Answers
Answer:
1.4 m/s
Explanation:
From the question given above, we obtained the following data:
Initial Displacement (d1) = 0.9 m
Final Displacement (d2) = 1.6 m
Initial time (t1) = 1.5 secs
Final time (t2) = 2 secs
Velocity (v) =..?
The velocity of an object can be defined as the rate of change of the displacement of the object with time. Mathematically, it can be expressed as follow:
Velocity = change of displacement /time
v = Δd / Δt
Thus, with the above formula, we can obtain the velocity of the car as follow:
Initial Displacement (d1) = 0.9 m
Final Displacement (d2) = 1.6 m
Change in displacement (Δd) = d2 – d1 = 1.6 – 0.9
= 0.7 m
Initial time (t1) = 1.5 secs
Final time (t2) = 2 secs
Change in time (Δt) = t2 – t1
= 2 – 1.5
= 0.5 s
Velocity (v) =..?
v = Δd / Δt
v = 0.7/0.5
v = 1.4 m/s
Therefore, the velocity of the car is 1.4 m/s
Related Questions
if a cart goes around a turn at 20 km/h ,what remains constant
1.position
2.velocity
3.direction
4.speed
Answers
Answer: 4.speed
Explanation:
In this case, we know that the cart remains at a constant 20km/h.
Now, one could say that "the velocity remains constant, because it always is 20km/h"
But remember that velocity is a vector, so this has a direction, and if the cart is going around a turn, then the direction of motion is changing, which tell us that there is acceleration.
But the module of the velocity, the speed, remains constant at 20km/h.
Then the correct option is 4, speed.
How does sleep affect your ability to handle stress?
Answers
Answer: Stress can adversely affect sleep quality and duration, while insufficient sleep can increase stress levels. Both stress and a lack of sleep can lead to lasting physical and mental health problems.
Explanation:
Many report that there stress increases when the length and quality of their sleep decreases. When you do not get enough sleep, 21 percent of adults report feeling more stressed.
Assuming it is a van der Waals gas, calculate the critical temperature, pressure and volume for CO2. (a = 3.610 atm L2 mol-2, b = 0.0429 L mol-1)
pc = ___ atm
Tc = ___ K
Vc = ___ L/mol
Answers
Answer
To get critical pressure
We use
Pc = a/(27b²)
So
= 3.610/(27 X 0.0429²)
We have
= 72.7 atm
Critical temperaturewe
We use
Tc = 8a/27Rb
= 8 x 3.610/(27 x 0.0812 x 0.0429)
= 307 K
Critical volume
We use
Vc =3b =
3 x 0.0429
= 0.129L/mol
A stone is thrown vertically upward with a speed of 28.0 m/s how much time is required to reach this height
Answers
Let’s review the 4 basic kinematic equations of motion for constant acceleration (this is a lesson – suggest you commit these to memory):
s = ut + ½ at^2 …. (1)
v^2 = u^2 + 2as …. (2)
v = u + at …. (3)
s = (u + v)t/2 …. (4)
where s is distance, u is initial velocity, v is final velocity, a is acceleration and t is time.
In this case, we know u = 17.0m/s, a = -g = -9.81m/s^2, s = 11.0m and we want to know v and t, so from equation (2):
v^2 = u^2 + 2as
v^2 = 17.0^2 -2(9.81)(11.0)
v = √73.18 = 8.55m/s
now from equation (3):
v = u + at
8.55 = 17.0 – 9.81t
t = (8.55 – 17.0)/(-9.81) = 0.86s
Which scientist determined that electrons had predicted zones orbiting the nucleus?
Answers
Answer:
Rutherford
Explanation:
Because
Schrödinger I just took the unit review
A ball of mass m moving with speed V collides with another ball of mass 2m (e= 1/2) in a horizontal smooth fixed circular tube of radius R (R is sufficiently large R>>>d). The time after which next collision will take place is:________
Answers
Answer:
[tex]$ \frac{4\pi R}{V}$[/tex]
Explanation:
Given :
Mass of ball 1 = m
Mass of ball 2 = 2m
Since, R>>>d, the collision is head on.
Therefore, we get
[tex]$ \frac{v_1 -v_2}{V}=\frac{1}{2}$[/tex]
[tex]$ \therefore \frac{\text{velocity of seperation}}{\text{velocity of approach}}= v_1-v_2 = \frac{V}{2}$[/tex]
Relative velocity is given by V/2. So, we get the time when the masses will again collide as
[tex]$ t = \frac{2\pi R}{\frac{V}{2}}=\frac{4\pi R}{V} $[/tex]
which water molecules have the greatest kinetic energy
Answers
A Lotus will travel 275 meters in 4.71 seconds. What is this car's average speed?
Answers
Your friend says, “chemical changes are caused by an input in energy. In physical changes, there is no transfer of energy” is your friend correct? Why or why not?
Answers
Answer:
Ok, let's suppose the simplest of the physical changes:
We have an object that is not moving (so it is not accelerated)
and there is change, now the object moves.
Because there was a change, means that there was an acceleration, and by the second Newton's law.
Force equals mass times acceleration:
F = m*a
There must be a force.
So suppose that you pushed the object, then some energy that you had, you transferred it to the object, that now is moving and now has kinetic energy.
Now, is kinda true that in a closed system the total energy is always constant, but it depends on what is our system.
So if we think in our system as you and the object, then in the whole system the energy does not change because the energy that you lost is now on the object, but again, there was a transfer of energy.
So no, your friend is not correct.
Both chemical and physical changes involve energy transfer, so your friend is not correct.
A chemical change occurs when a new substance is formed or created through a process of chemical reaction which is reversible.
The addition or removal of heat energy can affect the rate of chemical reaction. This addition or removal is known as energy transfer process.
Examples of chemical changes include;
Acid-base reaction.Rusting of iron in presence of moisture and oxygen. Cooking any food.A physical change on the other hand doesn't involve formation of new substance and it is can be reversible or irreversible.
The addition or removal of heat energy can affect the rate of physical changes.
Examples of such physical changes include;
vaporization of liquid (liquid to gas),freezing of liquid (liquid to solid), and condensation of gas (gas to liquid).Thus, both process (chemical and physical changes) involve energy transfer, so your friend is not correct.
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When charges qa, qb, and qc are placed respectively at the corners a, b, and c of a right triangle, the potential at the midpoint of the hypotenuse is 20 V. When the charge qa is removed, the potential at the midpoint becomes 15 V. When, instead, the charge qb is removed (qa and qc both in place), the potential at the midpoint becomes 12 V. What is the potential at the midpoint if only the charge qc is removed from the array of charges?
Answers
Answer:
8v
Explanation:
First we apply super position principle
Vt= v1 + v2+ v3
Remove qa
But vt= 20v
So V = v2+v3
V1= 20-15
= 5v
Remove qb
V= v1+v3
V=8v
So the potential when qa and qc are remove is the potential due to qb
Which is 8v
Explain why a Merry-Go-Round and a Ferris Wheel have a constant acceleration when they are moving?
Answers
Answer:
merry go round and Ferris wheel have a constant acceleration due to the change in direction at every point.
Answer:
A merry-go-round is accelerating. Acceleration is a change in speed, direction, or both. Even though the speed of the merry-go-round does not change, its direction constantly changes as it spins.
Explanation:
If one object (a) is moving at 60m/s^2, and the other object (b) is moving at 65m/s^2, at what time will the faster moving object be 10m ahead of the other object?
Answers
Answer:
a is moving at 60m and the other object
4. Lead has a density of 11.5g/cmº. A rectangular block of lead measures 7cm x5cmx2cm.
a) Find the volume of the block of lead.
b) Find the mass of the block of lead
Answers
Answer:
(a) 70cm³
(b) 805 grams
Explanation:
(a) V = L×B×H
= 7cm×5cm×2cm
= 35cm×2cm
= 70cm³
(b) Mass = Volume × Density
= 70cm³ × 11.5g/cm³
= 805 grams
A bungee jumper with mass 65.0 kg jumps from a high bridge. After reaching his lowest point, he oscillates up and down, hitting a low point eight more times in 38.0 s. After many oscillations, he finally comes to rest 25.0 m below the level of the bridge. Calculate the spring stiffness constant and the unstretched length of the bungee cord.
Answers
Explanation:
It is given that,
Mass of a bungee jumper is 65 kg
The time period of the oscillation is 38 s, hitting a low point eight more times.It means its time period is
[tex]T=\dfrac{38}{8}\\\\T=4.75\ s[/tex]
After many oscillations, he finally comes to rest 25.0 m below the level of the bridge.
For an oscillating object, the time period is given by :
[tex]T=2\pi \sqrt{\dfrac{m}{k}}[/tex]
k = spring stiffness constant
So,
[tex]k=\dfrac{4\pi ^2m}{T^2}\\\\k=\dfrac{4\pi ^2\times 65}{(4.75)^2}\\\\k=113.43\ N/m[/tex]
When the cord is in air,
mg=kx
x = the extension in the cord
[tex]x=\dfrac{mg}{k}\\\\x=\dfrac{65\times 9.8}{113.6}\\\\x=5.6\ m[/tex]
So, the unstretched length of the bungee cord is equal to 25 m - 5.6 m = 19.4 m
The spring stiffness constant is 116.7 N/m and the the unstretched length of the bungee cord is 19.54 m.
The given parameters;
mass of the bungee jumper, m = 65 kgtime of motion, t = 38 sdistance to come to rest, d = 25 mThe period of oscillation of the bungee jumper is calculated as follows;
[tex]T = \frac{t}{n} \\\\T = \frac{38}{8} \\\\T = 4.75 \ s[/tex]
The spring stiffness constant is calculated as follows;
[tex]T = 2\pi \sqrt{\frac{m}{k} } \\\\\sqrt{\frac{m}{k} } = \frac{T}{2\pi} \\\\k = m \times \frac{T^2}{4\pi^2} \\\\k = 65 \times \frac{(4.75)^2}{4\pi ^2} \\\\k = 116.7 \ N/m[/tex]
The extension of the cord is calculated as follows;
[tex]F = kx\\\\mg = kx\\\\x = \frac{mg}{k} \\\\x = \frac{65 \times 9.8}{116.7} \\\\x = 5.46 \ m[/tex]
The unstretched length of the bungee cord is calculated as;
[tex]\Delta x = l_2-l_1\\\\l_1 = l_2 - \Delta x\\\\l_1 = 25 - 5.46\\\\l_1 = 19.54 \ m[/tex]
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2. If a cyclist in the Tour de France traveled southwest a distance of 12,250 meters in one hour, what would the velocity of the cyclist be?
Answers
12,25 km/h
≈ 3,4 m/s
Explanation:v = d/t
= 12250m/h
= 12,25km/h
or
v = d/t
= 12250m/h
1h = 60m×60s = 3600s
= 12250m/3600s
≈ 3,4 m/s
Which statement best defines inertia? An object's motion is unaffected by any external forces acting upon it. An object responds to a force by tending to move in the direction of that force. An object opposes any motion, naturally returning to a state of rest on its own. An object opposes any change in its velocity, either to its direction or to its speed.
Answers
Answer:
An object opposes any change in its velocity, either to its direction or to its speed.
Explanation:
Edmentum Answer
The speed of a bus increases uniformly from
15 ms- to 60 ms in 20 seconds. Calculate
a. the average speed,
b. the acceleration,
C. the distance travelled during the entire
period The speed of a bus increases uniformly from
15 ms- to 60 ms in 20 seconds. Calculate
a. the average speed,
b. the acceleration,
C. the distance travelled during the entire
period
Answers
Explanation:
a. For constant acceleration:
v_avg = ½ (v + v₀)
v_avg = ½ (60 m/s + 15 m/s)
v_avg = 37.5 m/s
b. a = (v − v₀) / t
a = (60 m/s − 15 m/s) / 20 s
a = 2.25 m/s²
c. x = v_avg t
x = (37.5 m/s) (20 s)
x = 750 m
Gwen releases a rock at rest from the top of a 40-m tower. If g = 9.8 m/s2 and air resistance is negligible, what is the speed of the rock as it hits the ground?
Answers
Answer:
[tex]28\; \rm m \cdot s^{-1}[/tex].
Explanation:
Short ExplanationApply the SUVAT equation [tex]\left(v^2 - u^2) = 2\, a \, x[/tex], where:
[tex]v[/tex] is the final velocity of the object,[tex]u[/tex] is the initial velocity of the object, [tex]a[/tex] is the acceleration (should be constant,) and[tex]x[/tex] is the displacement of the object while its velocity changed from [tex]v[/tex] to [tex]u[/tex].Assume that going downwards corresponds to a positive displacement. For this question:
[tex]v[/tex] needs to be found.[tex]u = 0[/tex] because the rock is released from rest.[tex]a = g = 9.8 \; \rm m\cdot s^{-2}[/tex].[tex]x = 40\; \rm m[/tex].Solve this equation for [tex]v[/tex]:
[tex]\displaystyle v = \sqrt{2\, a\, x + u^2} = \sqrt{2\times 9.8 \times 40} = 28\; \rm m \cdot s^{-1}[/tex].
In other words, the rock reached a velocity of [tex]28\; \rm m\cdot s^{-1}[/tex] (downwards) right before it hits the ground.
ExplanationLet [tex]v[/tex] be the velocity (in [tex]\rm m \cdot s^{-1}[/tex]) of this rock right before it hits the ground. Under the assumptions of this question, it would take a time of [tex]t = (v / 9.8)[/tex] seconds for this rock to reach that velocity if it started from rest and accelerated at [tex]9.8\; \rm m \cdot s^{-2}[/tex].
Note that under these assumptions, the acceleration of this rock is constant. Therefore, the average velocity of this rock would be exactly one-half the sum of the initial and final velocity. In other words, if [tex]u[/tex] denotes the initial velocity of this rock, the average velocity of this rock during the fall would be:
[tex]\displaystyle \frac{u + v}{2}[/tex].
On the other hand, [tex]u = 0[/tex] because this stone is released from rest. Therefore, the average velocity of this rock during the fall would be exactly [tex](v / 2)[/tex].
The displacement of an object over a period of time is equal to the length of that period times the average velocity over that period. For this rock, the length of this fall would be [tex]t = (v / 9.8)[/tex], while the average velocity over that period would be [tex](v / 2)[/tex]. Therefore, the displacement (in meters) of the rock during the entire fall would be:
[tex]\displaystyle \left(\frac{v}{2}\right) \cdot \left(\frac{v}{9.8}\right) = \frac{v^2}{19.6}[/tex].
That displacement should be equal to the change in the height of the rock, [tex]40\; \rm m[/tex]:
[tex]\displaystyle \frac{v^2}{19.6} = 40[/tex].
Solve for [tex]v[/tex]:
[tex]v = 28\; \rm m \cdot s^{-1}[/tex].
Once again, the speed of the rock would be [tex]28\;\rm m \cdot s^{-1}[/tex] right before it hits the ground.
A paper airplane is thrown horizontally with a velocity of 20 mph. The plane is in the air for 7.63 s before coming to a standstill on the ground. What is the acceleration of the plane?
Answers
Answer:
-1.17 m/s²
Explanation:
Given:
v₀ = 20 mph = 8.94 m/s
v = 0 m/s
t = 7.63 s
Find: a
v = at + v₀
0 m/s = a (7.63 s) + 8.94 m/s
a = -1.17 m/s²
The acceleration of the plane will be:
"-1.17 m/s²".
Acceleration and VelocityAccording to the question,
Velocity, v₀ = 20 mph or,
= 8.94 m/s
and,
v = 0 m/s
Time, t = 7.63 s
We know the relation,
→ v = at + v₀
By substituting the values,
0 = a × 7.63 + 8.94
7.63a = - 8.94
a = -[tex]\frac{8.94}{7.63}[/tex]
= - 1.17 m/s²
Thus the response above is correct.
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An 85 kg skydiver is falling through the air at a constant speed of 195 km h-1. At what rate does air resistance remove energy from the skydiver?
Answers
Answer:
46041J
Explanation:
Using Energy lost= mgh
Changing to standard its we have
= 195*1000/3600=54.2m/s
So = 85*54.2*10= 46041J
Answer:
45167.15 J/s
Explanation:
mass of the man = 85 kg
The man's speed = 195 km/h = 195 x 1000/3600 = 54.167 m/s
The man's weight = mg
where
m is the mass
g is acceleration due to gravity = 9.81 m/s^2
weight = 85 x 9.81 = 833.85 N
The rate at which energy is removed from the man = speed x weight
==> 54.167 x 833.85 = 45167.15 J/s
Two 100kg bumper cars are moving towards eachother in oppisite directions. Car A is moving at 8 m/s and Car B at -10 m/s when they collide head on. If the resulting velocity of Car B after the collision is 8 m/s, what is the velocity of Car A after the collision
Answers
Answer:
[tex]-10 m/s[/tex]
Explanation:
When two cars collide then the momentum of two cars will remains conserved
Mass of two cars = 100 kg Speed of car A = 8 m/s Speed of car B = - 10 m/s After collision the speed of car B = +8 m/sBy momentum conservation equation
[tex]m1v1i+m2v2i=m1v1f + m2v2f[/tex]
[tex](100)(8)+(100)(-10)=(100v)+(100)(8)\\ v=-10 m/s[/tex]
A ball with a mass of 3.7 kg is thrown downward with an initial velocity of 8 m/s from a high building. How fast will it be moving after 3 seconds?
Answers
Answer:
v=37.4 m/s
Explanation:
It is given that,
Mass of a ball, m = 3.7 kg
Initial velocity of the ball is u = 8 m/s
We need to find its velocity after 3 seconds. It is moving downwards. The equation of motion is this case is
v=u+gt
[tex]v=8+9.8\times 3\\\\v=37.4\ m/s[/tex]
So, the velocity of the ball after 3 seconds is 37.4 m/s.
Two pounds of water vapor at 30 psia fill the 4-ft3 left chamber of a partitioned system. The right chamber has twice the volume of the left and is initially evacuated. Determine the pressure of the water after the partition has been removed and enough heat has been transferred so that the temperature of the water is 40oF.
Answers
Answer:
3.38atm
Explanation:
Using data from the steam table we have that
Moles of water vapour = 907.19 / 18
= 50.4 moles
So
p1 = 30 psi = 30 x 0.68 = 2.04 atm
v1 = 4ft³= 113.2 L
Then from
PV= nRT
Then to find T we use
T1 = p1 V1 / n R
= 2.04 x 113.2 / 50.4 x 0.0821
= 55.8 K
Then to find volume two
v2 = 2v1 + v1
So
3 v1 = 339.6 K
The pressure two we use
P2 = n R T2 / V2
= 50.4 x 0.0821 x 277.6 / 339.6
So we have
= 3.38 atm =
A -5.40nC point charge is on the x axis at x = 1.25m . A second point charge Q is on the x axis at -0.625m.
A) What must be the charge Q for the resultant electric field at the origin to be 50.0N/C in the +x direction?
B) What must be the charge Q for the resultant electric field at the origin to be 50.0N/C in the -x direction?
Answers
Answe
a) Q = 0.820 10⁻⁹ C , b) Q = -3.52 10⁻⁹ C
Explanation:
The electric field is given by the formula
E = k q / r²
where E is a vector quantity, so it must be added as a vector
E_total = E₁ + E₂
let's look for the two electric fields
E₁ = k q₁ / r₁²
E₁ = 9 10⁹ 5.4 10⁻⁹ / 1.25²
E₁ = 31.10 N / C
E2 = k Q / r₂²
E2 = 9 10⁹ Q / 0.625²
E2 = 23.04 10⁹ Q N / C (1)
now we can solve the two cases presented
a) The total field is
E_total = 50.0 N / C towards + x
since the test charge is positive the electric field E1 points to the right in the direction of the + x axis, so the equation is
E_total = E1 + E₂
E₂ = E_toal - E₁
E₂ = 50.0 -31.10
E2 = 18.9 N /C
With the value of the electric field we can calculate the charge (Q) using equation 1
E₂ = 23.04 10⁹ Q
Q = E₂ / 23.04 10⁹
Q = 18.9 / 23.04 10⁹
Q = 0.820 10⁻⁹ C
the charge on Q is positive
b) E_total = -50.0 N / C
E_total = E₁ + E₂
E₂ = E_total - E₁
E2 = -50.0 - 31.10
E2 = -81.10 N /C
we calculate the charge
Q = E2 / 23.04 10⁹
Q = -81.1 / 23.04 10⁹
Q = -3.52 10⁻⁹ C
for this case the charge is negative
What happens to the temperature as altitude increases in the exosphere? Does it increase or decrease the higher it goes?
Answers
Answer:
it gets colder the higher you go
A ski lift has a one-way length of 1 km and a vertical rise of 200 m. The chairs are spaced 20 m apart, and each chair can seat three people. The lift is operating at a steady speed of 10 km/h. Neglecting friction and air drag, and assuming that the average mass of each loaded chair is 250 kg, determine the power required to operate this ski lift. Also, estimate the power required to accelerate this ski lift in 17 s to its operating speed when it is first turned on.
Answers
Answer:
Explanation:
The question states that the chairs are spaced 20 m apart through a length of 1 km, or say, 1000 m.
It also does say that each chair weighs 250 kg, and as such the load is
M = 50 * 250
M = 12500.
Taking into consideration, the initial and final heights, we have
h1 = 0, h2 = 200 m
The work needed to raise the chairs,
W = mgh, where h = h2 - h1
W = 12500 * 9.81 * (200 - 0)
W = 2.54*10^7 J
The work is done at a rate of 10 km/h, and at a distance of 1 km, time taken would be
t = 1/10 = 0.1 h or say, 360 s
The power needed thus, is
P = W/t
P = 2.54*10^7 / 360
P = 68125 W, or 68 kW
Initial velocity, u = 0 m/s
Final velocity, v = 10 km/h = 2.78 m/s
Startup time, t is 17 s
Acceleration during the startup then is
a = (v - u)/t
a = 2.78/17
a = 0.163 m/s²
The power needed for the acceleration is
P = ½m [(v² - u²)/t]
P = ½ * 12500 * [2.78²/17]
P = 6250 * 0.455
P = 2844 W
A satellite dish has the shape of a parabola when viewed from the side. The dish is inches wide and inches deep. How far is the receiver from the bottom of the dish if the receiver is placed at the focus
Answers
Complete question is;
A satellite dish has the shape of a parabola when viewed from the side. The dish is 60 inches wide and 45 inches deep. How far is the receiver from the bottom of the dish if the receiver is placed at the focus?
Answer:
the receiver should be put 40 inches from the bottom of the dish on the concave side of the dish
Explanation:
The base of the dish would simply be the vertex of parabola.
Since we want to find how far the receiver is from the bottom, the place where we'll place the receiver is simply the focus of the parabola.
Now, for example, if this is a parabola that opens upward and has it's vertex at the origin, then half of the diameter at a height of 45 inches gives the two points (60, 22.5) and (-60, 22.5)
Standard form equation of parabola with vertex at origin and pointing upwards is given by;
x² = 4ay
Plugging in the values of x and y gives;
60² = 4a(22.5)
3600/90 = a
a = 40 inches
Thus, the receiver should be put 40 inches from the vertex on the concave side of the dish
A 10 kg object is dropped from rest. a. How far will it drop in 2s? b. How long will it take a 5 kg object to drop the same distance?
Answers
Answer:
Explanation:
Using the equation of motion S = ut + 1/2at² to get the height of drop where
u is the initial velocity of the object = 0m/s
a is the acceleration due to gravity = +9.81m/s² (downwards motion of object)
t is the time it takes the object to drop = 2secs
Substituting the given parameters into the formula to get the height of drop S, we will have;
S = 0(2)+1/2(9.81)(2)²
S = 0+9.81*2
S = 19.62m
Hence the object will drop at a distance of 19.62m
b.) To determine the time it will take a 5 kg object to drop the same distance, we will use the same formula S = ut+1/2at²
Since the 5kg object also drops at the same distance, then S = 19.62m
Substituting this values into the equation we will have;
19.62 = 0(t) + 1/2(9.81)t²
19.62 = 4.905t²
t² = 19.62/4.905
t² = 4
t =2secs
This shows that it will take 5kg object 2secs to fall from the same distance. This means that no matter the mass of the object, it will take them the same time to fall at the same distance because they are all falling under the same influence of gravity.
(7) A 2500 lbm car moving at 25 mi/hr is accelerated at a constant rate of 15 ft/s2 up to a speed of 50 mi/hr. What is the force and total time required?
Answers
Answer:
The force is [tex]F = 1164.6\ lbf[/tex]
The time is [tex]\Delta t = 2.44 \ s[/tex]
Explanation:
From the question we are told that
The mass of the car is [tex]m = 2500 \ lbm[/tex]
The initial velocity of the car is [tex]u = 25 \ mi/hr[/tex]
The final velocity of the car is [tex]v = 50 \ mi/hr[/tex]
The acceleration is [tex]a = 15 ft/s^2 = \frac{15 * 3600^2}{ 5280} = 36818.2 \ mi/h^2[/tex]
Generally the acceleration is mathematically represented as
[tex]a = \frac{v-u}{\Delta t}[/tex]
=> [tex]36818.2 = \frac{50 - 25 }{ \Delta t}[/tex]
=> [tex]t = 0.000679 \ hr[/tex]
converting to seconds
[tex]\Delta t = 0.0000679 * 3600[/tex]
=> [tex]\Delta t = 2.44 \ s[/tex]
Generally the force is mathematically represented as
[tex]F = m * a[/tex]
=> [tex]F = 2500 * 15[/tex]
=> [tex]F = 37500 \ \frac{lbm * ft}{s^2}[/tex]
Now converting to foot-pound-second we have
[tex]F = \frac{37500}{32.2}[/tex]
=> [tex]F = 1164.6\ lbf[/tex]
Finally, consider the expression (6.67 x 10^-11)(5.97 x 10^24)/(6.38 x 10^6)^2 Determine the values of a and k when the value of this expression is written in scientific rotation. Enter a and k, separated by commas.
Answers
Explanation:
We need to find the value of following expression :
[tex]\dfrac{(6.67 \times 10^{-11})(5.97 \times 10^{24})}{(6.38 \times 10^6)^2}[/tex]
Firstly, solving the numerator of the above expression :
[tex]=\dfrac{39.8199\times 10^{-11+24}}{40.7044\times 10^{12}}\\\\=\dfrac{39.8199\times 10^{13}}{40.7044\times 10^{12}}\\\\=9.7827[/tex]
Rounding off the result = 9.78
In scientific notation : [tex]9.78\times 10^0[/tex]
The value of a = 9.78 and k = 0.