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 firm wants to determine the amount of frictional torque in their current line of grindstones, so they can redesign them to be more energy efficient. To do this, they ask you to test the best-selling model, which is basically a diskshaped grindstone of mass 2.0 kg and radius 9.90 cm which operates at 720 rev/min. When the power is shut off, you time the grindstone and find it takes 41.9 s for it to stop rotating.
(a) What is the angular acceleration of the grindstone in rad/s^2? (b) What is the frictional torque exerted on the grindstone in N·m?
Answer:
a)- 1.799 rad/sec²
b)- 17.6 x 10ˉ³Nm
Explanation:
ω₀ = 720 rev/min x (1 min/60 sec) x (2π rad / 1 rev) = 24π rad/s
a) Assuming a constant angular acceleration, the formula will be
α = (ωf -ω₀) / t
As final state of the grindstone is at rest, so ωf =0
⇒ α = (0-24π) / 41.9 = - 1.799 rad/sec²
b)Moment of inertia I for a disk about its central axis
I = ½mr²
where m=2kg and radius 'r'= 0.099m
I = ½(2)(0.099²)
I = 9.8 x 10ˉ³ kgm²
Next is to determine the frictional torque exerted on the grindstone, that caused it to stop, applying the rotational equivalent of the Newton's 2nd law:
τ = I α =>(9.8 x 10ˉ³)(- 1.799)
τ = - 17.6 x 10ˉ³Nm
(The negative sign indicates that the frictional torque opposes to the rotation of the grindstone).
A. A PH202 student lives next to a construction site and sees a crane with a wrecking ball demolish the building next door. The wrecking ball swings along the wall between her house and the neighbor’s house. In an effort to determine the length of the cable on the wrecking ball the student builds a pendulum using a random rock and a string. Her pendulum turns out to be 0.500m long. While she plays with her pendulum she realizes that the wrecking ball swings back and forth in the same amount of time that it takes the rock to complete 5 full oscillations. What is the length of the cable on the wrecking ball?
Answer:
The length of cable is 12.5 m
Explanation:
Since, the wrecking ball completes 1 oscillation, in the same time, as it takes for the rock to complete 5 oscillations.
Therefore,
Time Period of Wrecking Ball = 5 (Time Period of Rock)
Since,
Time Period of Pendulum = 2π√(L/g)
Therefore,
2π√(L₁/g) = 5[2π√(L₂/g)]
√L₁ = 5√L₂
Squaring on both sides:
L₁ = 25 L₂
where,
L₁ = Length of Cable = ?
L₂ = Length of string = 0.5 m
Therefore,
L₁ = 25 (0.5 m)
L₁ = 12.5 m
A sensor mounted to a cantilever beam indicates beam motion with time. When the beam is deflected and released (step test), the sensor signal indicates that the beam oscillates as an underdamped second-order system with a ringing frequency of 10 rad/s. The maximum displacement amplitudes are measured at three different times corresponding to the 1st, 16th, and 32nd cycles and found to be 17, 9, and 5 mV, respectively. Estimate the damping ratio and natural frequency of the beam based on this measured signal, K
Answer:
The damping ratio= 2.93 × 10^-3 and the natural frequency is 10rad/s
Explanation:
Check attachment
does work done by the electric force depends on the path taken.
Answer:
yes but not sure
Explanation:
i don't know so much but I just guessed I'm not sure
Word ProNems:
1. A car goes from 4.47 m/s to 17.9 m/s in 3 seconds.
a. Did this car speed up or slow down?
b. Should your answer be positive or negative? Explain your reasoning.
c. Calculate the acceleration of the car.
Answer:
a:it speed up
b:it should be positive since final
velocity is larger than initial velocity
c:acceleration is approximately 4.5
m/s^2
Explanation:
initial velocity=u=4.47m/s
Final velocity=v=17.9m/s
Time=t=3 seconds
a:the car speed up since the velocity
increased
b:change in velocity is positive
because final velocity is larger than
initial velocity
17.9-4.47=13.43 m/s
c: acceleration=(v-u)/t
acceleration=(17.9-4.47)/3
acceleration=13.43/3
acceleration=4.5 m/s^2
If a car goes from a velocity u to a velocity v, the change in velocity is expressed as;
[tex]\triangle v =v-u[/tex]
Given
final velocity v = 17.9m/s
initial velocity u = 4.47m/s
[tex]\triangle v = 17.9-4.47\\\triangle v =13.43m/s[/tex]
a) Since the change in velocity is a positive value, this shows that the car speeds up.
b) The answer should be positive since the final velocity is greater than the initial velocity.
c) Acceleration is the change in velocity with respect to time
[tex]a=\frac{v-u}{t}\\a=\frac{17.9-4.47}{3} \\a=\frac{13.43}{3}\\a= 4.48m/s^2[/tex]
Hence the acceleration of the car is 4.48m/s²
Learn more here: https://brainly.com/question/24175401
What is the gravitational force related to the distance between two objects?
Answer:
Since gravitational force is inversely proportional to the square of the separation distance between the two interacting objects, more separation distance will result in weaker gravitational forces. So as two objects are separated from each other, the force of gravitational attraction between them also decreases.
answer: It is inversely proportional to the square of the distance.
Explanation:
for the team
I need some help!!!!!!!!!
Answer:
The Object will immediately begin moving toward the left
Explanation:
Because the force of thirteen is greater than ten and applied to the opposite side
the medium of a mechanical wave is the matter through wich it travels? tru or false
Answer:
it can only travel through matter
Explanation:
the mechanical wave can be any type of matter
Which of the following could possibly be predicted? Earthquakes Landslides Foreshocks Volcanic eruptions
Answer:
Volcanic Eruptions
Explanation:
The volcano can start showing signs that it may be about to explode.
Answer:
Volcanic eruptions.
Explanation:
I had the same question on an assignment and I got it right.
A bike travels 15.0 km in 45.0 min. Its average speed in km/h is .
The average speed of a bus traveling a distance of 15.0 km in 45.0 min is 20 km/hour.
What is speed?The speed of an object, also known as v in kinematics, is the size of the change in that object's position over time or the size of the change in that object's position per unit of time, making it a scalar quantity.
The distance travelled by an object in a time interval is divided by the length of the interval to determine its average speed.
Distance travelled by the bike = 15.0 km
Time taken by the object = 45.0 minute = (45.0 ÷ 60) hour = 0.75 hour
Hence, the average speed of the object = distance travelled / time taken
= 15.0 km/0.75 hour
= 20 km/hour,
Therefore, the average speed of a bus traveling a distance of15.0 km in 45.0 min is 20 km/hour.
Learn more about speed here:
https://brainly.com/question/28224010
#SPJ2
Alan leaves Los Angeles at 8:00 A.M. to drive to San Francisco 400 mi away. He manages to travel at a steady 50 mph in spite of traffic. Beth leaves Los Angeles at 9:00 A.M. and surprisingly manages to also drive at a constant speed, in this case 60 mph. (Knight 2.1) a. Who gets to San Francisco first? (Beth) b. How long does the first to arrive have to wait for the second? (20 minutes)
Answer:
a) Beth will reach before Alan
b)Beth has to wait 20 min for Alan to arrive
Explanation:
let 'd' be distance b/w Los Angeles and San Francisco i.e 400 mi
considering ,
Alan's speed [tex]v_A[/tex]=50mph
Beth's speed [tex]v_B[/tex]=60mph
->For Alan:
The time required [tex]t_A[/tex]= d/[tex]v_A[/tex]= 400/50 => 8h
-> For beth:
The time required [tex]t_B=\frac{d}{v_B} =\frac{400}{60} =>6\frac{2}{3} h[/tex] => 6h 40m
Alan will reach at 8:00 a.m +8h = 4:00p.m.
Beth will reach at 9:00 a.m +6h 40m= 3:40p.m.
a) Beth will reach before Alan
b)Beth has to wait 20 min for Alan to arrive
An RV is traveling 60 km/h along a highway at night. A boy sitting near the driver of the RV turns a flashlight on and shines it at another boy at the back end of the RV.
Answer:
300,000,000 and the second question is the same.
Explanation:
Edge
Answer:
What is the speed of the light relative to the boys?
✔ 300,000,000
What is the speed of light relative to a stationary observer on the side of the road?
✔ 300,000,000
Explanation: MARK ME BRAINLIEST CUHZZZZZ DA*N (cant say dam or they gone banned me again)
The motion of a set of particles moving along the x-axis is governed by the differential equation dx dt = t 3 - x3, where x1t2 denotes the position at time t of the particle. (a) If a particle is located at x = 1 when t = 2, what is its velocity at this time? (b) Show that the acceleration of a particle is
Answer:
a)V=7 m/s
b)a=3t²-3x² t³ +3 x ⁵
Explanation:
Given that
[tex]\dfrac{dx}{dt}=t^3-x^3[/tex]
a)
We know that velocity V is given as follows
[tex]V=\dfrac{dx}{dt}[/tex]
[tex]V=t^3-x^3[/tex]
At t= 2 s and x= 1 m
[tex]V=2^3-1^3=7 m/s[/tex]
V=7 m/s
b)
Acceleration a is given as follows
[tex]a=\dfrac{dV}{dt}[/tex]
[tex]a=3t^2-3x^2\dfrac{dx}{dt}[/tex]
Now by putting the values
[tex]a=3t^2-3x^2\times (t^3-x^3)[/tex]
a=3t²-3x² t³ +3 x ⁵
Therefore the acceleration of a particle will be 3t²-3x² t³ +3 x ⁵.
The graph shows the force and displacement of an object that is being pushed. How much work is done to push the object 0.12m?
Answer:
Work done, W = 0.72 J
Explanation:
From the given graph, it is clear that the force applied is 6 N if the extension is 0.12 m.
Work done is given by :
W = Fx
[tex]W=6\ N\times 0.12\ m\\\\W=0.72\ J[/tex]
So, 0.72 J of work is done to push the object 0.12 m
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
Archerfish are tropical fish that hunt by shooting drops of water from their mouths at insects above the water’s surface to knock them into the water, where the fish can eat them. A 65-g fish at rest just at the surface of the water can expel a 0.30-g drop of water in a short burst of 5.0 ms. High-speed measurements show that the water has a speed of 2.5 m/s just after the archerfish expels it. What is the speed of the archerfish immediately after it expels the drop of water?
(a) 0.0025 m/s
(b) 0.012 m/s
(c) 0.75 m/s
(d) 2.5 m/s.
Answer:
(b) 0.012 m/s
Explanation:
Given that:
mass (m₁) of the drop of water = 0.30 g = 3.0 × 10⁻⁴ kg
mass (M₂) of the fish = 65 g = 65 × 10⁻³ kg
speed (v₁) of the water = 2.5 m/s
speed (v₂) of the archerfish = ??
By conservation of momentum
m₁v₁ - M₂v₂ = 0
m₁v₁ = M₂v₂
v₂ = m₁v₁ / M₂
v₂ = ( 3.0 × 10⁻⁴ × 2.5 ) / 65 × 10⁻³
v₂ = 0.0115 m/s
v₂ ≅ 0.012 m/s
Therefore, the speed of the archerfish immediately after it expels the drop of water 0.012 m/s.
When a star explodes as a supernova, the expanding gas can pass through the interstellar medium triggering...
A.) an increase in variable star activity
B.) formation of new stars
C.) life to develop
D.) additional supernovas
B. Formation of new stars
I need help plz help me out 10 points!!!!!!!
Answer:
The answer is diffraction
Explanation:
Answer:
The answer is diffraction
Explanation:
I did the test! HOPE THIS HELPS!
