Answer:
Explanation:
s(t) = -16t2 + 8√t
A )
Average velocity
s(t) / t = (-16t2 + 8√t)/t
A(t)= -16t + 8 / √t
average velocity of the ball for t near 4.
A(t) = -16t + 8 / √t
Lt t⇒4
B )
Distance covered in 4 s
-16t2 + 8√t
= - 16 x 16 + 8 x 2
= - 240
Distance covered in 5 s
= - 16 x 25 + 8 √5
= -400 + 17.88
= -382.12
distance covered in duration from 4 to 5 sec
= -142.12
velocity = - 142.12 / 1 = - 142.12 m /s
Distance covered in 4.5 s
= -16 x 4.5² + 8√4.5
= -324 + 16.97
= -307
distance covered during 4 to 4.5
= 67
velocity during 4 to 4.5
= -67 / .5
- 134 m /s
distance covered in 4.05 s
-16 x 4.05² + 8√4.05
-262.44 + 16.1
-246.34
distance covered during 4 to 4.05
= -6.34
velocity during 4 to 4.05
= -6.34 / .05
- 126.8 m /s
C )
Instantaneous velocity at t = 4
= - 120 m /s
(A) As 't' approaches to 4s, the formula of average velocity is
[tex]v_{avg} _{|(t\rightarrow4)}=Lim_{|(t\rightarrow4)}[-16t+ \frac{8}{\sqrt{t}} ][/tex]
(B) The average velocity for the time interval starting at t = 4 with a duration of 1 second is [tex]142.11\,m/s[/tex].
The average velocity for the time interval starting at t = 4 with a duration of 0.5 seconds is [tex]-134.058\,m/s[/tex]
The average velocity for the time interval starting at t = 4 with a duration of 0.05 seconds is [tex]-126.8\,m/s[/tex].
(C) The instantaneous velocity of the ball at 4s is [tex]-126\,m/s[/tex].
The answers are explained as follows.
Given that the distance is a function of time.
[tex]s(t)=-16\,t^2\,+\,8\sqrt{t}[/tex]
(A) The average velocity can be given by,
[tex]v_{avg}=\frac{-16\,t^2\,+\,8\sqrt{t} }{t} =-16t+ \frac{8}{\sqrt{t}}[/tex]
As 't' approaches to 4s, the formula becomes;
[tex]v_{avg} _{|(t\rightarrow4)}=Lim_{|(t\rightarrow4)}[-16t+ \frac{8}{\sqrt{t}} ][/tex]
(B) We know that the average velocity or in this case speed is the total distance by the total time taken.
Distance covered in 4 s can be found by putting [tex]t=4s[/tex] in the distance formula.
[tex]s(4)=(-16\times16)+(8\times2)=-240\,m[/tex]Distance covered in 5 s can also be found by the same method
[tex]s(5)=(-16\times25)+(8\times\sqrt{5} )=-382.11\,m[/tex]Therefore, the distance covered in from 4 to 5 seconds is;
[tex]s(5) -s(4)=-382.11\,m-(-240\,m)=-142.11\,m[/tex]So, the average velocity here = [tex]\frac{-142.11\,m}{5\,s-4\,s}=-142.11\,m/s[/tex]Distance covered in 4.5 s is given by,
[tex]s(4.5)=(-16\times4.5^2)+(8\times\sqrt{4.5} )=-307.029\,m[/tex]Therefore, the distance covered in 4 to 4.5 seconds is;
[tex]s(4.5)-s(4)=-307.029\,m-(-240\,m)=67.029\,m[/tex]So, the average velocity here = [tex]\frac{-67.029\,m}{4.5\,s-4\,s}=-134.058\,m/s[/tex]Distance covered in 4.05 s is given by,
[tex]s(4.05)=(-16\times4.05^2)+(8\times\sqrt{4.05} )=-246.34\,m[/tex]Therefore, the distance covered in 4 to 4.05 seconds is;
[tex]s(4.05)-s(4)=-246.34\,m-(-240\,m)=-6.34\,m[/tex]So, the average velocity here = [tex]\frac{-6.34\,m}{4.05\,s-4\,s}=-126.8\,m/s[/tex](C) The instantaneous velocity of the ball can be found by differentiating the function [tex]s(t)[/tex].
[tex]v(t)=\frac{ds(t)}{dt} =\frac{d}{dt}(-16\,t^2\,+\,8\sqrt{t})=-32t+\frac{4}{\sqrt{t} }[/tex]For the instantaneous velocity of the ball at 4s, substitute [tex]t=4\,s[/tex] in the above equation.[tex]v(t)=(-32\times4)+\frac{4}{2}=-126\,m/s[/tex]Learn more about finding the velocity at a given time here:
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which is an accurate description of what occurs when a log burns
Answer:
charcoal
Explanation:
Answer:
charcoal
Explanation:
Mark as Brainliest
A model rocket accelerates upward from the ground with a constant acceleration, reaching a height of 50 m in 8 s.What is the speed (in m/s) at a height of 50 m?A: 1.07×101 B: 1.25×101 C: 1.46×101 D: 1.71×101 E: 2.00×101
Answer:
The value is [tex]v = 12.5 \ m/s[/tex]
Explanation:
From the question we are told that
The height is [tex]h = 50 \ m[/tex]
The time taken is [tex]t = 8 \ s [/tex]
From the equation of motion we have that
[tex]s = ut + \frac{1}{2} * a * t^2[/tex]
Here u = 0 because the rocket started at rest
[tex]50 = 0 + \frac{1}{2} * a * 8^2[/tex]
=> [tex]a = \frac{100}{64}[/tex]
=> [tex]a = 1.5625 \ m/s^2[/tex]
Also from the kinematic equation we have that
[tex]v = u + at[/tex]
=> [tex]v = 0 + (8 * 1.5625)[/tex]
=> [tex]v = 12.5 \ m/s[/tex]
If you want to replicate another scientist's experiment, which sections of their laboratory report will provide you with the most valuable information?
A sound wave is observed to travel through a liquid with a speed of 1400 m/s. The specific gravity of the liquid is 1.8.
Determine the bulk modulus for this fluid.
Answer:
21
Explanation:
I think its 21 because its 21
What two processes are involved in the formation of Earth’s magnetic field?
Answer:
i. Geodynamo
ii. Gravitational energy conversion
Explanation:
The outer core of the Earth is believed to be in molten form, consisting of mainly iron and nickel. The reaction (mixture) between these components or elements generates conventional currents which is maintained by the gravitational energy produced in the process. Thus, magnetic fields are generated continuously due to the continuous flow of electric current during the processes.
Two vectors of magnitudes |A| = 8 units and |B| = 5 units make an angle that can vary from 0° to 180°. The magnitude of the resultant vector A + B CANNOT have the value of:
The question is missing the alternatives. Here is the complete question.
Two vectors of magnitude |A| = 8units and |B| = 5units make an angle that can vary from 0° to 180°. The magnitude of the resultant vector A+B CANNOT have the value of:
A. 2 units
B. 5 units
C. 8 units
D. 12 units
Answer: A. 2 units
Explanation: Vector is an entity that has characteristics as magnitude and direction. Resultant vector is the "sum" of 2 or more vectors.
In this question, the vectors have magnitude and angle varies from 0° to 180°.
When angle between vectors A and B is 0°, they are parallel and pointing to the same direction, so:
[tex]V_{R} = |A| + |B|[/tex]
[tex]V_{R}=8+5[/tex]
[tex]V_{R}[/tex] = 13
When the angle is 180°, it means vectors are in opposing directions, so:
[tex]V_{R} = |A| - |B|[/tex]
[tex]V_{R} = 8-5[/tex]
[tex]V_{R}[/tex] = 3
From the calculations, we can conclude the magnitude of resultant vector varies between 3 and 13.
The least value is 3, so it cannot have a value of 2 units.
Maddie Cat can ride her tricycle at an amazing speed of 3.75 meters per second.
How far can she go in 5.0 seconds?
Distance = (speed) x (time)
Distance = (3.75 m/s) x (5.0 s)
Distance = 18.75 meters
Answer: 3.75 x 5.0= 18.75 meters
Explanation:
A flag pole is attached to the side of a building 5.00 ft off the ground and it hangs down 30 degrees from the vertical. If the pole is 6.00 ft in length, how high in feet is the top of the pole at its highest point (sticking out from building)? (Hint: draw a picture) E
Answer:
h ’= 0.6963 m
this is the height of the part of the post it supports.
Explanation:
For this exercise we can use the trigonometric relationships
Let's start by finding the vertical component of the height of the pole
cos 30 = x / L
x = L cos 30
x = 6 cps 30
x = 5.196 m
with this length, the post has the same height as the building, so there are
Aₓ = L -x
Δx = 6 -5.196
Δx = 0.804 ft
the height of this section is
cos 30 = h ’/ Dx
h ’= 0.804 cos30
h ’= 0.6963 m
this is the height of the part of the post it supports.
Tony completed a 720 km journey with an average speed of 90 km/h for the first 285 km he travelled at an average speed of 95 Km/h what was the average speed for the remaining journey
Answer:
[tex]Speed = 87 km/h[/tex]
Explanation:
Given
Total Distance = 720km
Average Speed = 90km/h
Average Speed (285km) = 95km/h
Required
Determine the average speed for the rest of the journey
First we need to determine the time taken to complete the journey using
[tex]Speed = \frac{Distance}{Time}[/tex]
[tex]90 = \frac{720}{Time_1}[/tex]
Make Time the subject of formula
[tex]Time_1 = \frac{720}{90}[/tex]
[tex]Time_1 = 8 hour[/tex]
Next, we determine the time taken to complete 285km of the journey using
[tex]Speed = \frac{Distance}{Time}[/tex]
[tex]95 = \frac{285}{Time_2}[/tex]
Make Time the subject of formula
[tex]Time_2 = \frac{285}{95}[/tex]
[tex]Time_2 = 3 hours[/tex]
The difference between these calculated times is the time taken to complete the rest of the journey
[tex]Time = Time_1 - Time_2[/tex]
[tex]Time = 8\ hour - 3\ hour[/tex]
[tex]Time = 5\ hour[/tex]
Also, we need to calculate the distance of the rest of the journey
[tex]Distance = 720\ km - 285\ km[/tex]
[tex]Distance = 435\ km[/tex]
The average speed can then be calculated;
[tex]Speed = \frac{Distance}{Time}[/tex]
[tex]Speed = \frac{435}{5}[/tex]
[tex]Speed = 87 km/h[/tex]
Hence, the average speed of the rest of the journey is 87km/h
If a gannet hits the water at 32 m/s, what height did it dive from? Assume that the gannet was motionless before starting its dive.
Answer:
The value is [tex]h = 52.2 \ m[/tex]
Explanation:
From the question we are told that
The velocity at which he hit the water is [tex]v = 32 \ m/s[/tex]
Generally from kinematic equation
[tex]v^2 = u^2 + 2gh[/tex]
Here the initial velocity u = 0 m/ given that the gannet started the jump from rest.
=> [tex]32^2 = 0^2 + 2*(9.8)h[/tex]
=> [tex]h = 52.2 \ m[/tex]
The fastest land animal is the cheetah. A cheetah can run 100 meter dash in 5.95 seconds, calculate the cheetahs average speed in meters per second
Answer:
16.80672269 m/s
Explanation:
100 divide by 5.95 seconds
Answer:
16.8m/s
Explanation:
S=D/T
=
[tex] \frac{100}{5.95} [/tex]
=16.80672269
=16.8
a 20 ft shipping container on a cargo ship has a mass of 24000 kg and a volume of 33.2m3. what is the density of the shipping container
Answer:
722.89
Explanation:
mass=24000kg
volume=33•2
density=?
now,
density=mass/volume
=24000/33•2
=722•89
density=722•89 kg/m^3
What net force is needed to accelerate a 1.25 kg book 5.00 m/s2?
Answer:
F = 6.25 N
Explanation:
Given that,
Mass of the book is 1.25 kg
Acceleration of the book is 5 m/s²
We need to find the net force needed to accelerate the book. Net force is given by :
F = ma
[tex]F=1.25\times 5\\\\F=6.25\ N[/tex]
So, 6.25 N of force is acting on the book.
Light of wavelength 600 nm illuminates a diffraction grating. The second order maximum is at an angle of 65 degrees. a) List your known variables. b) What is the spacing between slits in this grating? c) How many lines per millimeter does this grating have?
Answer:
A. Known variables include
Wavelength = 600nm
Theta= 65°
m= 2
B.
d= m x wavelength / sin theta
= 2 * 600*10^-9 /sin 65°
= 1.3*10^-6m
C.
N = 1/d
So N = 1/1.3*10^-6m
=76.9 lines per micro meter
How does society shape reality?
which sentences describes an object that has kinetic energy
Answer:
A boat sails across the Ocean
Explanation:
Out of all the options on Edmentum none of them where moving so it would only make sense that this is the only one with kinetic energy.
What are some superheroes that resemble neurotransmitter functions. Dopamine: Acetylcholine: Endorphins: GABA: Glutamates: Norepinephrine: Serotonin:
Answer:
Serotonin
___________
Regular exercise is positively related to wellnes t or f
Which is one physical property that all stars have? pick one They are made of gases. They shine very brightly. They have a triangle shape. They contain iron in their cores.
Answer: They are made of gases
The physical property that all stars have will be they are made of gases. option 1 is correct.
What is a star?A star is a heavenly body made up of a brilliant spheroid of plasma held together by gravity. The Sun is the closest star to Earth.
Many additional stars may be seen with the normal eye at night, but due to their great distance from Earth, they appear as stationary points of light in the sky.
The physical property that all stars have will be they are made of gases.
Hence option 1 is correct.
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A silver block of silver block of density 10.5 g/cm3 has a volume of 30 cm3. Which of the following is the correct mass of the block
➝ Density of block = 10.5 g/cm³
➝ Volume of block = 30 cm³
We have to find mass of block[tex].[/tex]
➠ Density is defined as mass of substance per unit volume[tex].[/tex]
[tex]\dag\:\boxed{\bf{Density=\dfrac{Mass}{Volume}}}[/tex]
[tex]:\implies\sf\:Mass=Density\times Volume[/tex]
[tex]:\implies\sf\:Mass=10.5\times 30[/tex]
[tex]:\implies\boxed{\boxed{\bf{\red{Mass=315\:g}}}}[/tex]
Starting from rest, a panther accelerates at 3 m/s/s for 10 seconds. What is the final velocity of the panther
Answer:
Final velocity = 30 m/s
Explanation:
We are given;
Initial velocity; u = 0 m/s
Acceleration; a = 3 m/s²
Time; t = 10 seconds
From Newton's first law of motion, the final velocity is given by the formula;
v = u + at
Plugging in the relevant values gives;
v = 0 + (3 × 10)
v = 30 m/s
Which is true for a car moving around a circular track with constant speed? 1. It has zero acceleration2. it has an acceleration directed toward the center of its path3. it has an acceleration directed away fromt he center of its path4. it has an acceleration with a direction that cannot be determined from the information given5. it has an acceleration component in the direction of its velocity.
Answer:
2. it has an acceleration directed toward the center of its path
Explanation:
For a car moving with constant velocity around a circular track, there is a a change in the direction of the velocity at every point on the path taken by the car around the track. This change in direction generates an acceleration on the car which will be directed towards the center of the track. This is why the inertia force on a person inside the car when it is driven around a circular track is usually away from the center of the track, since the inertia force acts in an opposite direction to the acceleration on a body.
Which of the following is the best way to make a conclusion? A. The experiment must be manipulated until the results show what you want B. Estimate results to where they should be C. Choosing the results you like best D. Comparing data from the experiment to
Answer:
D
Explanation:
You should compare your results and data with your original hypothesis. It is okay to have a hypothesis that was shown to be false. That's science. Things do not always work out. You can always discuss what you could have done differently and things that went wrong. There is no right answer in science!
If you do not have enough time to complete a full workout, you should:
A. Cut the cool-down portion short.
B. Skip the warm-up and cool-down portions altogether.
C. Cut the warm-up portion short.
D. Cut the exercise portion short.
Answer:
the answer to this question would be cut the exercise portion down (D)
this is because warm ups and cooling down during exercise are more important than the exercise itself
Explanation:
the four strings of a bass guitar are 0.865 m long and are tuned to the notes g (98 hz), d (73.4 hz), a (55 hz), and e (41.2 hz). in one bass guitar, the g and d strings have a linear mass density of 4.8 g/m, and the a and e strings have a linear mass density of 29.8 g/m.what is the total force exerted by the strings on the neck?
Answer: Total force = 636,554.55N
Explanation: To determine tension of strings, wave speed on a string is necessary. Speed is found by:
v = f.λ
f is frequency
λ is wavelength
For the strings, wavelength equals to:
[tex]\lambda = 2L[/tex]
L is the length of the bass guitar string
Then, wave speed:
[tex]v=f.2L[/tex]
Tension on a string is
[tex]v=\sqrt{\frac{F_{T}}{\mu} }[/tex]
[tex]v^{2}=\frac{{F_{T}} }{\mu}[/tex]
[tex]F_{T} = v^{2}\mu[/tex]
[tex]F_{T} = (2f\lambda)^{2}\mu[/tex]
[tex]F_{T} = 4(f\lambda)^{2}\mu[/tex]
μ is linear mass density
For g string:
[tex]F_{T} = 4(98.0.865)^{2}.4.8[/tex]
[tex]F_{T}[/tex] = 137970.3N
For d string:
[tex]F_{T} = 4(73.4.0.865)^{2}.4.8[/tex]
[tex]F_{T}=[/tex] 77397.25N
For a string:
[tex]F_{T} = 4(55.0.865)^{2}.29.8[/tex]
[tex]F_{T}=[/tex] 269795N
For e string:
[tex]F_{T} = 4(41.2.0.865)^{2}.29.8[/tex]
[tex]F_{T}=[/tex] 151392N
Total force = 137,970.3 + 77,397.25 + 269,795 + 151,392
Total force = 636,554.55N
Total force exerted on the neck by the strings is 636,554.55N.
This question involves the concepts of the tension force in strings, linear mass density, and frequency.
The total force exerted by strings on the guitar is "359.4 N".
The tension force exerted by each string is given as:
[tex]F_T=v^2\mu[/tex]
where,
F_T = tension force = ?
v = speed = (frequency)(wavelength) = fλ
μ = linear mass density
Therefore,
[tex]F_T=f^2\lambda^2\mu[/tex]
but for strings in this case:
[tex]\lambda = 2(Length of string) = 2(0.865\ m)=1.73\ m[/tex]
Therefore,
[tex]F_T=f^2(1.3\ m)^2\mu[/tex]
For string g:
[tex]F_{Tg}=(98\ Hz)^2(1.3\ m)^2(4.8\ x\ 10^{-3}\ kg/m)\\F_{Tg}=77.9\ N[/tex]
For string d:
[tex]F_{Td}=(73.4\ Hz)^2(1.3\ m)^2(4.8\ x\ 10^{-3}\ kg/m)\\F_{Td}=43.7\ N[/tex]
For string a:
[tex]F_{Ta}=(55\ Hz)^2(1.3\ m)^2(29.8\ x\ 10^{-3}\ kg/m)\\F_{Ta}=152.3\ N[/tex]
For string e:
[tex]F_{Te}=(41.2\ Hz)^2(1.3\ m)^2(29.8\ x\ 10^{-3}\ kg/m)\\F_{Te}=85.5\ N[/tex]
So, the total force will be the sum of all tension forces:
[tex]F=F_{Tg}+F_{Td}+F_{Ta}+F_{Te}[/tex]
F = 77.9 N + 43.7 N + 152.3 N + 85.5 N
F = 359.4 N
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a) An electric circuit is rates, 240V, 13Ω calculate the amount of current that can be used by this appliance.
Answer:
18.4615385 amps
Explanation:
The voltage V in volts (V) is equal to the current I in amps (A) times the resistance R in ohms (Ω):
who is known as father of science?
GalileoGalilei is known as the father of science
Calculate the displacement traveled (in m) from start to finish.
Answer:
nnkn
Explanation:
nkn
How does the Coriolis effect impact ocean currents in the Northern and Southern Hemispheres?
Now let’s pretend you’re standing at the North Pole. When you throw the ball to your friend, it will again to appear to land to the right of him. But this time, it’s because he’s moving faster than you are and has moved ahead of the ball.
Everywhere you play global-scale "catch" in the Northern Hemisphere, the ball will deflect to the right.
A child on a bridge throws a rock straight down to the water below. The point where the child released the rock is 74 m above the water and it took 2.7 s for the rock to reach the water. Determine the rock's velocity (magnitude & direction) at the moment the child released it. Also determine the rock's velocity (magnitude & direction) at the moment it reached the water. Ignore air drag.
The rock's altitude y at time t, thrown with initial velocity v, is given by
[tex]y=74\,\mathrm m+vt-\dfrac12gt^2[/tex]
where [tex]g=9.80\frac{\rm m}{\mathrm s^2}[/tex] is the acceleration due to gravity.
After t = 2.7 s, the rock reaches the water (0 altitude), so
[tex]0=74\,\mathrm m+v(2.7\,\mathrm s)-\dfrac12g(2.7\,\mathrm s)^2[/tex]
[tex]\implies v=-\dfrac{74\,\mathrm m-\frac g2(2.7\,\mathrm s)^2}{2.7\,\mathrm s}\approx-14.177\dfrac{\rm m}{\rm s}[/tex]
so the rock was thrown with a velocity with magnitude 14 m/s and downward direction.
Its velocity at time t is [tex]v-gt[/tex] (with no horizontal component), so that at the moment it hits the water, its velocity is
[tex]v-g(2.7\,\mathrm s)\approx-40.637\dfrac{\rm m}{\rm s}[/tex]
That is, its final velocity has an approximate magnitude of 41 m/s, also directed downward.