Answer:
The work done by the force is 820.745 joules.
Explanation:
Let suppose that changes in potential energy can be neglected. According to the Work-Energy Theorem, an external conservative force generates a change in the state of motion of the object, that is a change in kinetic energy. This phenomenon is describe by the following mathematical model:
[tex]K_{1} + W_{F} = K_{2}[/tex]
Where:
[tex]W_{F}[/tex] - Work done by the external force, measured in joules.
[tex]K_{1}[/tex], [tex]K_{2}[/tex] - Translational potential energy, measured in joules.
The work done by the external force is now cleared within:
[tex]W_{F} = K_{2} - K_{1}[/tex]
After using the definition of translational kinetic energy, the previous expression is now expanded as a function of mass and initial and final speeds of the object:
[tex]W_{F} = \frac{1}{2}\cdot m \cdot (v_{2}^{2}-v_{1}^{2})[/tex]
Where:
[tex]m[/tex] - Mass of the object, measured in kilograms.
[tex]v_{1}[/tex], [tex]v_{2}[/tex] - Initial and final speeds of the object, measured in meters per second.
Now, each speed is the magnitude of respective velocity vector:
Initial velocity
[tex]v_{1} = \sqrt{v_{1,x}^{2}+v_{1,y}^{2}}[/tex]
[tex]v_{1} = \sqrt{\left(12\,\frac{m}{s} \right)^{2}+\left(22\,\frac{m}{s} \right)^{2}}[/tex]
[tex]v_{1} \approx 25.060\,\frac{m}{s}[/tex]
Final velocity
[tex]v_{2} = \sqrt{v_{2,x}^{2}+v_{2,y}^{2}}[/tex]
[tex]v_{2} = \sqrt{\left(16\,\frac{m}{s} \right)^{2}+\left(29\,\frac{m}{s} \right)^{2}}[/tex]
[tex]v_{2} \approx 33.121\,\frac{m}{s}[/tex]
Finally, if [tex]m = 3.5\,kg[/tex], [tex]v_{1} \approx 25.060\,\frac{m}{s}[/tex] and [tex]v_{2} \approx 33.121\,\frac{m}{s}[/tex], then the work done by the force is:
[tex]W_{F} = \frac{1}{2}\cdot (3.5\,kg)\cdot \left[\left(33.121\,\frac{m}{s} \right)^{2}-\left(25.060\,\frac{m}{s} \right)^{2}\right][/tex]
[tex]W_{F} = 820.745\,J[/tex]
The work done by the force is 820.745 joules.
A skier goes down a slope and detaches from the ground moving in the horizontal direction with a speed of 25m / s. The slope has an inclination of 35 °
a) At what point does the skier make contact again with the ground?
Answer:
107 m down the incline
Explanation:
Given:
v₀ₓ = 25 m/s
v₀ᵧ = 0 m/s
aₓ = 0 m/s²
aᵧ = -10 m/s²
-Δy/Δx = tan 35°
Find: d
First, find Δy and Δx in terms of t.
Δy = v₀ᵧ t + ½ aᵧ t²
Δy = (0 m/s) t + ½ (-10 m/s²) t²
Δy = -5t²
Δx = v₀ₓ t + ½ aₓ t²
Δx = (25 m/s) t + ½ (0 m/s²) t²
Δx = 25t
Substitute:
-(-5t²) / (25t) = tan 35°
t/5 = tan 35°
t = 5 tan 35°
t ≈ 3.50 s
Now find Δy and Δx.
Δy ≈ -61.3 m
Δx ≈ 87.5 m
Therefore, the distance down the incline is:
d = √(x² + y²)
d ≈ 107 m
Please Help!!! I WILL GIVE BRAINLIEST!!!! An electron is in motion at 4.0 × 10^6 m/s horizontally when it enters a region of space between two parallel plates, starting at the negative plate. The electron deflects downwards and strikes the bottom plate. The magnitude of the electric field between the plates is 4.0 x 10^2 N/C and separation between the charged plates is 2.0 cm. a.) Determine the horizontal distance traveled by the electron when it hits the plate. b.)Determine the velocity of the electron as it strikes the plate.
Answer:
Explanation:
Given that
speed u=4*10^6 m/s
electric field E=4*10^3 N/c
distance b/w the plates d=2 cm
basing on the concept of the electrostatices
now we find the acceleration b/w the plates
acceleration a=qE/m=1.6*10^-19*4*10^3/9.1*10^-31=0.7*10^15 =7*10^14 m/s
now we find the horizantal distance travelled by electrons hit the plates
horizantal distance X=u[2y/a]^1/2
=4*10^6[2*2*10^-2/7*10^14]^1/2
=3*10^-2=3 cm
now we find the velocity f the electron strike the plate
v^2-(4*10^6)^2=2*7*10^14*2*10^-2
v^2=16*10^12+28*10^12
v^2=44*10^12
speed after hits =>V=6.6*10^6 m/s
_____________ friction is the interlocking of surfaces due to irregularities on the surfaces preventing those surfaces from moving/sliding against each other. For surfaces moving/sliding on each other, ___________ friction overwhelms kinetic friction to that movement/sliding. Kinetic friction is alway larger than ____________ friction. Kinetic friction is alway equal to _________ friction.
Answer:
STATIC, STATIC
KINETIC friction is less than static friction
Explanation:
In this exercise you are asked to complete the sentences with the correct words.
STATIC friction prevents the relative movement of two surfaces in contact.
For moving surfaces the friction is STATIC is greater than the kinetic friction.
For the last two sentences I think they are misspelled, the correct thing is
KINETIC friction is less than static friction
A clarinet behaves like a tube closed at one end. If its length is 1.0 m, and the velocity of sound is 344 m/s, what is its fundamental frequency (in Hz)
Answer:
Fundamental frequency = 1376Hz
Fundamental frequency of it is 86 Hz.
What is frequency?In physics, frequency is the number of waves that pass a fixed point in a unit of time as well as the number of cycles or vibrations that a body in periodic motion experiences in a unit of time.
After moving through a sequence of situations or locations and then returning to its initial position, a body in periodic motion is said to have experienced one cycle or one vibration.
Given parameters:
Length of the one end closed tube; L = 1 m.
Velocity of sound; v = 344 m/s.
We have to find, fundamental frequency of it: f₀ = ?
We know that frequency of one end closed tube is given by: f = (2n−1)v/4L where; n = 1, 2, 3, 4, .....
So, for n = 1; fundamental frequency is given by: f₀ = v/4L = 344/(4×1) Hz. = 86 Hz.
Learn more about frequency here:
https://brainly.com/question/5102661
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At a particular instant, a moving body has a kinetic energy of 295 J and a momentum of magnitude 25.1 kg · m/s.(a)What is the speed (in m/s) of the body at this instant?m/s(b)What is the mass (in kg) of the body at this instant?kg
Answer:
a) 23.51 m/s
b) 1.07 kg
Explanation:
Parameters given:
Kinetic energy, K = 295 J
Momentum, p = 25.1 kgm/s
a) The kinetic energy of a body is given as:
[tex]K = \frac{1}{2} mv^2[/tex]
where m = mass of the body and v = speed of the body
We know that momentum is given as:
p = mv
Therefore:
K = 1/2 * pv
=> v = 2K / p
v = (2 * 295) / 25.1 = 23.51 m/s
The velocity of the body at that instant is 23.51 m/s.
b) Momentum is given as:
p = mv
=> m = p / v
m = 25.1 / 23.51 = 1.07 kg
The mass of the body at that instant is 1.07 kg
A parallel-plate capacitor C is charged up to a potential V0 with a charge of magnitude Q0 on each plate. It is then disconnected from the battery, and the plates are pulled apart to twice their original separation.
A. What is the new capacitance in terms of C?
B. How much charge is now on the plates in terms of Q_0 ?
C. What is the potential difference across the plates in terms of V_0
Answer:
A. The new capacitance is half in term of C
B. The charge remains the same in terms of [tex]Q_{0}[/tex]
C. The potential difference is double in terms of [tex]V_{0}[/tex]
Explanation:
The battery with a voltage of [tex]V_{0}[/tex] is used to charge the plates, giving it a capacitance of C.
The charging process leaves a charge of magnitude [tex]Q_{0}[/tex] on the plate
The battery is disconnected (this will leave it with a constant charge [tex]Q_{0}[/tex])
the relationship between the charge, voltage and capacitance of the plate is
[tex]Q_{0}[/tex] = C[tex]V_{0}[/tex] ......... equ 1
A. The relationship between capacitance and the distance of the plate is given as
C = Aε/d ......... equ 2
where A is the area of the plate,
ε is the permeability of free space,
d is the distance between the plates
The area of the plate does not change, and permeability of free space is a constant. The combination of all these means that if the distance is doubled, then the capacitance will be halved. This is from equ 2 when the distance becomes 2d, then we have
C' = Aε/2d
==> C' = C/2
B. Since the battery is disconnected, and the capacitor is not discharged, the charge on the plate will remains the same as [tex]Q_{0}[/tex]. This is due to the conservation of charges.
C. Since the charge remains constant, and the capacitance is halved, then from equ 1, the new potential difference V will become double of the initial potential difference [tex]V_{0}[/tex]
==> V = 2[tex]V_{0}[/tex]
A 1.5-kg mass attached to spring with a force constant of 20.0 N/m oscillates on a horizontal, frictionless track. At t = 0, the mass is released from rest at x = 10.0 cm. ( That is, the spring is stretched by 10.0 cm.) (a) Determine the frequency of the oscillations. (b) Determine the maximum speed of the mass. Where dos the maximum speed occur? (c) Determine the maximum acceleration of the mass. Where does the maximum acceleration occur? (d) Determine the total energy of teh oscillating system. (e) Express the displacement as a function of time.
Answer:
(a) f = 0.58Hz
(b) vmax = 0.364m/s
(c) amax = 1.32m/s^2
(d) E = 0.1J
(e) [tex]x(t)=0.1m\ cos(2\pi(0.58s^{-1})t)[/tex]
Explanation:
(a) The frequency of the oscillation, in a spring-mass system, is calulated by using the following formula:
[tex]f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}[/tex] (1)
k: spring constant = 20.0N/m
m: mass = 1.5kg
you replace the values of m and k for getting f:
[tex]f=\frac{1}{2\pi}\sqrt{\frac{20.0N/m}{1.5kg}}=0.58s^{-1}=0.58Hz[/tex]
The frequency of the oscillation is 0.58Hz
(b) The maximum speed is given by:
[tex]v_{max}=\omega A=2\pi f A[/tex] (2)
A: amplitude of the oscillations = 10.0cm = 0.10m
[tex]v_{max}=2\pi (0.58s^{-1})(0.10m)=0.364\frac{m}{s}[/tex]
The maximum speed of the mass is 0.364 m/s.
The maximum speed occurs when the mass passes trough the equilibrium point of the oscillation.
(c) The maximum acceleration is given by:
[tex]a_{max}=\omega^2A=(2\pi f)^2 A[/tex]
[tex]a_{max}=(2\pi (0.58s^{-1}))(0.10m)=1.32\frac{m}{s^2}[/tex]
The maximum acceleration is 1.32 m/s^2
The maximum acceleration occurs where the elastic force is a maximum, that is, where the mass is at the maximum distance from the equilibrium point, that is, the acceleration.
(d) The total energy of the system is:
[tex]E=\frac{1}{2}kA^2=\frac{1}{2}(20.0N/m)(0.10m)^2=0.1J[/tex]
The total energy is 0.1J
(e) The displacement as a function of time is:
[tex]x(t)=Acos(\omega t)=Acos(2\pi ft)\\\\x(t)=0.1m\ cos(2\pi(0.58s^{-1})t)[/tex]
n electric motor rotating a workshop grinding wheel at 1.10 102 rev/min is switched off. Assume the wheel has a constant negative angular acceleration of magnitude 1.92 rad/s2. (a) How long does it take the grinding wheel to stop?
Answer:
Time taken to stop = 6 sec
Explanation:
Given:
Rotation of wheel = 1.10 × 10² rev/min
Final velocity (v) = 0
Angular acceleration (a) = - 1.92 rad/s²
Find:
Time taken to stop
Computation:
Initial velocity (u) = (1.10 × 10² × 2π rad) / 60 sec
Initial velocity (u) = 11.52 rad / sec
We know that,
V = U +at
t = (v-u)a
t = (0 - 11.52) / (-1.92)
Time taken to stop = 6 sec
11. A seesaw sits in static equilibrium. A child with a mass of 30 kg sits 1 m away from a pivot point. Another child sits 0.75 m away from the pivot point on the opposite side. The second child's mass is _____ kg.
Answer:
40 kgExplanation:
Find the diagram relating to the question for proper explanation of the question below.
Using the principle of moment
Sum of clockwise moments = Sum of anticlockwise moments
Moment = Force * perpendicular distance
For anti-clockwise moment:
Since the 30 kg moves in the anticlockwise direction according to the diagram
ACW moment = 30 * 1 = 30 kgm
For clockwise moment
If another child sits 0.75 m away from the pivot point on the opposite side, moment of the child in clockwise direction = M * 0.75 = 0.75M (M is the mass of the unknown child).
Equating both moments we have;
0.75M = 30
M = 30/0.75
M = 40 kg
The second child's mass is 40 kg
Help yet again :) A hockey player is skating on the ice at 15km/h. He shoots the puck at 138 km/h according to a radar gun on the side of the ice. From the hockey player frame of reference how fast did he shoot the puck (in km/h)?
Answer:
speed of puck acc. to the radar gun = 138 km/h
speed of player = 15 km/h
since the player is in motion when he shoots, the speed of the puck will be the sum of the speed of the player and the speed at which he shot. so,
speed of puck = speed of player + speed of puck acc. to player
138 = 15 + speed of puck acc. to player
speed of puck acc. to player = 138 -15
speed of puck acc. to player = 123 km/h
Brainly this answer if you think it deserves it
A machinist is required to manufacture a circular metal disk with area 1300 cm2. (a) What radius produces such a disk
Answer:
Radius r = 20.34 cm
The radius that can produces such a disk is 20.34 cm
Explanation:
Area of a circle;
A = πr^2
A = area
r = radius
Making r the subject of formula;
r = √(A/π) ........1
Given;
A = 1300 cm^2
Substituting into the equation 1;
r = √(1300/π)
r = 20.34214472564 cm
r = 20.34 cm
The radius that can produces such a disk is 20.34 cm
a fly undergoes a displacement of - 5.80 while accelerating at -1.33 m/s^2 for 4.22 s. what was the initial velocity of the fly?
Answer:
[tex]v_i = 1.44\frac{m}{s}[/tex]
Explanation:
The computation of the initial velocity of the fly is shown below:-
But before that we need to do the following calculations
For 4.22 seconds
[tex]\bar v = \frac{-5.80 m}{4.22 s}[/tex]
[tex]= -1.37\frac{m}{s}[/tex]
For uniform acceleration
[tex]\bar v = \frac{v_i +v_f}{2}[/tex]
[tex]= v_i + v_f[/tex]
[tex]= -2.74\frac{m}{s}[/tex]
With initial and final velocities
[tex]= -1.33\frac{m}{s^2}[/tex]
[tex]= \frac{v_i +v_f}{4.22s}[/tex]
[tex]= -v_i + v_f[/tex]
[tex]= -5.61\frac{m}{s}[/tex]
So, the initial velocity is
[tex]v_i = 1.44\frac{m}{s}[/tex]
We simply applied the above steps to reach at the final solution i.e initial velocity
There are two nearby point sources, each emitting a light wave of frequency f. When the frequency f is increased, how will the distance between troughs of constructive interference change?
Answer:
An increase in frequency will cause an increase in the number of lines per centimeter and a smaller distance between each consecutive line. That is the distance between each trough
Explanation:
This is because higher frequency light source should produce an interference pattern with more lines per centimeter in the pattern and a smaller spacing between lines.
A force of 720 Newton stretches a spring 4 meters. A mass of 45 Kilograms is attached to the spring and is initially released from the equilibrium position with an upward velocity of 6 meters per second. Find an equation of the motion.
Answer:
x(t) = -3sin2t
Explanation:
Given that
Spring force of, W = 720 N
Extension of the spring, s = 4 m
Attached mass to the spring, m = 45 kg
Velocity of, v = 6 m/s
The proper calculation is attached via the image below.
Final solution is x(t) = -3.sin2t
In a series RC circuit, the resistor voltage is 124 V and the capacitor voltage is 167 V. What is the total voltage
Answer:
208 V
Explanation:
resistor voltage = Vr = 124 V
capacitor voltage Vc = 167 V
the total voltage in the RC circuit is the resultant voltage of the resistor and the capacitor
total voltage i= [tex]\sqrt{Vr^{2} + Vc^{2} }[/tex]
==> [tex]\sqrt{124^{2} + 167^{2} } =[/tex] 208 V
What will be the volume and density of stone if mass of stone is 10 gram .please tell the answer fast it's very urgent I will mark as a brain me answer if you will answer it correct.
Answer:
[tex]\large \boxed{\text{3.3 cm}^{3}}[/tex]
Explanation:
Assume the stone consists of basalt, which has a density of 3.0 g/cm³.
[tex]\rho = \text{10 g}\times\dfrac{\text{1 cm}^{3}}{\text{3.0 g}} = \text{3.3 cm}^{3}\\\\\text{The volume of the stone is $\large \boxed{\textbf{3.3 cm}^{3}}$}[/tex]
a pendulum hanging from the ceiling of a train traveling with constant speed deviates 37 ° from the vertical when the train describes a curve of 60 m radius determines the train's rapids
Answer:
21 m/s
Explanation:
There are three forces on the pendulum:
Weight force mg pulling down,
Vertical tension component Tᵧ pulling up,
and horizontal tension component Tₓ pulling towards the center of the curve.
Sum of forces in the y direction:
∑F = ma
Tᵧ − mg = 0
T cos 37° = mg
T = mg / cos 37°
Sum of forces in the centripetal direction:
∑F = ma
Tₓ = mv²/r
T sin 37° = mv²/r
(mg / cos 37°) sin 37° = mv²/r
g tan 37° = v²/r
v = √(gr tan 37°)
v = √(9.8 m/s² × 60 m × tan 37°)
v = 21 m/s
Suppose the same magnitude force is applied at the same point as in the example, and the torque is found to have the same magnitude but in the opposite direction of the torque found there. What are the components of the force?
Answer:
-i - 7j
Explanation:
The computation of components of the force is shown below:-
torque T = r cross F
T = (4.00 i + 5.00 j + 0 k) X (1.00 i + 7.00 j)
Now we will cross multiplying vectorilly
T = 4 × (iXi) + 28 × (jXi) + 0 + 5 × (iXj)35 × (jXj) + 0
T = 4 × 0 + 28 × (-k) + 5 × (k) + 35 × 0
T = 28 × (-k) + 5 × (k) = -23k
net torque |T| = 23 N - m
direction >> negative k
Or Simply we can do by the below method
r × f
28 - 5 = 23 K
-i - 7j
Describe the relationship between the density of electric field lines and the strength of the electric field?
Answer:
The greater the density of the electric field lines the stronger the electric field and vice versa
Explanation:
Electric field can be defined as the region where an electric force is experienced by a charged body. A charged body experiences a force whenever it is positioned close to another charged body.
An electric field may be described in terms of lines of force which represent the direction of a small positive charge placed at that point assuming that the charge is so small that it does not change appreciably in the presence of another charge. Arrows on the lines of force indicate the direction of the electric field.
The lines of force are indicated in such a way that the strength of the electric field is shown by the number or density of electric field lines crossing a unit area perpendicular to the lines. Hence, the greater the density of the electric field lines the stronger the the electric field and vice versa
A solid conducting sphere is placed in an external uniform electric field. With regard to the electric field on the sphere's interior, which statement is correct
Complete question:
A solid conducting sphere is placed in an external uniform electric field. With regard to the electric field on the sphere's interior, which statement is correct?
A. the interior field points in a direction parallel to the exterior field
B. There is no electric field on the interior of the conducting sphere.
C. The interior field points in a direction perpendicular to the exterior field.
D. the interior field points in a direction opposite to the exterior field.
Answer:
B. There is no electric field on the interior of the conducting sphere.
Explanation:
Conductors are said to have free charges that move around easily. When the conductor is now placed in a static electric field, the free charges react to attain electrostatic equilibrium (steady state).
Here, a solid conducting sphere is placed in an external uniform electric field. Until the lines of the electric field are perpendicular to the surface, the free charges will move around the spherical conductor, causing polarization. There would be no electric field in the interior of the spherical conductor because there would be movement of free charges in the spherical conductor in response to any field until its neutralization.
Option B is correct.
There is no electric field on the interior of the conducting sphere.
2. A pair of narrow, parallel slits sep by 0.25 mm is illuminated by 546 nm green light. The interference pattern is observed on a screen situated at 1.3 m away from the slits. Calculate the distance from the central maximum to the
Answer:
for the first interference m = 1 y = 2,839 10-3 m
for the second interference m = 2 y = 5,678 10-3 m
Explanation:
The double slit interference phenomenon, for constructive interference is described by the expression
d sin θ = m λ
where d is the separation between the slits, λ the wavelength and m an integer that corresponds to the interference we see.
In these experiments in general the observation screen is L >> d, let's use trigonometry to find the angles
tan θ = y / L
with the angle it is small,
tan θ = sin θ / cos θ = sin θ
we substitute
sin θ = y / L
d y / L = m λ
the distance between the central maximum and an interference line is
y = m λ L / d
let's reduce the magnitudes to the SI system
λ = 546 nm = 546 10⁻⁹ m
d = 0.25 mm = 0.25 10⁻³ m
let's substitute the values
y = m 546 10⁻⁹ 1.3 / 0.25 10⁻³
y = m 2,839 10⁻³
the explicit value for a line depends on the value of the integer m, for example
for the first interference m = 1
the distance from the central maximum to the first line is y = 2,839 10-3 m
for the second interference m = 2
the distance from the central maximum to the second line is y = 5,678 10-3 m
A converging lens has a focal length of 14.0cm. For an object to the left of the lens, at distances of 18.0cm and 7.00cm, determine:
a. The image position
b. The magnification
c. Whether the image is real or virtual
d. Whether the image is upright or inverted
Answer:
Explanation:
A converging lens id also known as a convex lens. A convex lens has a positive focal length.
Using the lens formula to determine the image distance from the lens for each object distance.
1/f = 1/u + 1/v
f = focal length
u = object distance
v = image distance
For an object placed at distance of 18.0cm with focal length 14.0cm,
1/v = 1/f-1/u
1/v = 1/14 - 1/18
1/v = 9-7/126
1/v = 2/126
v = 126/2
v = 63cm
Since the image distance is positive for an object 18cm from the lens, the image formed by the object at this distance is a real and inverted image.
The magnification = v/u = 63/18 = 3.5
Similarly for an object placed at distance of 7.0cm with focal length 14.0cm,
v = uf/u-f
v = 7(14)/7-14
v= -14.0 cm
Since the image distance is negative for an object placed 7.0 cm from the lens, the image formed by the object at this distance is a virtual and upright image.
The magnification = v/u = 14/7 = 2
Note that the negative value is not taken into account when calculating magnification. The negative value only tells us the nature of the image formed.
what is the preferred method of using percentage data by using a circle divided into sections
Answer:
A pie chart is a type of graph in which a circle is divided into sectors that each represents a proportion of the whole
Explanation:
pie charts are a useful way to organize data in order to see the size of components relative to the whole.
Consider the momentum of a small ball during the projectile motion. Assume that there is no air friction. Is the momentum of the ball conserved
Answer:
Only the horizontal component of a projectile’s momentum is conserved. Where as The vertical component of the momentum is not conserved, because the net vertical force Fy–net is not zero
Explanation:
At t = 0, a battery is connected to a series arrangement of a resistor and an inductor. If the inductive time constant is 36.0 ms, at what time is the rate at which energy is dissipated in the resistor equal to the rate at which energy is stored in the inductor's magnetic field?
Answer:
The rate at which energy is dissipated in the resistor is equal to the rate at which energy is stored in the inductor's magnetic field in 24.95 ms.
Explanation:
The energy stored in the inductor is given as
E₁ = ½LI²
The rate at which energy is stored in the inductor is
(dE₁/dt) = (d/dt) (½LI²)
Since L is a constant
(dE₁/dt) = ½L × 2I (dI/dt) = LI (dI/dt)
(dE₁/dt) = LI (dI/dt)
Rate of Energy dissipated in a resistor = Power = I²R
(dE₂/dt) = I²R
When the rate at which energy is dissipated in the resistor equal to the rate at which energy is stored in the inductor's magnetic field
(dE₁/dt) = (dE₂/dt)
OK (dI/dt) = I²R
L (dI/dt) = IR
Current in a this kind of series setup of inductor and resistor at any time, t, is given as
I = (V/R) (1 - e⁻ᵏᵗ)
k = (1/time constant) = (R/L)
(dI/dt) = (kV/R) e⁻ᵏᵗ = (RV/RL) e⁻ᵏᵗ = (V/L) e⁻ᵏᵗ
L (dI/dt) = IR
L [(V/L) e⁻ᵏᵗ] = R [(V/R) (1 - e⁻ᵏᵗ)
V e⁻ᵏᵗ = V (1 - e⁻ᵏᵗ)
e⁻ᵏᵗ = 1 - e⁻ᵏᵗ
2 e⁻ᵏᵗ = 1
e⁻ᵏᵗ = (1/2) = 0.5
e⁻ᵏᵗ = 0.5
In e⁻ᵏᵗ = In 0.5 = -0.69315
- kt = -0.69315
kt = 0.69315
k = (1/time constant)
Time constant = 36.0 ms = 0.036 s
k = (1/0.036) = 27.78
27.78t = 0.69315
t = (0.69315/27.78) = 0.02495 = 24.95 ms
Hope this Helps!!!
Muons are elementary particles that are formed high in the atmosphere by the interactions of cosmic rays with atomic nuclei. Muons are radioactive and have average lifetimes of about two-millionths of a second. Even though they travel at almost the speed of light, they have so far to travel through the atmosphere that very few should be detected at sea level - at least according to classical physics. Laboratory measurements, however, show that muons in great number do reach the earth's surface. What is the explanation?
Answer:
Muons reach the earth in great amount due to the relativistic time dilation from an earthly frame of reference.
Explanation:
Muons travel at exceedingly high speed; close to the speed of light. At this speed, relativistic effect starts to take effect. The effect of this is that, when viewed from an earthly reference frame, their short half life of about two-millionth of a second is dilated. The dilated time, due to relativistic effects on time for travelling at speed close to the speed of light, gives the muons an extended relative travel time before their complete decay. So in reality, the muon do not have enough half-life to survive the distance from their point of production high up in the atmosphere to sea level, but relativistic effect due to their near-light speed, dilates their half-life; enough for them to be found in sufficient amount at sea level.
Alternating Current In Europe, the voltage of the alternating current coming through an electrical outlet can be modeled by the function V 230 sin (100t), where tis measured in seconds and Vin volts.What is the frequency of the voltage
Answer:
[tex]\frac{50}{\pi }[/tex]Hz
Explanation:
In alternating current (AC) circuits, voltage (V) oscillates in a sine wave pattern and has a general equation as a function of time (t) as follows;
V(t) = V sin (ωt + Ф) -----------------(i)
Where;
V = amplitude value of the voltage
ω = angular frequency = 2 π f [f = cyclic frequency or simply, frequency]
Ф = phase difference between voltage and current.
Now,
From the question,
V(t) = 230 sin (100t) ---------------(ii)
By comparing equations (i) and (ii) the following holds;
V = 230
ω = 100
Ф = 0
But;
ω = 2 π f = 100
2 π f = 100 [divide both sides by 2]
π f = 50
f = [tex]\frac{50}{\pi }[/tex]Hz
Therefore, the frequency of the voltage is [tex]\frac{50}{\pi }[/tex]Hz
How fast must a 2500-kg elephant move to have the same kinetic energy as a 67.0-kg sprinter running at 15.0 m/s
Answer:
2.45 m/s
Explanation:
kinetic energy = 1/2 * m * v^2
then, 0.5 * 2500 * x^2 = 0.5 * 67 * 15^2
by solving for x, X = 2.45 m/s
A tennis player tosses a tennis ball straight up and then catches it after 1.91 s at the same height as the point of release. (a) What is the acceleration of the ball while it is in flight? magnitude m/s2 direction ---Select--- (b) What is the velocity of the ball when it reaches its maximum height? magnitude m/s direction (c) Find the initial velocity of the ball. m/s upward (d) Find the maximum height it reaches. m
Explanation:
(a) The acceleration is 9.8 m/s² downwards.
(b) At the maximum height, the velocity is 0 m/s.
(c) v = at + v₀
0 m/s = (-9.8 m/s²) (1.91 s) + v₀
v₀ = 18.7 m/s
(d) Δy = vt − ½ at²
Δy = (0 m/s) (1.91 s) − ½ (-9.8 m/s²) (1.91 s)²
Δy = 17.9 m
A trolley going down an inclined plane has an acceleration of 2cm/s^2 What will be its velocity
3s after the start.
Answer:
[tex]V_{f}[/tex] = 6 cm/s
Explanation:
Given:
Acceleration = a = 2 cm/s²
Time = t = 3s
Initial Velocity = [tex]V_{i}[/tex] = 0 cm/s
Required:
Velocity = [tex]V_{f}[/tex] = ?
Formula:
a = [tex]\frac{V_{f}-V_{i}}{t}[/tex]
Solution:
2 = [tex]\frac{V_{f}-0}{3}[/tex]
=> [tex]V_{f}[/tex] = 2*3
=> [tex]V_{f}[/tex] = 6 cm/s