the circuit is 1600 ohms, and the potential (V) at the other end of the wire is 21 volts.
To calculate the resistance of the copper wire (R), we use the formula R = (ρ * L) / A, where ρ represents the resistivity of the material, L is the length of the wire, and A is the cross-sectional area. In this case, the length of the copper wire is 8 mm (0.008 m), and the cross-sectional area is 0.3 mm^2 (0.3 * 10^(-6) m^2). With the resistivity of copper being 6 * 10^7 ohm^(-1) m^(-1), we can calculate the resistance as follows: R = (6 * 10^7 ohm^(-1) m^(-1) * 0.008 m) / (0.3 * 10^(-6) m^2), which gives us 1600 ohms. Since the copper wire and the carbon resistor are connected in series in the circuit, the potential difference (V) across each component is the same. In the case of a thick, high-conductivity copper wire, the small drift speed of electrons requires only a very small electric field, resulting in a negligible potential difference along the wire. As a result, the potential throughout the thick copper wire remains practically uniform.
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A helicopter's speed increases from 25 m/s to 60 m/s in 5 seconds. What is the acceleration of this helicopter?
7 m/s
2-2.3 m/s
2-1.9 m/s
21.67 m/s2
Given, Initial speed of the helicopter = 25 m/s. Final speed of the helicopter = 60 m/s, Time taken by the helicopter to change its speed from 25 m/s to 60 m/s is 5 seconds. Therefore, the acceleration of the helicopter is 7 m/s².Hence, option (A) is correct.
To find the acceleration of the helicopter, we can use the formula:
Acceleration= (Final velocity - Initial velocity)/Time,
Acceleration = (60 m/s - 25 m/s)/5 s,
Acceleration = 35 m/s/5 s,
Acceleration = 7 m/s²
Therefore, the acceleration of the helicopter is 7 m/s².
We know that,
Acceleration = change in velocity / time taken,
Acceleration can be defined as the rate at which an object changes its velocity.
It can be measured in units such as m/s², cm/s², etc.
Here, initial velocity = 25 m/s, Final velocity = 60 m/s time = 5 s.
Hence,
Acceleration= (Final velocity - Initial velocity)/Time
Acceleration = (60 m/s - 25 m/s)/5 s
Acceleration= 35/5
Acceleration= 7 m/s²
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Which of the following correctly defines the speed of a wave?
O A. v=1
B. A = vf
O C. v= f 2
OD. =v2
Answer:
V = f x λ
Explanation:
The options are confusing
  do batteries produce or make energy
Answer:
Batteries are intended to store chemical energy and then it converts into electricidal energy. So overall, batteries produce energy.
Explanation:
Is this right. Please help me ITS SOCIOLOGY
Answer:
Yes
Explanation:
sorry if im wrong
Light and Polarization: Learn from Einstein the properties of light Virtual Lab
Light is an electromagnetic wave that consists of oscillating electric and magnetic fields. It exhibits various properties, including wavelength, frequency, speed, and polarization.
Polarization refers to the orientation of the electric field vector of a light wave. Unpolarized light consists of electric field vectors oscillating in all possible directions perpendicular to the direction of propagation. Polarized light, on the other hand, has its electric field vectors confined to a specific orientation. Polarization can be achieved through various mechanisms, such as reflection, scattering, or passing light through certain materials. Polarizers, such as polarizing filters, can selectively transmit or block light waves based on their polarization orientation. The study of light and its properties, including polarization, has contributed to numerous advancements in various fields, such as optics, telecommunications, and technology.
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a thermodynamic cycle is a series of steps that ultimately returns to its beginning point. compute the total work performed around the thermodynamic cycle of quasi-static processes in the figure 7.
The total work performed around the thermodynamic cycle in Figure 7 is 350 Joules.
In a thermodynamic cycle, the total work performed is equal to the net area enclosed by the cycle on a pressure-volume (PV) diagram. From the given figure, we can observe that the cycle consists of two quasi-static processes: Process 1-2 and Process 2-3.
In Process 1-2, the system undergoes an expansion at constant pressure, represented by the horizontal line between points 1 and 2 on the PV diagram. As the volume increases, work is done by the system, and the work done in this process is given by the equation W_1-2 = PΔV = 2 * 50 = 100 Joules (where P is the constant pressure and ΔV is the change in volume).
In Process 2-3, the system undergoes compression at constant volume, represented by the vertical line between points 2 and 3 on the PV diagram. Work is done on the system during this process, and the work done in this process is given by the equation W_2-3 = -PΔV = -3 * 50 = -150 Joules (where P is the constant pressure and ΔV is the change in volume).
The total work performed around the cycle is the sum of the work done in each process, i.e., 100 Joules + (-150 Joules) = -50 Joules.
The total work performed around the thermodynamic cycle in Figure 7 is -50 Joules.
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Rita is a salon owner. She notices that her salon charged one of her clients, Linda, extra for a service that the clent did not request. What do you think Rita should do? Α. Remaln qulet about the extra money. B. Distribute the money equally among the staff. C. Call the client and Inform her that she was incorrectly charged. D. Try overcharging the next client too and check if it goes unnoticed.
Answer:
C hope it helps
call the client and inform her that she was incorrectly charged.
Answer:
The answer is C. call the client and inform her that she was incorrectky charged
Explanation:
A 72.0 kg ice skater is moving at 3.1 m/s on frictionless ice throws a 0.21 kg snowball horizontally at a speed of 28.0 m/s. What is the final velocity of the skater?
Answer:
the final velocity of the skater after throwing the snowball is 3.17 m/s.
Explanation:
Given;
mass of the ice skater, m₁ = 72 kg
initial velocity of the ice skater, u₁ = 3.1 m/s
mass of the snowball, m₂ = 0.21 kg
initial speed of the snowball, u₂ = 28.0 m/s
Let the final velocity of the skater after throwing the snowball = v
Apply the principle of conservation of linear momentum to determine v;
m₁u₁ + m₂u₂ = v(m₁ + m₂)
72 x 3.1 + 0.21 x 28 = v(72 + 0.21)
229.08 = v(72.21)
v = 229.08 / 72.21
v = 3.17 m/s
Therefore, the final velocity of the skater after throwing the snowball is 3.17 m/s.
Use the drop-down menus to complete the statements. Outlook allows users to insert symbols and characters not located on the , but they can have shortcut keys. Inserting horizontal lines in the message body breaks up different sections and mostly benefits people using Outlook on .
Answer:
Keyboard and Mobile devices
Explanation:
got it right on edge 2021
Answer:
Outlook allows users to insert symbols and characters not located on the
keyboard
, but they can have shortcut keys.
Inserting horizontal lines in the message body breaks up different sections and mostly benefits people using Outlook on
mobile devices
.
Explanation:
A metal rod is 25.000 cm long at 25.0 degrees Celsius. When heated to 102.0 degrees Celsius, it is 25.054 cm long. What is the coefficient of linear expansion for this metal.
n2 Find a linear homogeneous recurrence relation with constant coefficients that the sequence An = (n-3) (2") + NEN 2n satisfies. Do not forget to specify initial values.
The linear homogeneous recurrence relation with constant coefficients that the sequence is: An = -(-1)n + (3)n or An = (1)n + (3)n
The sequence An = (n-3)(2n) + n(2n) satisfies the linear homogeneous recurrence relation with constant coefficients. In order to prove this, we need to first find the general formula of the sequence.
The formula can be found by replacing n with n+1 as follows:
An+1 = (n+1-3)(2n+2) + (n+1)(2n+2)
An+1 = n(2n+4) + (n+1)(2n+2)
An+1 = 2n² + 4n + 2n² + 4n + n + 2n + 2
An+1 = 4n² + 7n + 2
The characteristic equation of the linear homogeneous recurrence relation is given by:
r² - 2r - 3 = 0
Solving this quadratic equation, we get:r = -1 or r = 3
Hence, the general formula of the sequence is given by:An = A(-1)n + B(3)n
Now, we need to find the values of A and B using the initial values of the sequence. The first two terms of the sequence are given by:
A0 = -6 and A1 = 2
Substituting these values in the general formula, we get:
-6 = A + B2 = -A + 3B
From the above two equations, we can solve for A and B to get:
A = -1 and B = 1
Hence, the linear homogeneous recurrence relation with constant coefficients that the sequence An = (n-3)(2n) + n(2n) satisfies is given by:
An = -(-1)n + (3)n or An = (1)n + (3)n
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planet a exerts a force on planet b. what can be said about the magnitude and direction of the gravitational force planet b exerts on planet a?
Planet A exerts a force on planet B, the magnitude and direction of the gravitational force planet B exerts on planet A the gravitational force exerted by planet B on planet A is the same as the magnitude of the gravitational force exerted by planet A on planet B
Newton's third law states that if object A exerts a force on object B, then object B exerts an equal and opposite force on object A. Hence, if planet A exerts a gravitational force on planet B, then planet B exerts an equal and opposite gravitational force on planet A.The magnitude of the gravitational force exerted by planet B on planet A is the same as the magnitude of the gravitational force exerted by planet A on planet B, this is according to the law of universal gravitation,
This law states that the force of gravity between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers. The direction of the gravitational force exerted by planet B on planet A is towards planet B's center, just as the direction of the gravitational force exerted by planet A on planet B is towards planet A's center. Therefore, we can say that the magnitude and direction of the gravitational force planet B exerts on planet A is equal and opposite to the gravitational force planet A exerts on planet B
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an automobile engine slows down from 4,087 rpm to 1,830 rpm in 1,419 revolutions. calculate the magnitude of its angular acceleration in rad/s2. (assume a uniform deceleration.)
An automobile engine slows down from 4,087 rpm to 1,830 rpm in 1,419 revolutions the magnitude of the angular acceleration of the automobile engine is approximately 77.75 rad/s².
To calculate the magnitude of the angular acceleration, we can use the formula:
Angular acceleration (α) = (ω2 - ω1) / (t2 - t1)
where:
ω1 and ω2 are the initial and final angular velocities, respectively, and
t1 and t2 are the initial and final times, respectively.
Initial angular velocity ω1 = 4087 rpm
Final angular velocity ω2 = 1830 rpm
Number of revolutions (n) = 1419
First, we need to convert the angular velocities from rpm to radians per second (rad/s):
ω1 = (4087 rpm) * (2π rad/1 min) * (1 min/60 s) ≈ 426.97 rad/s
ω2 = (1830 rpm) * (2π rad/1 min) * (1 min/60 s) ≈ 191.46 rad/s
Next, we can calculate the time interval (t2 - t1) using the number of revolutions and the initial and final angular velocities:
t2 - t1 = (n / ω2) - (n / ω1)
t2 - t1 = (1419 / 191.46) - (1419 / 426.97) ≈ 3.3 s
Finally, we can calculate the magnitude of the angular acceleration:
α = (ω2 - ω1) / (t2 - t1)
α = (191.46 rad/s - 426.97 rad/s) / (3.3 s)
α ≈ -77.75 rad/s²
Therefore, the magnitude of the angular acceleration of the automobile engine is approximately 77.75 rad/s².
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A 13 500 N car traveling at 50.0 km/h rounds a curve of radius 2.00 × 102 m. Find the following: a. the centripetal acceleration of the car b. the centripetal force c. the minimum coefficient of static friction between the tires and the road that will allow the car to round the curve safely
Answer:
a. 0.947 m/s^2
b. 1304.54 N
c. 0.0966
Explanation:
mass of car = 13500 N = 13500/9.8 = 1377.55 kg
velocity = 50 km/h = 50,000 m/h = 13.9 m/s
raidus = 204 m
a. centripetal acceleartion = v^2/r = 13.9^2/204 = 0.947 m/s^2
b. centripetal force = m*centripetal acceleration = 1377.55 * 0.947 = 1304.54 N
c. In order for the car to round the curve safely, static friction = centripetal force
static friction = coefficient of friction (mu) * mg = mu* 1377.55*9.8 = 13500mu
13500mu = 1304.54
mu = 1304.54/13500 = 0.0966
The acceleration, force and coefficient of friction is required.
Centripetal acceleration is [tex]0.965\ \text{m/s}^2[/tex]
Centripetal force is [tex]1328\ \text{N}[/tex]
Coefficient of friction is [tex]0.1[/tex]
N = Weight of car = 13500 N
v = Velocity = [tex]50=\dfrac{50}{3.6}=13.89\ \text{m/s}[/tex]
r = Radius = [tex]2\times 10^2\ \text{m}[/tex]
m = Mass of car = [tex]\dfrac{N}{g}[/tex]
g = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]
Centripetal acceleration is
[tex]a_c=\dfrac{v^2}{r}\\\Rightarrow a_c=\dfrac{13.89^2}{2\times 10^2}\\\Rightarrow a_c=0.965\ \text{m/s}^2[/tex]
Force is given by
[tex]F_c=ma_c\\\Rightarrow F_c=\dfrac{N}{g}a_c\\\Rightarrow F_c=\dfrac{13500}{9.81}\times 0.965\\\Rightarrow F_c=1328\ \text{N}[/tex]
Coefficient of friction is given by
[tex]\mu=\dfrac{F_c}{N}\\\Rightarrow \mu=\dfrac{1328}{13500}\\\Rightarrow \mu=0.098\approx 0.1[/tex]
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some diversification benefits can be achieved by combining securities in a portfolio as long as the correlation between the securities is _____________.
A. 1 B. Less than 1
C. Between 0 and 1 D. Less than or equal to 0
Some diversification benefits can be achieved by combining securities in a portfolio as long as the correlation between the securities isThe correct answer is C. Between 0 and 1.
Diversification benefits can be achieved by combining securities in a portfolio as long as the correlation between the securities is between 0 and 1. Correlation measures the degree to which two securities move in relation to each other. A correlation of 1 indicates a perfect positive correlation, where the securities move in perfect tandem. A correlation of less than 1 indicates a less than perfect positive correlation, where the securities move somewhat together but not completely. When the correlation is between 0 and 1, it means that the securities have some degree of independence and tend to move differently from each other. This allows for diversification benefits as losses in one security may be offset by gains in another, reducing overall portfolio risk.
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a 10 kg mass slides down a flat hill that makes an angle of 10 degrees with the horizontal. If friction is negligible, what is the resultant force on the sled?
a) 98N
b) 1.7N
c) 97N
d) 17N
A 10 kg mass slides down a flat hill that makes an angle of 10 degrees with the horizontal. Therefore, the resultant force on the sled is option (c) 97N.
If friction is negligible, the resultant force on the sled will be calculated below:
We know that gravitational force can be broken into two components - force parallel to the slope and force perpendicular to the slope.
The parallel component is given by
Fg * sin θ = 10*9.8*sin10 = 16.87 N.
The perpendicular component is given by
Fg * cos θ = 10*9.8*cos10 = 96.94 N.
The total force acting on the sled is the vector sum of the two components: Resultant force = √(16.87² + 96.94²) = 97 N.
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What is specific heat capacity?
A. The amount of energy required for a unit mass of a substance to
undergo a phase change from a liquid to a gas.
B. The amount of energy required to raise the temperature of 1 kg of
a substance 1 °C.
C. The amount of energy required for a unit mass of a substance to
undergo a phase change from solid to liquid.
O D. A physical change from one form (or phase) of matter to another.
A wagon wheel is a ring (hoop) of mass 77.1 kg and radius 0.630 m. The wagon is moving at 1.22 m/s. What is the TOTAL kinetic energy of the wheel? [?] Remember v = wr Enter
Kinetic energy is the energy that an object possesses due to its motion. Hence, the correct option is [109.5 J].
The formula to calculate kinetic energy is KE = 1/2 mv², where m is the mass of the object and v is its velocity or speed. In this question, the wagon wheel is a ring of mass 77.1 kg and radius 0.630 m. The wagon is moving at 1.22 m/s. We have to calculate the total kinetic energy of the wheel.The velocity of the wagon can be converted into the angular velocity of the wheel by using the formula v = wr. The angular velocity w is calculated as:w = v/rw = 1.22 m/s ÷ 0.630 m ≈ 1.936 rad/sNow that we know the angular velocity of the wheel, we can calculate its total kinetic energy using the formula KE = 1/2 Iw², where I is the moment of inertia of the wheel. For a ring-shaped object, the moment of inertia is given by I = mr².KE = 1/2 Iw²KE = 1/2 (mr²) (w²)KE = 1/2 (77.1 kg) (0.630 m)² (1.936 rad/s)²KE ≈ 109.5 J. Therefore, the total kinetic energy of the wagon wheel is approximately 109.5 J. Note: As per the question, the total kinetic energy of the wheel is required, not of the wagon.
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what is the difference between malleability and ductile ??
Please don't post invalid answer please..!!
Please post correct answer please..!!
Answer:
A malleable material is one in which a thin sheet can be easily formed by hammering or rolling. ... In contrast, ductility is the ability of a solid material to deform under tensile stress.
A 95 kg falling trunk strikes the ground with a speed of 40 m/s. Assuming that there is no loss of energy due to air resistance, what is the height from which the trunk falls?
Answer:
80 m or 81.3 m
Explanation:
Initially it had no velocity(as at rest). Let the height be h
Using conservation of energy:
Initial KE + PE = final KE + PE
(1/2) m(0)² + mgh = (1/2) m(40)² + mg(0)
mgh = (1/2) m (1600)
2gh = 16000
h = 1600/(2*10) {for g = 10}
h = 80
80 For g = 9.8, h = 81.3 m
Using Newton's eqⁿ of motion:
v² = u² + 2aS
40² = 0² + 2(g)S
1600/2g = S
For g = 10, S = 80 m
g = 9.8, S = 81.33 m
What is the maximum flow rate of water in a smooth pipe 8.0 cm diameter if the flow is to be laminar
Answer:
0.05 m/s
Explanation:
We start by finding the average velocity of water in the pipe. This is done by saying
R(e) = ρv(avg)d/μ
Where,
R(e) = Reynolds number, and that's 2000
ρ = Density of water, 1000 kg/m³
μ = Viscosity of water, 10^-3
d = diameter of pipe
v(avg) = average velocity
Since we're interested in average velocity, we make v(avg) the subject of formula. So that
V(avg) = R(e).μ/ρ.d
V(avg) = 2000 * 10^-3 / 1000 * 0.08
V(avg) = 2 / 80
V(avg) = 0.025 m/s
The maximum flow rate of water in the pipe usually is twice the average velocity, and as such
V(max) = 2 * V(avg)
V(max) = 2 * 0.025
V(max) = 0.05 m/s
1) longer storms should print in the correct order, with 200ms between each event in the array and 400ms between each flash! and boom!
For accurate printing of longer storms, it is important to ensure that the events in the array are arranged in the correct order, with a 200ms delay between each event and a 400ms delay between each flash and boom.
When printing longer storms, it is crucial to maintain the correct order of events in the array. By arranging the events in the correct sequence, the storm will be printed accurately, providing a realistic representation. To achieve this, a delay of 200ms should be implemented between each event in the array. This delay ensures that each event is printed with the appropriate timing, creating a smooth and coherent storm simulation.
Additionally, it is necessary to introduce a 400ms delay between each "flash" and "boom" in the storm. This delay creates a distinct gap between these two elements, mimicking the natural occurrence of a lightning flash followed by the accompanying thunder. By allowing sufficient time between the flash and boom, the printed storm will convey a more realistic and immersive experience.
In summary, to accurately print longer storms, it is essential to maintain the correct order of events in the array and introduce appropriate delays. A 200ms delay between each event ensures accurate timing, while a 400ms delay between each flash and boom replicates the natural occurrence of lightning and thunder. Following these guidelines will result in a more realistic representation of storms when printing them.
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Eee A student conducts an investigation to determine how the force of gravity affects different objects dropped from different heights. The student tests each object one time and announces that all objects experienced gravity the same way. What is wrong with the student's reasoning?
Answer:
For which the reasoning of the boy is correct for small heights, but as height increases his analysis is not correct.
Explanation:
The force of gravity comes from Newton's second law with the force the universal attraction
F = ma
F = [tex]G \frac{m_1 M}{(R_e +h)^2}[/tex]
we substitute
[tex]G \frac{m_1 M}{ (R_e+ h)^2}[/tex] = m₁ a
where Re is the radius of the Earth 6.37 106 m
a = [tex]G\frac{M}{R_e^2} \ ( 1 + \frac{h}{R_e})^{-2}[/tex]
In general, the height is much less than the radius of the earth, therefore the term ha / Re is very small and we can use a series expansion leaving only the first fears.
(1 + x)⁻² = 1 -2x + [tex]\frac{2 \ 1}{2!}[/tex] x²
we substitute
a = g₀ ([tex]1 - 2 \frac{h}{R_e}[/tex] )
with
g₀ = [tex]G \frac{M}{R_e^2}[/tex]
let's launch the expression.
* For small height compared to the radius of the earth we can neglect the last term
g = g₀
* For height comparable to the radius of the Earth
g = g₀ [tex](1 - \frac{2h}{Re} )[/tex]
We see that the acceleration of gravity is decreasing.
For which the reasoning of the boy is correct for small heights, but as height increases his analysis is not correct.
The student's reasoning gone wrong when the analysis is undertaken for the increasing heights, to drop the object.
The given problem is based on the concept of gravity and gravitational force. The force of gravity comes from Newton's second law with the force the universal attraction as,
F = ma
[tex]F=G\dfrac{mM}{(R+h)^{2}}\\\\\\ma = G\dfrac{mM}{(R+h)^{2}}[/tex]
Here, a is the linear acceleration, m is the mass of object, M is the mass of Earth, R is the radius of Earth and h is the height from where the objects will be dropped. Then,
[tex]a = \dfrac{GM}{R^{2}} \times(1+h/R)^{-2}[/tex]
In general, the height is much less than the radius of the earth, therefore the term h/ R is very small, hence can be neglected.
[tex]a = \dfrac{GM}{R^{2}}\\\\a=g = \dfrac{GM}{R^{2}}[/tex]
g is the gravitational acceleration.
For small height compared to the radius of the earth we can neglect the last term as,
a = g
And for the height comparable to radius of Earth,
[tex]a = \dfrac{GM}{R^{2}} \times(1+h/R)^{-2}\\\\a=g \times(1+h/R)^{-2}[/tex]
Clearly, the acceleration of gravity is decreasing, for which the reasoning of the boy is correct for small heights, but as height increases his analysis is not correct.
Thus, we can conclude that the student's reasoning gone wrong when the analysis is undertaken for the increasing heights, to drop the object.
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What is the magnitude of the electrostatic force between two electrons each having a charge of 1.6 x 10-19 C separated by a distance of 1.00 × l0– 8 meter?
A)
2.56 × 10–22 N
B)
2.30 × 10–20 N
C)
2.30 × 10–12 N
D)
1.44 × 10–1 N
Answer:
The correct option is (C).
Explanation:
The distance between two charges, [tex]d=10^{-8}\ m[/tex]
We need to find the magnitude of the electrostatic force between two electrons. The formula for the electric force between charges is given by :
[tex]F=\dfrac{kq^2}{r^2}\\\\=\dfrac{9\times 10^9\times (1.6\times 10^{-19})^2}{(10^{-8})^2}\\\\=2.304\times 10^{-12}\ N[/tex]
So, the magnitude of force is equal to [tex]2.30\times 10^{-12}\ N[/tex].
How are light waves used to bring far away objects into view and how does the eye translate them?
HELPPP
Answer:
When focused light is projected onto the retina, it stimulates the rods and cones. The retina then sends nerve signals are sent through the back of the eye to the optic nerve. The optic nerve carries these signals to the brain, which interprets them as visual images.
Explanation:
Hope it will help u
Remember that the Sun produces 400 trillion trillion joules per second via the proton-proton chain, and that the Sun has a mass of 2 million trillion trillion kilograms. How many years would the Sun have to shine before one percent of its mass is converted to energy? Your answer should be significant to only one or two digits.
The Sun would have to shine for approximately 2 billion years for one percent of its mass to be converted into energy.
To calculate the time required for the Sun to convert one percent of its mass into energy, we need to determine the total energy the Sun can produce and then calculate the amount of mass that would be converted.
Energy produced per second by the Sun via the proton-proton chain = 400 trillion trillion joules (4 x 10²⁶joules)
Mass of the Sun = 2 million trillion trillion kilograms (2 x 10³⁰ kilograms)
First, we need to calculate the total energy the Sun can produce per year:
Energy produced per year = Energy produced per second × Number of seconds in a year
Number of seconds in a year = 365 days × 24 hours × 60 minutes × 60 seconds
Number of seconds in a year = 31,536,000 seconds
Energy produced per year = 4 x 10²⁶joules/second × 31,536,000 seconds/year
Now, we can calculate the mass that would be converted to energy:
Mass converted to energy per year = Energy produced per year / (Speed of light)²
The speed of light (c) is approximately 3 x 10⁸meters per second.
Mass converted to energy per year = (4 x 10²⁶ joules/second × 31,536,000 seconds/year) / (3 x 10^8 meters/second)²
Now, let's calculate the mass converted to energy as a percentage of the Sun's total mass:
Mass percentage = (Mass converted to energy per year / Mass of the Sun) × 100
Mass percentage = ((4 x 10²⁶ joules/second × 31,536,000 seconds/year) / (3 x 10⁸ meters/second)²) / (2 x 10³⁰ kilograms) × 100
Finally, we can calculate the number of years required for one percent of the Sun's mass to be converted into energy:
Years = 1% / Mass percentage
Years = 1% / ((4 x 10²⁶ joules/second × 31,536,000 seconds/year) / (3 x 10⁸ meters/second)²) / (2 x 10³⁰kilograms) × 100
After performing the calculations, the result is approximately 2 billion years.
The Sun would have to shine for approximately 2 billion years for one percent of its mass to be converted into energy.
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