Answer:
It will take 15 seconds
With all his gear, Neil Armstrong weighed 360 pounds on Earth. When he landed on the Moon, he weighed 60 pounds. Why?
Answer:C: The gravity on the moon is less than the gravity on Earth.
Explanation:
Calculate the total distance traveled (in m)
Answer:
Distance is 13m and Displacement is 9m
Explanation:
A sled starts from rest,
and slides 10.0 m down a 28.0°
frictionless hill. What is its
acceleration?
(Unit = m/s2)
Answer:
The acceleration of the sled is [tex]4.6\ m/s^2[/tex].
Explanation:
It is given that,
Initial speed of sled is 0 because it was at rest.
It is placed at an angle of 28° on a frictionless hill.
We need to find the acceleration of the sled. It is placed at an incline. It means that the acceleration of the sled is given by :
[tex]a=g\sin\theta\\\\a=9.8\times \sin(28)\\\\a=4.6\ m/s^2[/tex]
So, the acceleration of the sled is [tex]4.6\ m/s^2[/tex].
Scientists might use a diagram to model the water cycle. What are two
benefits of this model?
Answer:
B, and D
Explanation:
The two benefits of this model is it specify a process that is very complex. It can represent changes that occur very slowly. Thus option B and D is correct.
What is water cycle?The water cycle is defined as a cycle of events that involves precipitation as rain and snow, drainage in streams and rivers, and return to the atmosphere by evaporation and transpiration. Water moves through this cycle between the earth's oceans, atmosphere, and land.
It can also be defined as the route all water takes as it travels around Earth in various conditions.
There are basically six stages of water cycle.
EvaporationSublimationCondensationPrecipitationInfiltrationRunoffThus, the two benefits of this model is it specify a process that is very complex. It can represent changes that occur very slowly. Thus option B and D is correct.
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A fireworks shell is accelerated from rest to a velocity of 68.0 m/s over a distance of 0.230 m.
A) Calculate the acceleration.
B) How long did the acceleration last?
Answer:
(A) 10052.2 m/s²
(B) 0.00678 seconds
Explanation:
From the question,
(A) Applying
V² = U²+2as..................... Equation 1
Where V = Final velocity, U = Initial velocity, a = acceleration due to gravity, s = distance.
make a the subject of the equation
a = (V²-U²)/2s........................ Equation 2
Given: U = 0 m/s( from rest), V = 68 m/s, s = 0.230 m
Substitute these values into equation 2
a = (68²-0²)/(2×0.230)
a = 10052.2 m/s²
(B) Using,
a = (V-U)/t......................... Equation 3
Where t= time.
make t the subject of the equation
t = (V-U)/a......................... Equation 4
Given: V = 68 m/s, U = 0 m/s, a = 10052.2
Substitute into equation 4
t = (68-0)/10052.2
t = 0.00678 seconds
The results of a dart game were precise but not accurate.The accepted value of the game was the center of the dartboard.Which correctly describes the results.
A. All of the darts hit the center of the board
B.Some of the darts hit the center of the board
C.The darts hit the same general area of the board
D.The darts hit very different areas of the board
Answer:
The darts hit the same general area of the board.
Answer: C. The darts hit the same general area of the board.
Explanation: The results of a dart game were precise but not accurate. The accepted value of the game was the center of the dartboard.
A car accelerates at a rate of 5ft/s/s for a time of 9 seconds. How far does the car go?
The car will move in a speed of 45 meter per second
Answer: The car will move in a speed of 45 meter per second
Explanation:
26.0 g of copper pellets are removed from a 300∘C oven and immediately dropped into 120 mL of water at 21.0 ∘C in an insulated cup.What will the new water temperature be?
Answer:
The new water temperature is 26.4 °C
Explanation:
Given;
mass of copper, [tex]M_{cu}[/tex] = 26 g = 0.026 kg
temperature of copper, t = 300 °C
volume of water, V = 120 mL = 0.12 L
temperature of water, t = 21 °C
density of water, ρ = 1 kg/L
mass of water = density x volume
mass of water = (1 kg/L) x 0.12 L = 0.12 kg
heat lost by copper = heat gained by water
Both copper and water reach final temperature, T
Heat gained by water, [tex]Q_w[/tex] = [tex]m_w[/tex]cΔθ = [tex]m_w C(T - t)[/tex]
[tex]Q_w = m_w C(T - t)\\\\Q_w = 0.12*4200(T-21)\\\\Q_w = 504(T-21)[/tex]
Heat lost by copper is given by;
[tex]Q_{cu} = m_{cu}C(300-T)\\\\Q_{cu} = 0.026*385(300-T)\\\\Q_{cu} = 10.01(300 - T)[/tex]
[tex]Q_{cu} = Q_w[/tex]
504(T- 21) = 10.01(300 - T)
504 T - 10584 = 3003 - 10.01 T
504 T + 10.01 T= 3003 + 10584
514.01 T = 13587
T = (13587) / 514.01
T = 26.4 °C
Therefore, the new water temperature is 26.4 °C
The new water temperature using the given parameters is;
26.49°C
We are given;
Mass of copper; m_cu = 26 g = 0.026 kg
Temperature; T_cu = 300 °C
Volume of water; V = 120 mL = 0.12 L
Temperature of water, T_w = 21 °C
Density of water; ρ = 1 kg/L
Let us find the mass of water from the formula;
m_w = ρ × V
m_w = 1 × 0.12
m_w = 0.12 kg
Now, from the principle of conservation of energy, we can say that the total heat lost by a hot body is equal to the total heat gained by a cold body.
The hot body here is copper while the cold body is water. Thus;
Heat lost by copper = Heat gained by water
Formula for heat lost by copper is;
Q_cu = m_cu * c_cu * (T_f - T_cu)
Formula for heat gained by water is;
Q_w = m_w * c_w * (T_f - T_w)
Where;
T_f is final temperature reached by copper and water
c_cu is specific heat capacity of copper = 385 J/kg.°C
c_w is specific heat capacity of water = 4200 J/kg.°C
Thus;
Q_cu = 0.026 × 385 × (300 - T_f)
Q_cu = 3030 - 10.01T_f
similarly;
Q_w = 0.12 × 4200 × (T_f - 21)
Q_w = 504T_f - 10584
Thus;
3030 - 10.01T_f = 504T_f - 10584
3030 + 10584 = 504T_f + 10.01T_f
13614 = 514.01T_f
T_f = 13614/514.01
T_f = 26.49°C
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The average speed during any time interval is equal to the total distance of travel divided by the total time. Let d represent the distance between A and B. Let t1 be the time for which she has the higher speed of 5.15 m/s = d/t1. Let t2 represent the longer time for the return trip at 2.80 m/s =d/t2. Then the times are t1 = d/5.15 5.15 m/s and t2 = d/2.80 2.8 m/s. The average speed vavg is given by the following equation. vavg = Total distance/Total time = d + d/t1 + t2.
Answer:
Average speed = 3.63 m/s
Explanation:
The average speed during any time interval is equal to the total distance travelled divided by the total time.
That is,
Average speed = distance/ time
Let d represent the distance between A and B.
Let t1 be the time for which she has the higher speed of 5.15 m/s. Therefore,
5.15 = d/t1.
Make d the subject of formula
d = 5.15t1
Let t2 represent the longer time for the return trip at 2.80 m/s . That is,
2.80 = d/t2.
Then the times are t1 = d/5.15 5 and
t2 = d/2.80.
The average speed vavg is given by the following equation.
avg speed = Total distance/Total time
Avg speed = d + d/t1 + t2
Where
Total distance = 2d
Total time = t1 + t2
Total time = d/5.15 + d/2.80
Total time = (2.8d + 5.15d)/14.42
Total time = 7.95d/14.42
Total time = 0.55d
Substitute total distance and time into the formula above.
Avg speed = 2d / 0.55d
Avg Speed = 3.63 m/s
A man standing on the top of a hill and sees a flagpole he knows is 45 feet high. The angle of depression to the bottom of the pole is 12 degrees, and the angle of elevation to the top of the pole is 16 degrees. Find his distance from the pole
Answer:
160.44 feetExplanation:
check the attachment for the diagram of the set up for proper clarification.
From the diagram, the man's distance from the pole is labelled AB.
To get AB =, we need to get side CB first which is equal to ED i.e ED = CB.
From ΔCDE, opposite side = 45 feet and the adjacent is ED.
USing TOA, according to trig function SOH, CAH, TOA;
tan 16° = opp/adj = BD/ED
tan 16° = 45/ED
ED = 45/tan 16°
ED = 156.93 feet
Since ED = CB, hence CB = 156.93 feet
From ΔABC, adj = CB and hypotenuse = AB. According to CAH;
cos 12° = adj/hyp = CB/AB
cos 12° = 156.93/AB
AB = 156.93/cos 12°
AB = 160.44 feet
Hence the man's distance from the pole is 160.44 feet
Convert 7 (gcm^2)/(min^2) into a value in standard S.I. units. Be sure to use scientific notation if necessary. You do not need to answer units.
The required value is required in SI units.
The required answer is [tex]1.94\times10^{-10}\ \text{kg m}^2/\text{s}^2[/tex]
SI unitsThe SI unit of mass, length and time is kg, m and s respectively.
In order to convert one unit into another it has to be multiplied or divided by the conversion factors.
A definite magnitude which has some quantity which is defined by convention or law is called a unit.
The conversion factors are
[tex]1\ \text{g}=10^{-3}\ \text{kg}[/tex]
[tex]1\ \text{cm}=10^{-2}\ \text{m}[/tex]
[tex]1\ \text{cm}^2=10^{-4}\ \text{m}^2[/tex]
1 min = 60 s
[tex]1\ \text{min}^2=60\times60\ \text{s}^2[/tex]
So,
[tex]7\ \text{g cm}^2/\text{min}^2=7\times \dfrac{10^{-3}\times 10^{-4}}{60\times 60}\\ =1.94\times10^{-10}\ \text{kg m}^2/\text{s}^2[/tex]
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What's the name of the theory that explains things better than newton's laws ?
Which lists the elements in order from most conductive to least conductive?
potassium (K), selenium (Se), germanium (Ge)
germanium (Ge), potassium (K), selenium (Se)
selenium (Se), germanium (Ge), potassium (K)
potassium (K), germanium (Ge), selenium (Se)
Answer:
Answer: potassium (K) germanium (Ge) selenium (Se)
Explanation:
I just took the test, the farther to the right on the periodic table you go the less conductive it gets.
Answer:
Answer: potassium (K) germanium (Ge) selenium (Se)
Explanation:
The most conductive to least conductive is from left to right
If a cheetah could maintain its top speed of 120 km/h for 20 minutes, how far would it run?
The acceleration of a piston is given by : rw^2 cos(wt-pi/4) find amplitude.
Given :
The acceleration of a piston is given by :
[tex]a=r\omega^2cos\ (\omega t-\dfrac{\pi}{4})[/tex]
To Find :
The amplitude of the acceleration of piston .
Solution :
Now , acceleration is :
[tex]a=r\omega^2cos\ (\omega t-\dfrac{\pi}{4})[/tex]
Now , amplitude is maximum of any function .
So , maximum value of [tex]cos\ \theta[/tex] is equal to 1 .
Therefore , amplitude of the acceleration of piston is :
[tex]A=r\omega^2[/tex]
Hence , this is the required solution .
Question 1
When you stand on a floor the electrons surround your shoes and the floor. What is the force that prevents you from falling through a floor?
a) atomic direction
b) potential energy
c) atomic speed
d) electrical repulsion
Answer:
d) electrical repulsion
Explanation:
When you stand on a floor, the two different bodies i.e your shoes and the floor on which you stand, are in a very close contact, the electrons that surround your shoes and that of the floor exert a repulsive force on each other, also known as Coulomb's force of repulsion or electrical repulsive force.
This electrical repulsive force prevents you from falling through the floor.
A car traveling due east at 20 m/s reverses its direction over a period of 10 seconds so that it is now traveling due west at 20 m/s. What is the direction of the car's average acceleration over this period?a- The car's average acceleration points due west.b- The car's average acceleration is zero.c- The car's average acceleration points due east.d- The direction of the car's average acceleration cannot be determined from the given information.
Answer:
The car's average acceleration points due west.
Explanation:
Resolving the acceleration will give a resultant due west
A velocity time graph shows how____ changes over time
Answer: A velocity-time graph shows how velocity changes over time
Explanation:
Answer:velocity
Explanation:
What is correct regarding an ideal isotropic antenna? a. An ideal isotropic antenna is a highly efficient antenna used extensively in today’s communication systems b. An ideal isotropic antenna is a specialized antenna used to direct EM signal energy towards a specific direction c. An ideal isotropic antenna is a theoretical antenna that does not exist in practice, but is useful in explaining power density and unguided EM signal attenuation d. None of the above are correct statements
Answer:
C An ideal isotropic antenna is a theoretical antenna that does not exist in practice, but is useful in explaining power density and unguided EM signal attenuation
Explanation:
Because practically speaking there is no ideal isotropic antenna it is only imaginary radiating in all directions and use as an arbitrary point for antenna gain
Why is the endocrine system important in psychology
the endocrine system releases hormones, and hormones impact the systems that cause changes in our behavior, including changes in what we might call biological motivations.
Examine the resistor network. The answers to each of the questions can be either "none" or the numbers of one or more resistors. Which resistors are connected in parallel with resistor 2? Which resistors are connected in parallel with resistor 9? Which resistors are connected in parallel with resistor 11? Which resistors are not connected in parallel with any other resistors?
Answer:
parallel with 2: 1parallel with 9: 5, 6, 8parallel with 11: 4, 7not in a parallel branch: 3Explanation:
The attachments show the labeling of the network nodes. We found it convenient to do this so we could identify the specific nodes any given branch was attached to. Then the connections for each resistor were identified, and the list sorted to make it easier to see parallel connections.
Resistors 1 and 2 are in parallel
Resistors 5, 6, 8, 9 are in parallel
Resistors 4, 7, 11 are in parallel
No resistor has a parallel connection to resistor 3.
Explain the difference between the justice system laws and scientific laws
A pendulum built from a steel sphere with radius r cm 5 and density stl kg m S 3 7800 is attached to an aluminum bar with length l m 1 thickness t cm 0 8. and width w cm 4 and density . al kg m S 3 2820 a. Calculate the mass moment of inertia of the pendulum about its center of mass, . cm I b. Calculate the mass moment of inertia of the pendulum about its pivot point, o
Answer:
1) I_ pendulum = 2.3159 kg m² , 2) I_pendulum = 24.683 kg m²
Explanation:
In this exercise we are asked to calculate the moment of inertia of a physical pendulum, let's start by calculating the center of mass of each elements of the pendulum and then the center of mass of the pendulum
Sphere
They indicate the density of the sphere roh = 37800 kg / m³ and its radius
r = 5 cm = 0.05 m
we use the definition of density
ρ = M / V
M = ρ V
the volume of a sphere is
V = 4/3 π r³
we substitute
M = ρ 4/3 π r³
we calculate
M = 37800 4/3 π 0.05³
M = 19,792 kg
Bar
the density is ρ = 32800 kg / m³ and its dimensions are 1 m,
0.8 cm = 0.0008 m and 4cm = 0.04 m
The volume of the bar is
V = l w h
m = ρ l w h
we calculate
m = 32800 (1 0.008 0.04)
m = 10.496 kg
Now we can calculate the center of mass of the pendulum, we use that the center of mass of the sphere is its geometric center, that is, its center and the center of mass of the bar is where the diagonals intersect, in this case it is a very bar. long and narrow, whereby the center of mass is about half the length. It's mass scepter of the pendulum is
r_cm = 1 / M (M r_sphere + m r_bar)
M = 19,792 + 10,496 = 30,288 kg
r_cm = 1 / 30,288 (10,496 0.5 + 19.792 (1 + 0.05))
r_cm = 1 / 30,288 (5,248 + 20,7816)
r_cm = 0.859 m
This is the center of mass of the pendulum.
1) Now we can calculate the moment of inertia with respect to this center of mass, for this we can use the theorem of parallel axes and that the moments of inertia of the bodies are:
Sphere I = 2/5 M r2
Bar I = 1/12 m L2
parallel axes theorem
I = I_cm + m D²
where m is the mass of the body and D is the distance from the body to the axis of rotation
Sphere
m = 19,792 ka
the distance D is
D = 1.05 -0.85
D = 0.2 m
we calculate
I_sphere = 2/5 19.792 0.05 2 + 19.792 0.2 2 = 0.019792 +0.79168
I_sphere = 0.811472 kg m²
Bar
m = 10.496 kg
distance D
D = 0.85 - 0.5
D = 0.35 m
I_bar = 1/12 10.496 0.5 2 + 10.496 0.35 2 = 0.2186 + 1.28576
I_bar = 1.5044 kg m²
The moment of inertia is a scalar quantity whereby the moment of inertia of the body is the sum of the moment of the parts
I_pendulum = I_sphere + I_bar
I_pendulum = 0.811472 +1.5044
I_ pendulum = 2.3159 kg m²
this is the moment of inertia of the pendulum with respect to its center of mass located at r = 0.85 m
2) The moment is requested with respect to the pivot point at r = 0 m
Sphere
D = 1.05 m
I_sphere = 2/5 M r2 + M D2
I_sphere = 2/5 19.792 0.05 2 + 19.792 1.05 2 = 0.019792 +21.82
I_sphere = 21.84 kg m²
Bar
D = 0.5 m
I_bar = 1/12 10.496 0.5 2 + 10.496 0.5 2 = 0.21866 + 2.624
I_bar = 2,84266 kg m 2
The pendulum moment of inertia is
I_pendulum = 21.84 +2.843
I_pendulum = 24.683 kg m²
This moment of inertia is about the turning point at r = 0 m
The mass of a high speed train is 4.5×105 kg, and it is traveling forward at a velocity of 8.3×101 m/s. Given that momentum equals mass times velocity, determine the values of m and n when the momentum of the train (in kg⋅m/s) is written in scientific notation.
Answer:
The value of m = 3.735 and the value of n = 7.
Explanation:
The equation for the momentum of the train is,
P = mv
Here, m is the mass of the train and v is the speed of the train.
Substitute [tex]4.5 \times {10^5}{\rm{ kg}} , 8.3 \times {10^1}{\rm{ m/s}}[/tex] for m and v respectively in above equation.
[tex]\begin{array}{c}\\P = \left( {4.5 \times {{10}^5}{\rm{ kg}}} \right)\left( {8.3 \times {{10}^1}{\rm{ m/s}}} \right)\\\\ = 37.35 \times {10^6}{\rm{ kg}} \cdot {\rm{m/s}}\left( {\frac{{{{10}^1}{\rm{kg}} \cdot {\rm{m/s}}}}{{10\,{\rm{kg}} \cdot {\rm{m/s}}}}} \right)\\\\ = 3.735 \times {10^7}{\rm{ kg}} \cdot {\rm{m/s}}\\\end{array}[/tex]
According to the scientific notation, here the value of m is 3.735 and the value of n is 7 in the final answer of the momentum.
The value of m = 3.735 and the value of n = 7.
.A cart rolling down an incline for 5.0 seconds has an acceleration of 1.6 m/s2. If the cart has a beginning speed of 2.0 m/s, and its final velocity of 10 m/s, what was the car's displacement?
Use the formula,
[tex]\Delta x=v_it+\dfrac12at^2[/tex]
where [tex]\Delta x[/tex] is the cart's displacement (from the origin), [tex]v_i[/tex] is its initial speed, [tex]a[/tex] is its acceleration, and [tex]t[/tex] is time.
[tex]\Delta x=\left(2.0\dfrac{\rm m}{\rm s}\right)(5.0\,\mathrm s)+\dfrac12\left(1.6\dfrac{\rm m}{\mathrm s^2}\right)(5.0\,\mathrm s)^2[/tex]
[tex]\implies\boxed{\Delta x=30\,\mathrm m}[/tex]
Alternatively, since acceleration is constant, we have
[tex]\dfrac{v_f+v_i}2=\dfrac{\Delta x}t[/tex]
That is, we have these two equivalent expressions for average velocity, where [tex]v_f[/tex] is the cart's final velocity. Solve for [tex]\Delta x[/tex]:
[tex]\dfrac{10\frac{\rm m}{\rm s}+2.0\frac{\rm m}{\rm s}}2=\dfrac{\Delta x}{5.0\,\mathrm s}[/tex]
[tex]\implies\boxed{\Delta x=30\,\mathrm m}[/tex]
What is the difiniton of absolute value in science
Answer: The absolute value of a number is symbolized by two vertical lines as 3 or -3
Please pick me as brainliest please
Answer: the magnetic of a real number without regard to its sign
Explanation: In mathematics, the absolute value or modulus of a real number x, denoted |x| or, is the non-negative value of x without regard to its sign. Namely, |x| = x if x is positive, and |x| = −x if x is negative, and |0| = 0. For example, the absolute value of 3 is 3, and the absolute value of −3 is also 3
The earth is about 93,000,000 miles from the sun and traverses its orbit, which is nearly circular, every 365.25 days. What is the linear velocity of the earth in its orbit, in miles per hour? \
Answer:
10609.17miles per hr is the linear velocity
Explanation:
To arrive at Ans in miles per hr we first convert 365.25 days to hours
Which is 365.25 x24hrs per day= 8766hrs
So velocity= distance/time
So
93000000/8766= 10609.17miles per hr
3. According to the Guinness Book of Records the heaviest baby ever born weighed 29 lbs 4 oz. (29.25 lbs).
What was the baby's mass in kg? (Historical Note: The birth occurred in Effingham IL in 1939 and due to
respiratory problems the baby died two hours later. The heaviest babies to survive weighed 22.5 lbs and were born
in 1955 and 1982.)
Answer: 13.2678 kg
Explanation: Well, 13.2678 kg is 29.25 in mass and kg.
The deck of a bridge is suspended 235 feet above a river. If a pebble falls off the side of the bridge, the height, in feet, of the pebble above the water surface after t seconds is given by y = 235 − 16t^2.
Required:
a. Find the average velocity (in ft/s) of the pebble for the time period beginning when t = 2 and lasting the following amount of time.
1. 0.1 sec:________
2. 0.05 sec:_______
3. 0.01 sec:_______
b. Estimate the instantaneous velocity (in ft/s) of the pebble after 3 seconds. ft/s.
Answer:
(a) 1, average velocity = -65.6 m/s
2, average velocity = -64.8 m/s
3, average velocity = -64.16 m/s
(b) The instantaneous velocity is -96 m/s
Explanation:
(a)
Average velocity is given by;
[tex]y(t_2,t_1) = \frac{y(t_2) - y(t_1)}{t_2-t_1}[/tex]
(1)
[tex]y(2.1,2) = \frac{(235-16*2.1^2) - (235-16*2^2)}{2.1-2}\\\\ y(2.1,2) = -65.6 \ m/s[/tex]
(2)
[tex]y(2.05,2) = \frac{(235-16*2.05^2) - (235-16*2^2)}{2.05-2}\\\\ y(2.05,2) = -64.8 \ m/s[/tex]
(3)
[tex]y(2.01,2) = \frac{(235-16*2.01^2) - (235-16*2^2)}{2.01-2}\\\\ y(2.01,2) = -64.16 \ m/s[/tex]
b. y = 235 - 16t²
The instantaneous velocity is given by;
v = dy /dt
dy / dt = -32t
when t = 3 s
v = -32(3)
v = -96 m/s
(a)The average velocity for the 0.1 sec,0.05 sec,0.01 sec will be -65.6 m/s, -64.8 m/s and -64.16 m/s respectively.
What is the average velocity?The total displacement traveled by an object divided by the total time taken is the average velocity.
Average velocity is given by;
[tex]\rm y(t_2,t_1)=\frac{y(t_2)-y(t_1)}{t_2-t_1} \\\\[/tex]
The average velocity for case 1;
[tex]\rm y(2.1,2)=\frac{(235-16\times 2\times 1^2)-(235-16 \times 2^2)}{2.1-2} \\\\ \rm y(2.1,2)=\-65.6 \ m/sec[/tex]
The average velocity for case 2;
[tex]\rm y(2.015,2)=\frac{(235-16\times 2.05^2)-(235-16 \times 2^2)}{2.05-2} \\\\ \rm y(2.1,2)=\-64.8 \ m/sec[/tex]
The average velocity for case 3;
[tex]\rm y(2.01,2)=\frac{(235-16\times 2.01^2)-(235-16 \times 2^2)}{2.01-2} \\\\ \rm y(2.1,2)=\-64.16 \ m/sec[/tex]
Hence the average velocity for the 0.1 sec,0.05 sec,0.01 sec will be -65.6 m/s, -64.8 m/s and -64.16 m/s respectively.
(b) The instantaneous velocity will be -96 m/s.
The given equation in the problem is;
[tex]\rm y = 235 - 16t^2[/tex]
The instantaneous velocity is given as;
[tex]v = \frac{dy}{dt} \\\\ \frac{dy}{dt} = -32t\\\\ t = 3 s\\\\ v = -32\times 3\\\\ v = -96 m/s[/tex]
Hence the instantaneous velocity will be -96 m/s.
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Determine the slope of this graph from zero seconds to five seconds.
Explanation:
(m2-m1)/t
25-0/5
25/5
5m/s