Answer:
The electric and magnetic fields gradually increase.
Explanation:
Time constant in RL circuits, denoted by τ, is equal to the value of L / R which is the value of the inductor over the resistor. It is used to calculate the point when the current will reach the maximum value in the steady state of the circuit. Because of this behavior of the circuit, the magnetic field and the electric field gradually increase to their maximum values.
I hope this answer helps.
PIUDICITIS CONSECulvely and Circle your aliswers. Lilyo
proper significant digits.
53. When you turn on your CD player, the turntable accelerates from zero to 41.8 rad/s in
3.0 s. What is the angular acceleration?
or
Answer:
The angular acceleration of CD player is [tex]13.93\ rad/s^2[/tex].
Explanation:
Initial angular speed of a CD player is 0 and final angular speed is 41.8 rad/s. Time to change the angular speed is 3 s.
It is required to find the angular acceleration. The change in angular speed of the CD player divided by time taken is called its angular acceleration. It can be given by :
[tex]a=\dfrac{\omega_f-\omega_i}{t}\\\\a=\dfrac{41.8-0}{3}\\\\a=13.94\ rad/s^2[/tex]
So, the angular acceleration of CD player is [tex]13.93\ rad/s^2[/tex].
Block A, with a mass of 4 kg, is moving with a speed of 2 m/s while Block B, with a mass of 8.4 kg, is moving in the opposite direction with a speed of 6.1 m/s. The center of mass of the two block system is moving with a velocity of ____ m/s. Round your answer to the nearest tenth. Assume Block A is moving in the positive direction.
Answer:
The center of mass move with the velocity of -3.487 m/s.
Explanation:
Given values of block A.
Mass of block A, (M1) = 4 kg
Speed of block A, (V1) = 2 m/s
Given values of block B.
Mass of block B, (M2) = 8.4 kg
Speed of block B, (V2) = -6.1 m/s
Below is the formula to find the velocity of center of mass.
[tex]Velocity = \frac{M1V1 + M2V2}{M1 + M2} \\[/tex]
[tex]= \frac{4 \times 2 + 8.4 \times (-6.1) }{4 + 8.4} \\[/tex]
[tex]= \frac{- 43.24}{12.4}\\[/tex]
[tex]= - 3.487 m/s[/tex]
A plane electromagnetic wave varies sinusoidally at 90.0 MHz as it travels along the x direction. The peak value of the electric field is 200 V/m, and it is directed along the y direction. Find the wavelength, the period and the maximum value of the magnetic field. Write expressions in SI units for the space and time variations of the electric field and of the magnetic field. Include numerical values, and include subscripts to indicate coordinate directions. Find the average power per unit area that this wave propagates through space.
Answer:
Explanation:
frequency n = 90 x 10⁶ Hz .
time period T = 1 / n
= 1 / 90 x 10⁶
= 1.11 x 10⁻⁸ S.
wavelength = velocity of light / frequency
= 3 x 10⁸ / 90 x 10⁶
= 3.33 m
maximum value of the magnetic field. ( B₀ )
E₀ / B₀ = c where E₀ and B₀ are maximum electric and magnetic field .
E₀ / c= B₀
200/ 3 x 10⁸
= 66.67 x 10⁻⁸ T .
expressions in SI units for the space and time variations of the electric field
[tex]E=E_{0y}sin(2\pi nt - \frac{2\pi x}{\lambda} )[/tex]
[tex]E=200sin(180\times 10^6\pi t - \frac{2\pi x}{\lambda} )[/tex] N/C
[tex]B=B_{0z}sin(2\pi nt - \frac{2\pi x}{\lambda} )[/tex]
[tex]B=66.67\times 10^{-8}sin(180\times 10^6\pi t - \frac{2\pi x}{\lambda} )[/tex] T
How do most black holes form?
A 25 kg rock resting on the bottom of a lake must be moved from the paths of boats. The rock has a density of 2350 kg/m^3. What force is needed to lift the rock while under water?
Answer:
The force needed is the weight of the rock minus the buoyant force.
Explanation:
Which term BEST describes the movement of air from the ocean toward the land in the daytime? (AKS 4b DOK 1) *
1 point
Sea breeze
Land Breeze
Valley Breeze
Current Breeze
Answer:
Option A, Sea Breeze
Explanation:
Ssea breeze is a wind that blows from the ocean or any water body to the nearby land mass. This breeze is cold as compared to the air on land. The water in water bodies has high specific heat capacity and hence takes longer time to cool as compared to the surrounding objects. The warmer air over the land rises upward thereby reducing the pressure on land and hence the sea breeze starts flowing from region of high pressure (i.e above the water body) towards the low pressure region that is the land.
Hence, option A is correct
The greater the distance between two objects in space, the _______ their gravitational
Answer is Weaker. If it is talking about the objects' gravitational forces.
1. A block of metal of mass 2kg is resting on
a frictionless floor. It is struck by a jet
releasing water at the rate of 1kg/sec at a
speed of 5ms-1. What will be the initial
acceleration of the block?
Answer:
The acceleration is [tex]a = 2.5 \ m/s^2[/tex]
Explanation:
From the question we are told that
The mass of the metal block is [tex]m_b = 2 \ kg[/tex]
The mass flow rate of the water is [tex]\r m = 1\ kg/s[/tex]
The speed of the water of the water release is [tex]v_w = 5 m/s[/tex]
Generally according to the law of conservation of linear momentum
[tex]p_i = p_f[/tex]
Now [tex]p_i[/tex] is the initial momentum of the system which mathematically represented as
[tex]p_i = m_w * v_w + m_b * v_b[/tex]
Now [tex]m_w[/tex] is the mass of water at the point of contact with the block which can be deduced as [tex]m_w = 1 \ kg[/tex]
Now since at initial the block is at rest
[tex]v_b = 0 \ m/s[/tex]
So
[tex]p_i = 1 * 5[/tex]
[tex]p_i = 5 \ kgm/ s[/tex]
And [tex]p_f[/tex] is the final momentum of the system which mathematically represented as
[tex]p_f = m_w * v__{fw} } + m_b * v__{fb}}[/tex]
So [tex]v__{fw} }[/tex] is the final velocity of water which is zero due to the fact that when the water hits the block it losses its momentum and eventually the velocity becomes zero
So
[tex]5 = 2 * v__{fb }[/tex]
Thus [tex]v__{fb }} = \frac{5}{2}[/tex]
[tex]v__{fb }} = 2.5 \ m/s[/tex]
Thus
[tex]p_f = 2.5 * 2[/tex]
[tex]p_f = 5 \ kgm /s[/tex]
Now the average momentum change is
[tex]p_a = \frac{p_i +p_f}{2}[/tex]
[tex]p_a = \frac{5+5}{2}[/tex]
[tex]p_a =5 kgm/s[/tex]
Now the force acting on the block is
[tex]F = \frac{p_a }{t}[/tex]
and from the question the initial movement of the block took 1s as it is a mass of water moving at a rate of 1kg/s that caused the first movement of the block
So
[tex]F= \frac{5}{1}[/tex]
[tex]F= 5 \ N[/tex]
Now the acceleration is
[tex]a = \frac{F}{m_b}[/tex]
=> [tex]a = \frac{5}{2}[/tex]
[tex]a = 2.5 \ m/s^2[/tex]
A 46.0-kg box is being pushed a distance of 8.80 m across the floor by a force P whose magnitude is 171 N. The force P is parallel to the displacement of the box. The coefficient of kinetic friction is 0.250. Determine the work done on the box by (a) the applied force, (b) the friction force, (c) the normal force, and (d) by the force of gravity. Be sure to include the proper plus or minus sign for the work done by each force.
Answer:
a) 1504.8 J
b) 991.76 J
c) 0J
d) 0J
Explanation:
(a) The work done by the force P on the box is given by the following formula:
[tex]W_P=Px[/tex]
P: applied force = 171N
x: distance in which the for P is applied = 8.80m
you replace the values of P and x and obtain:
[tex]W_P=(171N)(8.80m)=1504.8J[/tex]
(b) The work don by the friction force is:
[tex]W_f=F_fx=\mu N x=\mu Mg x[/tex]
μ = coefficient of kinetic friction = 0.250
M: mass of the box = 46.0kg
g: gravitational constant = 9.8 m/s^2
[tex]W_f=(0.250)(46.0kg)(9.8m/s^2)(8.80m)=991.76J[/tex]
(c) The Normal force is
[tex]N=Mg=(46.0kg)(9,8m/s^2)=450.8N[/tex]
but this force does not do work on the box because the direction is perpendicular to the direction of the force P.
[tex]W_N=0J[/tex]
(d) the same as before:
[tex]W_g=0J[/tex]
Arm OA rotates counterclockwise with a constant angular velocity of ω = 5 rad/s. As the arm passes the horizontal position, a 6 kg ball is placed at the end of the arm. As the arm moves upward, the ball begins to roll, with negligible rolling resistance, towards the pivot O. It is noted that at θ = 30 ◦ , the ball is 0.9 meters from the pivot and moving towards O along the length of the arm. The ball moves with a speed of 0.4 m/s along the bar. What is the normal force that the arm applies to the ball at this instant? Please, indicate which principle you are applying and explain why.
Answer:
26.92 N
Explanation:
The normal reaction of the ball is due to two force component acting on it.
The normal reaction as a result of the weight of the ballThe normal reaction due to the component of the acceleration of the ball with the rod.However ; the acceleration is in polar coordinate which is given by the relation:
[tex]a^ { ^ \to} = (r- r \omega^2) \hat {e_r} + ( r \theta + 2 r \omega ) \hat {e_ \theta}[/tex]
[tex]a_{\theta} = r \theta + 2 r \omega[/tex]
Given that :
ω = 5 rad/s
mass m = 6 kg
θ = 30 ◦
r = 0.9 m
speed v = 0.4 m/s
[tex]a_{\theta} = 0 + 2(-0.4)*5[/tex]
[tex]a_{\theta}= -4 \ m/s[/tex]
The normal force reaction (N) that the arm applies to the ball at this instant is :
N = mg cos θ + [tex]ma_{\theta}[/tex]
N = (6 × 9.8× cos 30) + (6 ×(-4))
N = 26.92 N
A light bulb dissipates 100 Watts of power when it is supplied a voltage of 220 volts.
a) What is the current flowing through this light bulb?
b) What is the resistance of the light bulb?
Given Information:
Power = P = 100 Watts
Voltage = V = 220 Volts
Required Information:
a) Current = I = ?
b) Resistance = R = ?
Answer:
a) Current = I = 0.4545 A
b) Resistance = R = 484 Ω
Explanation:
According to the Ohm’s law, the power dissipated in the light bulb is given by
[tex]P = VI[/tex]
Where V is the voltage across the light bulb, I is the current flowing through the light bulb and P is the power dissipated in the light bulb.
Re-arranging the above equation for current I yields,
[tex]I = \frac{P}{V} \\\\I = \frac{100}{220} \\\\I = 0.4545 \: A \\\\[/tex]
Therefore, 0.4545 A current is flowing through the light bulb.
According to the Ohm’s law, the voltage across the light bulb is given by
[tex]V = IR[/tex]
Where V is the voltage across the light bulb, I is the current flowing through the light bulb and R is the resistance of the light bulb.
Re-arranging the above equation for resistance R yields,
[tex]R = \frac{V}{I} \\\\R = \frac{220}{0.4545} \\\\R = 484 \: \Omega[/tex]
Therefore, the resistance of the bulb is 484 Ω
Answer:
bulb will burn out!
Explanation:
what do hydroelectric plants use to generate electrical energy?
Answer:
A. falling water
Explanation:
I got it right on Edgenuity. Good luck on your quiz.
In hydroelectric plants, water falls on turbine and makes it rotate. In generator, this mechanical energy transforms into electrical energy.
What is hydroelectric power?Hydroelectric power is generated by turbines that turn the potential energy of falling or swiftly flowing water into mechanical energy, which is then used to power generators. The most popular renewable energy source in the early 21st century was hydroelectricity, which in 2019 accounted for more than 18% of the world's total power producing capacity.
Water is gathered or stored at a higher elevation during the production of hydroelectric power and then transported through substantial pipes or tunnels (penstocks) to a lower elevation; the difference between these two elevations is referred to as the head. The falling water turns turbines as it nears the bottom of the pipelines. In turn, the turbines power generators, which transfer the mechanical energy of the turbines into electricity.
Learn more about hydroelectric power here:
https://brainly.com/question/15228003
#SPJ2
A brick is dropped from a high scaffold.
a. How far does the brick fall during this time?
Answer:
a: after 1 seconds it will have fallen 0.2452
after 2 seconds it will have fallen 0.981
after 3 seconds it will have fallen 2.2072
after 4 seconds it will have fallen 3.924
Explanation:
the formula for acceleration due to gravity is (ignoring friction I think)
g = G*M/R^2
earths gravitational constant is about 9.807
g = 9.807*M/R^2
The average weight of a brick is 5 pounds and I'm going to say it's 10 feet off the ground.
g = 9.807*5/10^2. g = 0.4905 so every second the brick will go 0.4905 fps faster. (fps means feet per second.)
after 1 seconds it will have fallen 0.2452
after 2 seconds it will have fallen 0.981
after 3 seconds it will have fallen 2.2072
after 4 seconds it will have fallen 3.924
If you apply a force of 130 N to the lever, how much force is applied to lift the
crate?
Answer:171 N
Explanation:
Answer:
171 N.
Explanation:
A rectangular painting measures 1.0 m tall along the y' axis and 3.0 m wide along the
x' axis. The painting is hung on the side wall of a spaceship which is moving passed
the Earth at a speed of 0.9c. Assume that the spaceship is moving along the (x, x')
direction.
a) What are the dimensions of the picture according to the captain of the
spaceship?
b) What are the dimensions of the picture as seen by an observer on the Earth?
Answer:
a) 1 m tall, 3 m wide
b) 1 m tall, 1.31 m wide
Explanation:
According to the captain of the spaceship, the dimensions of the picture is the same i.e 1.0 m tall along the y' axis and 3.0 m wide along the x' axis.
b) The dimensions of the picture as seen by an observer on the Earth along the y axis will remain the same, 1.0 m tall, for the direction of the y axis is perpendicular to the spaceship movement.
The dimensions of the picture as seen by an observer on the Earth along the x axis will reduce if we are to go by the Lorentz contraction:
L(x) = L(x)' * √[1 - (v²/c²)]
where
L(x)' = the dimensions of the picture along the x axis on the spaceship,
v² = the speed of the spaceship and c² = the speed of light in the vacuum.
On substituting, we have
L(x) = 3 * √[1 - (0.81c²/c²)]
L(x) = 1.31 m
3. The current in a flashlight powered by 4.5 Volts is 0.5 A. What is the power delivered to the flashlight?
4.If the flashlight in the previous problem is left on for 3 minutes, how much electric energy is delivered to the bulb?
Answer:
Question 3: 2.25 watts
Question 4: 405 joules
Explanation:
question 3:
Current =0.5 amps
Voltage =4.5 volts
Power= current x voltage
Power=0.5 x 4.5
power=2.25 watts
Question 4
Current =0.5 amps
Voltage =4.5v
Time=3 minutes
Time =3x60
Time =180 seconds
Energy=current x voltage x time
Energy =0.5 x 4.5 x 180
Energy =405 joules
Consider a circular vertical loop-the-loop on a roller coaster. A car coasts without power around the loop. Determine the difference between the normal force exerted by the car on a passenger with a mass of mm at the top of the loop and the normal force exerted by the car on her at the bottom of the loop. Express your answer in terms of mmm and the acceleration due to gravity ggg.
Answer:
Explanation:
Let v₁ and v₂ be velocities at lowest and topmost position . Let r be the radius of the circle .
Let N₁ and N₂ be the normal reaction force .
At the top position
centripetal force = N₂ + mg ; so
N₂ + mg = m v₂² / r
At the bottom position
centripetal force = N₁ - mg ; so
N₁ - mg = m v₁² / r
subtracting these two equations
N₁ - mg - N₂ - mg = m v₁² / r - m v₂² / r
N₁ - N₂ - 2mg = 1/r (m v₁² - m v₂² )
N₁ - N₂ - 2mg = 1/r x mg x 2r ( loss of potential energy = gain of kinetic energy )
N₁ - N₂ = 2mg + 2mg
= 4 mg .
Calculate potential energy of a 5 kg object sitting on 3 meter ledge
Answer:147 joules
Explanation:
Mass=m=5kg
Acceleration due to gravity=g=9.8m/s^2
Height=h=3 meter
Potential energy=m x g x h
Potential energy=5 x 9.8 x 3
Potential energy=147 joules
A pendulum is swinging back and forth with no non-conservative forces acting on it. At the highest points of its trajectory, the kinetic energy of the pendulum bob is instantaneously equal to zero joules. At the lowest point of its trajectory, the potential energy is instantaneously equal to zero joules. Which one of the following expressions describes the kinetic and potential energies at the point mid-way between to the highest and lowest points?
A. K = 0, U = Umax
B. K = U
C. K < U
D. K > U
E. U = 0, K = Kmax
Answer:
K = U ( b )
Explanation:
The expression that describes the kinetic and potential energies at the point mid-way between to the highest and lowest points is K = U
this is because at the midpoint between the highest point and the lowest point the height is expressed as( h/2) therefore potential energy at that point is expressed as m*g*h/2 therefore the remaining energy at this point will be considered the kinetic energy which will be = m*g*h/2 as well hence at midpoint Kinetic energy = potential energy
Two atomic particles approach each other in a head-on collision. Each particle has a mass of 2.97 × 10-25 kg. The speed of each particle is 2.19 × 108 m/s when measured by an observer standing in the laboratory. (a) What is the speed of one particle as seen by the other particle? (b) Determine the relativistic momentum of one particle, as it would be observed by the other.
Answer:
a) [tex]\vec{v}_{12}=2.86*10^{8} m/s[/tex]
b) [tex]p=2.81*10^{-16} kg*m/s[/tex]
Explanation:
a) When we have two particles traveling in parallel directions, the formula for relative velocity is:
[tex]\vec{v}_{12}=\frac{\vec{v}_{1}-\vec{v}_{2}}{1-\frac{\vec{v}_{1}\vec{v}_{2}}{c^{2}}}[/tex]
Here we have that v(1) = -v(2), the speed of the of the second particle is the negative of the first one.
If we use these equivalence we have:
[tex]\vec{v}_{12}=\frac{2\vec{v}_{1}}{1+\frac{\vec{v}_{1}^{2}}{c^{2}}}[/tex]
[tex]\vec{v}_{12}=\frac{2*2.19*10^{8}}{1+\frac{2.19*10^{16}}{3*10^{16}}}[/tex]
[tex]\vec{v}_{12}=2.86*10^{8} m/s[/tex]
And, [tex]\vec{v}_{21}=-2.86*10^{8} m/s[/tex]
b) The relativistic momentum equation to one particle observed by the other particle, is:
[tex]p=\gamma mv[/tex]
Where gamma is:
[tex]\gamma = \frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}[/tex]
v is the speed of the first particle relative to the second particle ([tex]2.86*10^{8}[/tex])m is the mass of the particle [tex]2.97*10^{-25} kg[/tex]Then gamma will be:
[tex]\gamma=\frac{1}{\sqrt{1-\frac{(2.86*10^{8})^{2}}{(3*10^{8})^{2}}}}[/tex]
[tex]\gamma=3.31[/tex]
Finally, the value of the momentum will be:
[tex]p=3.31*2.97*10^{-25}*2.86*10^{8}[/tex]
[tex]p=2.81*10^{-16} kg*m/s[/tex]
I hope it helps you!
Part A - At what angle does it leave?
Part B - At what distance x does it exit the field?
Answer:
Total internal reflection (TIR) is the phenomenon that involves the reflection of all the incident light off the boundary. TIR only takes place when both of the following two conditions are met: the light is in the more dense medium and approaching the less dense medium.
Explanation: Hope i helped!!!
If Jim could drive a Jetson's flying car at a constant speed of 440 km/hr across oceans and space, approximately how long (in millions of years, in 106 years) would he take to drive to a nearby star that is 12.0 light-years away? Use 9.461 × 1012 km/light-year and 8766 hours per year (365.25 days).
Answer:
t = 2.94 x 10⁶ years
Explanation:
The equation used in the case of constant speed is:
s = vt
t = s/v
where,
s = distance = 12 light years
s = (12 light years)(9.461 x 10¹² km/light year) = 113.532 x 10¹² km
v = speed = 440 km/hr
t = time passed = ?
Therefore,
t = (113.532 x 10¹² km)/(440 km/hr)
t = 2.58 x 10¹¹ hr
Now, converting it to years:
t = (2.58 x 10¹¹ hr)(1 year/8766 hr)
t = 2.94 x 10⁶ years
The strength of the force of friction depends on which two factors?
Answer:
coefficient of friction (μ) and normal force (N)
Answer: How hard the surfaces push together and the types of surfaces involved
Explanation:
Time Warner Cable's leadership development program that spanned over 30 days and included weekly videos, practice exercises, and a two hour webinar was discussed in the Sed as an example of which of the following?
a) tracking training through a leaming records store LRS)
b) using big data to analyze training compliance
c) using gamification to enhance learning
d) an application of advances in neuroscience to training
Answer: A.
tracking training through a leaming records store LRS.
Explanation:
An LRS uses xAPI to collect learner data, or experiences, from both online and offline sources. These experiences are reported back to the LRS in the form of xAPI statements, where they are stored. These statements can then be retrieved for reporting and interpretation of the learner data.
Vocabulary Matching
The specialized equipment used to conduct research and repair
damaged equipment
Instruments
Space Station
Space Suit
Accomodations
Answer:
instruments
Explanation:
Refracted light rays...
A- are bent as they pass into a different medium
B- are absorbed by an object
C- are reflected from an object at a variety of angles
D- bounce off a medium
Answer:
A.
Explanation:
Refraction light rays are bent as they pass into a different medium where its speed is different. As refracted light rays pass from a fast medium to a slow medium, the light ray bends toward the normal to the boundary between the two medium. Light refracts as it travels at an angle into a medium with a different refractive index.
What must x be so that the handle end of the bat remains at rest as the bat begins to move? (Hint: Consider the motion of the center of mass and the rotation about the center of mass. Find x so that these two motions combine to give v=0 for the end of the bat just after the collision. Also, note that integration of equation ∑τ⃗ =dL⃗ dt gives ΔL=∫t1t2(∑τ)dt. )
Complete Question
The complete question is shown on the first uploaded image
Answer:
Explanation:
From the question we are told that
The mass of the bat is [tex]m_b = 0.800 \ kg[/tex]
The bat length is [tex]L_b = 0.900 \ m[/tex]
The distance of the bat's center of mass to the handle end is [tex]z_c = 0.600 \ m[/tex]
The moment of inertia of the bat is [tex]I = 0.0530 \ kg \cdot m^2[/tex]
The objective of the solution is to find x which is the distance from the handle of the bat to the point where the baseball hit the bat
Generally the velocity change at the end of the bat is mathematically represented as
[tex]\Delta v_e = \Delta v_c - \Delta w* z_c[/tex]
Where [tex]\Delta v_c[/tex] is the velocity change at the center of the bat which is mathematically represented as
[tex]\Delta v_c = \frac{Impulse}{m_b }[/tex]
We are told that the impulse is J so
[tex]\Delta v_c = \frac{J}{m_b }[/tex]
And [tex]\Delta w[/tex] is the change in angular velocity which is mathematically represented as
[tex]\Delta w = \frac{J (z -z_c)}{I}[/tex]
Now we have that
[tex]\Delta v_e = \frac{J}{m_b } - [\frac{J (x- z_c)}{I} ] * z_c[/tex]
Before a swing when the bat is at rest the velocity change a the end of the bat handle is zero and the impulse will be 1
So
[tex]0 = \frac{1}{m_b } - [\frac{J (x- z_c)}{I} ] * z_c[/tex]
=> [tex]x = \frac{I}{m_b z_c} + m_b[/tex]
substituting values
[tex]x = \frac{0.530}{0.800 * 0.600} + 0.600[/tex]
[tex]x = 0.710 \ m[/tex]
An astronaut visiting Jupiter's satellite Europa leaves a canister of 1.20 mol of nitrogen gas (28.0 g/mol) at 50.0 ∘C on the satellite's surface. Europa has no significant atmosphere, and the acceleration due to gravity at its surface is 1.30 m/s2. The canister springs a leak, allowing molecules to escape from a small hole. Neglect the interaction with surrounding atmosphere. (a) What is the maximum height (in km) above Europa's surface that is reached by a nitrogen molecule whose speed equals the rms speed? Assume that the molecule is shot straight up out of the hole in the canister, and ignore the variation in g with altitude. (b) The escape speed from Europa is 2025 m/s. Can any of the nitrogen molecules escape from Europa and into space?
Answer:
the answer is a
Explanation:
The instantaneous speed of a particle moving along one straight line is v(t) = ate−6t, where the speed v is measured in meters per second, the time t is measured in seconds, and the magnitude of the constant a is measured in meters per second squared. What is its maximum speed, expressed as a multiple of a? (Do not include units in your answer.)
Answer:
v_max = (1/6)e^-1 a
Explanation:
You have the following equation for the instantaneous speed of a particle:
[tex]v(t)=ate^{-6t}[/tex] (1)
To find the expression for the maximum speed in terms of the acceleration "a", you first derivative v(t) respect to time t:
[tex]\frac{dv(t)}{dt}=\frac{d}{dt}[ate^{-6t}]=a[(1)e^{-6t}+t(e^{-6t}(-6))][/tex] (2)
where you have use the derivative of a product.
Next, you equal the expression (2) to zero in order to calculate t:
[tex]a[(1)e^{-6t}-6te^{-6t}]=0\\\\1-6t=0\\\\t=\frac{1}{6}[/tex]
For t = 1/6 you obtain the maximum speed.
Then, you replace that value of t in the expression (1):
[tex]v_{max}=a(\frac{1}{6})e^{-6(\frac{1}{6})}=\frac{e^{-1}}{6}a[/tex]
hence, the maximum speed is v_max = ((1/6)e^-1)a
A 0.009 kg bullet fired through a door enters at 803 m/s and leaves at 617 m/s. If the door material is known to exert an average resistive force of 5620 N on bullets of this type at usual speeds, find the thickness of the door.
Answer:
The thickness of the door is 0.4230 m
Explanation:
Given;
mass of bullet, m = 0.009 kg
initial velocity of the bullet, u = 803 m/s
final velocity of the bullet, v = 617 m/s
average resistive force of the door on the bullet, F = 5620 N
Apply Newton's second law of motion;
Force exerted by the door on the bullet = Force of the moving bullet
F = ma
where;
F is applied force
m is mass
a is acceleration
Also, Force exerted by the door on the bullet = Force of the moving bullet
[tex]F =ma, \ But \ a =\frac{dv}{dt} = \frac{u-v}{t} \\\\F = \frac{m(u-v)}{t}[/tex]
where;
v is the final velocity of the bullet
u is initial velocity of the bullet
t is time
We need to calculate the time spent by the bullet before it passes through the door.
[tex]t = \frac{m(u-v)}{F} \\\\t = \frac{0.009(803-617)}{5620} = 0.0002979 \ s[/tex]
Distance traveled by the bullet within this time period = thickness of the door
This distance is equivalent to the product of average velocity and time
[tex]S = (\frac{u+v}{2}) t[/tex]
where;
s is the distance traveled
[tex]S = (\frac{u+v}{2}) t\\\\S = (\frac{803+617}{2}) 0.0002979\\\\S = 0.4230 \ m[/tex]
Therefore, the thickness of the door is 0.4230 m
