The following data were collected during a short race between two friends. Velocity (m/s) 0 0.5 1 1.5 2 2 4 6 2 0 Time (s) 0 2 4 6 8 10 12 14 16 18 a) Describe the different sections of the graph. b) Determine the acceleration over the first eight seconds. c) Determine the maximum acceleration. d) Using the graph calculate the displacement: i) over the first eight seconds ii) the total race. e) Find the maximum velocity reached by the runner.

Answers

Answer 1

The characteristics of the kinematics allow to find the results for the questions about the movement of the body are:

a)  we have four sections;

0 to 8 s The body is accelerating. 8 to 10 s The body goes at a constant speed, the acceleration is zero. 10 to 14 Body accelerating. 14 to 18 Body slowing down.

b)  The acceleration is the first 8 s is:  a = 0.25 m / s²

c) The maximum acceleration is:    a = 1 m / s²

d) The displacement   is:  i) d₁ =  8m,     ii)  [tex]d_{total}[/tex]= 16 m

e) maximum speed  is:      v = 6 m / s

Kinematics studies the movement of bodies by finding relationships between the position, speed and acceleration of bodies.

        v = v₀ + a t

        y = v₀ t + ½ a t²

where v and v₀ is the current and initial velocity, respectively, a is the acceleration and t is time.

In many circumstances graphs are made for their analysis, in a graph of speed versus time when we have a horizontal line the speed is constant, the acceleration is zero and in the case of a slope there is an acceleration, we have two cases:

Positive slope the body is accelerating and the speed is increasing. Negative slope the body is stopping, the speed decreases.

Let's answer the different questions about the system.

a) in the attached we have a graph of the velocity versus time, each section corresponds to a change in the slope of the graph, we have four sections;

0 to 8 s The body is accelerating. 8 to 10 s The body goes at a constant speed, the acceleration is zero. 10 to 14 Body accelerating. 14 to 18 Body slowing down.

b) The acceleration is the first 8 s

          v = v₀ + a t

          [tex]a = \frac{v-v_o}{\Delta t}[/tex]  

          [tex]a = \frac{2-0}{8-0}[/tex]  

          a = 0.25 m / s²

c) The maximum acceleration is when the slope is maximum.

          [tex]a = \frac{6-2}{ 14-10}[/tex]  

          a = 1 m / s²

Therefore the acceleration is maximum in the section between 10 and 14 s

d) The total displacement is the sum of the displacements of each section.

         [tex]d_{total } = d_1 +d_2 + d_3 +d_4[/tex]  

We look for every displacement.

       d₁ = v₀ + ½ a₁ Δt²

       d₁ = 0 + ½ 0.25 8²

       d₁ = 8 m

In the second section the velocity is constant

         d₂ = v₂ Δt₂

         d₂ = 2 (10-8)

         d₂ = 4 m

The third section.

    d₃ = v₀ + ½ a t²

    d₃ = 2 + ½ 1 (14-10) ²

    d₃ = 10 m

The distance of the fourth section.

       

we look for acceleration

          a₄ = [tex]\frac{v-v_o}{\Delta t}[/tex]  

          a₄ = [tex]\frac{0-6}{18-14}[/tex]  

          a₄ = -1.5 m / s²

     

          d₄ = 6 + ½ (-1.5) (1814) ²

          d₄ = -6 m

The total displacement is;

          [tex]d_{total}[/tex] = 8 + 4 + 10 -6

          [tex]d_{total}[/tex] = 16 m

e) The maximum speed is the highest point in the graph of speed versus time that in the attachment we can see corresponds to

          v = 6 m / s

In conclusion using the characteristics of kinematics we can find the results for the questions about the motion of bodies are:

  a)  we have four sections;

0 to 8 s The body is accelerating. 8 to 10 s The body goes at a constant speed, the acceleration is zero. 10 to 14 Body accelerating. 14 to 18 Body slowing down.

b)  The acceleration is the first 8 s is:  a = 0.25 m / s²

c) The maximum acceleration is:    a = 1 m / s²

d) The displacement   is:  i) d₁ =  8m,     ii)  [tex]d_{total}[/tex]= 16 m

e) maximum speed  is:      v = 6 m / s

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The Following Data Were Collected During A Short Race Between Two Friends. Velocity (m/s) 0 0.5 1 1.5

Related Questions

What is number 10 for this?

Answers

Answer:

The wavelength of WFNX’s radio waves with the given speed and frequency is 2.95m.

Given data in the question;

Speed of wave;

Frequency of wave;

wavelength;

To determine the wavelength of the radio wave, we use the expression for the relations between wavelength, frequency and speed.

Where  is wavelength, f is frequency and c is the speed.

We substitute our given values into the equation

Therefore, the wavelength of WFNX’s radio waves with the given speed and frequency is 2.95m.

Explanation:

can somebody explain it to me please?​

Answers

Range be R and height be h

[tex]\boxed{\sf R=\dfrac{u^2sin2\theta}{g}}[/tex]

[tex]\boxed{\sf h=\dfrac{u^2sin^2\theta}{2g}}[/tex]

u=initial velocity

theta is angle of projection.

g=acceleration due to gravity

ATQ

[tex]\\ \sf\longmapsto R=2h[/tex]

[tex]\\ \sf\longmapsto \dfrac{u^2sin2\theta}{g}=\dfrac{2u^2sin^2\theta}{2g}[/tex]

Cancelling required ones

[tex]\\ \sf\longmapsto sin^2\theta=sin2\theta[/tex]

sin2O=2sinOcosO

[tex]\\ \sf\longmapsto sin^2\theta=2sin\theta cos\theta [/tex]

[tex]\\ \sf\longmapsto \dfrac{sin^2\theta}{sin\theta cos\theta=2[/tex]

[tex]\\ \sf\longmapsto \dfrac{sin\theta}{cos\theta}=2[/tex]

[tex]\\ \sf\longmapsto tan\theta=2[/tex]

[tex]\\ \sf\longmapsto \theta=tan^{-1}(2)[/tex]

[tex]\\ \sf\longmapsto \theta=63.4°[/tex]

Done

Option B is correct

Which is the correct answer

Answers

Answer:

C

Explanation:

All of the other options are affecting the gravitational force and opposing force that pushed upwards, the diagram represents an object that is not moving up or down.

What does it mean when a wave’s amplitude increases?
The wave’s wavelength gets longer.
The wave is moving through a denser medium.
The wave is carrying more energy.
The wave’s frequency also increases.

Answers

Answer:

the wave is carrying more energy

Explanation:

trust me broski

Answer:

a

Explanation:

what is the prmary source of energy inside of the earth

Answers

The correct answer is the energy of the sun
The primary source would be the sun

Two students are sitting on a see-saw. The length of the board is 2 meters with the
pivot point being in the very center. One student sitting on the end of the left side
has a mass of 34 kg. If the student sitting on the right has a mass of 42.5 kg, where
should she sit for the see-saw to be in equilibrium?

Answers

To solve the problem it is necessary to apply the concepts of Torque and equilibrium.

The Torque is defined as:

Where

F= Force

d = Distance

In this particular case, the force is caused by the weight of both children.

In turn, when there is equilibrium, the two torques must be equal therefore

Replacing with our values

Re-arrange to find  

Therefore the distance that the second kid should sit to balance the see-saw is 1.8m from the pivot.

according to newton, doubling the distance between two interacting objects:

Answers

Answer:

So as two objects are separated from each other, the force of gravitational attraction between them also decreases. If the separation distance between two objects is doubled (increased by a factor of 2), then the force of gravitational attraction is decreased by a factor of 4 (2 raised to the second power).

Explanation:

hope his helps

please help me !! i’ll mark brainliest if you’re right!

Answers

Option 2nd is the answer

A 2-column table with 5 rows. The first column has entries empty, time of trial number 1 (seconds), time of trial number 2 (seconds), time of trial number 3 (seconds), average time (seconds). The second column labeled one quarter checkpoint has entries 2. 15, 2. 05, 02. 02, 02. 7. Use the table to answer the questions. What is the fastest time trial for the first quarter checkpoint? seconds What is the slowest time trial for the first quarter checkpoint? seconds What is the range of times measured for this checkpoint? seconds.

Answers

The fastest time trial for the first quarter checkpoint is 2.02 s.

The slowest time trial for the first quarter checkpoint is 2.7 s.

The range of the times measured for the checkpoint is 0.68 s.

The given parameters;

Time for quarter checkpoint, = 2.15, 2.05, 2.02, 2.7

The fastest time trial for the first quarter checkpoint is the least measured time value.

fastest time trial = least time measured

fastest time trial =  2.02 s

The slowest time trial for the first quarter checkpoint is the highest measured time value.

slowest time trial = 2.7 s

The range of the times measured for the checkpoint is difference between the fastest time and slowest time.

Range = fastest time - slowest time

Range = 2.7 - 2.02

Range = 0.68 s

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Answer:

2.02, 2.15, 0.13

Explanation:

A 2-column table with 5 rows. The first column has entries empty, time of trial number 1 (seconds), time of trial number 2 (seconds), time of trial number 3 (seconds), average time (seconds). The second column labeled one quarter checkpoint has entries 2.15, 2.05, 2.02, 2.07.

Use the table to answer the questions.

What is the fastest time trial for the first quarter checkpoint?

2.02

seconds

What is the slowest time trial for the first quarter checkpoint?

2.15 seconds

What is the range of times measured for this checkpoint?

0.13 seconds

can someone explain it with steps?

A car was moving on a road at a constant speed of 15 m/s when suddenly the car driver saw some animal on the road at a distance of 21 m from the car, so he applied the brakes after a response time of 0.4 s and stopped before hitting the animal by 1 m. What was the deceleration of the car?

a-7.5 m/s^2

b-5.2 m/s^2

c-8.0 m/s^2

d-5.6 m/s^2​

Answers

Answer:

Option C is the correct answer

Explanation:

Distance travelled by car during reaction time

[tex]=15\times0.4\\\\=6m[/tex]

The car stopped before hitting the animal by [tex]1 m[/tex]

Distance travelled during deceleration is [tex]21-6-1=14m[/tex]

Hence by [tex]v^2=u^2+2as[/tex]

We have

[tex]0^2=15^2+2 \cdot a \cdot 14\\\\a=\frac{-225}{28} \\\\=-8.03m/s^2[/tex]

Option C is the correct answer

Initial velocity=15m/s=uFinal velocity=0deacceleration=a

Distance traveled during reaction time

15(0.4)=6m

Total distance

21-6-1=14m

[tex]\\ \sf\longmapsto v^2-u^2=2as[/tex]

[tex]\\ \sf\longmapsto -(15)^2=2(14)a[/tex]

[tex]\\ \sf\longmapsto -225=28a[/tex]

[tex]\\ \sf\longmapsto a=-8.0m/s^2[/tex]

Which type of border shows the division between Sonora and Chihuahua?

National border
State border
Physical boundary
Natural boundary

Answers

The answer is the national border

The mass of a sample of sodium bicarbonate is 2. 1 kilograms (kg). There are 1,000 grams (g) in 1 kg, and 1 Times. 109 nanograms (ng) in 1 g. What is the mass of this sample in ng? 2. 1 Times. 103 ng 2. 1 Times. 106 ng 2. 1 Times. 109 ng 2. 1 Times. 1012 ng.

Answers

2.1 kg of sodium bicarbonate is equal to the 2.1 x 10¹² ng of sample. Option D is correct.

Mass is the quantity of the substance in the body or object. The SI unit of mass is Kilogram.

There are other units of measure,

Milligram: 1 g is equal to the [tex]\bold {10^3 \ mg}[/tex]Micro-gram: 1 g is equal to [tex]\bold {10^{6} \ \mu g}[/tex]Nano-gram: 1 g is  is equal to[tex]\bold {10^{9} \ ng}[/tex]

First convert kg to gram,

Since, 1 Kg = 1000 g

2.1 kg = grams of sample

So,

Do the cross multiplication,

[tex]\rm mass\ of\ sample = \dfrac {2.1\ kg \times 1000\ g }{ 1 kg}\\\\\rm mass\ of\ sample =2100 g[/tex]

Now, convert 2100 g to nano-grams

Since, 1 g = 1 x 10⁹ ng

2100 g = ng of sample

So,

Do the cross multiplication,

[tex]\rm mass\ of\ sample = \dfrac {2100 g \times 1 \times 10^9 ng }{1\ g}\\\\\rm mass\ of\ sample = 2.1 \times 10^1^2 ng[/tex]

Therefore, 2.1 kg of  sodium bicarbonate is equal to the 2.1 x 10¹² ng of sample.

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Which would ba another example of newtons first law?

Answers

For example, a book resting on a table applies a downward force equal to its weight on the table. According to the third law, the table applies an equal and opposite force to the book.

define nuclear energy​

Answers

Answer:

Nuclear energy is the energy stored in atoms that can produce electricity.

Hope that helps. x

PLEASE HELP!!
5. A 700 kg race car travels
around the track at 65 m/s. The track has a radius of 75 m.

a. What is the centripetal acceleration?

b. What is the centripetal force?

c. Is the net force on the car less than, equal to, or greater than the centripetal force? Why?

Answers

Hi there!

a.

The equation for centripetal acceleration is as follows:

[tex]\large\boxed{a_c = \frac{v^2}{r}}}[/tex]

Plug in the given values to solve:

[tex]\large\boxed{a_c = \frac{(65)^2}{75} = 56.33 m/s^2}[/tex]

b.

According to Newton's Second Law:

[tex]\large\boxed{\Sigma F = ma}}[/tex]

The acceleration is v²/r, so the net force is:

[tex]\large\boxed{\Sigma F = m(\frac{v^2}{r})}}[/tex]

Multiply by the given mass:

[tex]\large\boxed{\Sigma F = 700(56.33) = 39433.33 N}}[/tex]

c.

There is NO net force in the vertical direction since the object is NOT accelerating in the vertical direction (normal force and weight cancel).

Thus, the ONLY net force experienced by the object is in the horizontal direction and is EQUAL to the centripetal force.

in what direction will the seesaw rotate and what will the sign of the angular acceleration be?

Answers

Answer:

It can rotate in any direction. The sign of the angular acceleration depends on how you set the reference system, it can be both negative or positive.

a 3 kg rock is falling from a rock ledge in the absence of air resistance how much force will the rock strikes the ground with

Answers

Answer:

papi sus

Explanation:

I desperately need help with my Physics Exam I am failing this class will mark Brainliest to whoever helps the most: PART 1 (I don't want to overwhelm anyone so I'm posting my questions in parts) Also the questions marked are incorrect, so just work with the 3 remaining answers.
10000000 thank you's

Answers

Option 2nd is correct

last option is correct

how much force would be needed to push a box weighing 30 N up a ramp that ahas an ideal mechanical advantage of 3

Answers

Answer:

60n would be the needed force

The force needed to push a box weighing 30 N up a ramp that has an ideal mechanical advantage of 3 is equal to 10 N.

What is the mechanical advantage?

The mechanical advantage can be described as the ratio of the input force to the output force. The mechanical advantage of any machine can be determined by the ratio of the forces involved to do the work.

The ratio of the resistance force to the effort is called the actual mechanical advantage which will be comparatively less. The efficiency of a machine is always determined by equating the ratio of its output to its input.

The efficiency of the machine is equal to the ratio of the actual mechanical advantage (M.A.) and theoretical mechanical advantage. Mechanical advantage can be defined as the force produced by a machine to the force applied to it.

Given the load = 30 N and the ideal mechanical advantage = 3

Mechanical advantage = Load/ Effort

Input force or effort = Load/ M.A.

Force = 30/3

Input Force = 10 N

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Why is kinetic energy lost in an inelastic collision?

Answers

Answer:

This is because some kinetic energy had been transferred to something else.

Explanation:

An inelastic collision is a collision in which there is a loss of kinetic energy. While momentum of the system is conserved in an inelastic collision, kinetic energy is not.


An airplane initially travels at 12 m/s when passing the "acceleration line." The airplane then
accelerates to 9 m/s2 until reaching its take off velocity of 40.0 m/s. What is the displacement of
the plane during the acceleration?

Answers

The displacement of  the plane during the acceleration is equal to 80.89 meters.

Given the following data:

Initial velocity = 12 m/sFinal velocity = 40 m/sAcceleration = 9 [tex]m/s^2[/tex]

To determine the displacement of  the plane during the acceleration, we would use the third equation of motion;

[tex]V^2 = U^2 + 2aS[/tex]

Where:

V is the final speed.U is the initial speed.a is the acceleration.S is the displacement traveled.

Substituting the given parameters into the formula, we have;

[tex]40^2 =12^2 + 2(9)S\\\\1600=144+18S\\\\18S=1600-144\\\\18S=1456\\\\S=\frac{1456}{18}[/tex]

Displacement, S = 80.89 meters.

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Which of the following is an example of physical weathering?
O calcium carbonate in limestone changes to calcium hydrogen carbonate
O flow water carves erosion channels in a hillside
Ostalactites precipitate in a cave
O acid rain corrodes a monument

Answers

Answer:

acid rain corrodes a monument. d.

Explanation:it only makes sense because its doing something to a physical object.and the other ones aren't.

If the road becomes wet or crowded, you should ____. slow down and increase your following distance All choices are incorrect. maintain your speed and following distance speed up and decrease your following distance Submit answer

Answers

Answer:

The first one.

If the road becomes wet or crowded, you should ___ slow down and increase your following distance_

What is meant by surface distance ?

The safe distance between vehicles traveling in column specified by the command in light of safety requirements

The slower speed will help you save gas and avoid potential accidents. You can easily eliminate chances of rear end collision by maintaining a safe distance between your car and the vehicles ahead.

hence , a) slow down and increase your following distance is a correct option

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A 7.5 kg beaver dives horizontally off a 50 kg log at a speed of 4 m/s. What is the speed of the log?

Answers

Answer:

[tex]0.6\; \rm m\cdot s^{-1}[/tex], assuming that drag of the water on the log is negligible.

Explanation:

The momentum [tex]p[/tex] of an object is equal to the product of mass [tex]m[/tex] and velocity [tex]v[/tex]. That is, [tex]p = m\, v[/tex].

If the drag of water on the log is negligible, momentum of the beaver and the log, combined, would be preserved.

The momentum of the beaver and the log, combined, was initially [tex]0\; \rm kg\cdot m \cdot s^{-1}[/tex].

The momentum of the beaver right after the dive would be [tex]7.5 \; {\rm kg} \times 4\; {\rm m \cdot s^{-1}} = 30\; {\rm kg \cdot m \cdot s^{-1}}[/tex].

The sum of the momentum of the beaver and the log is conserved and should continue to be [tex]0\; {\rm kg\cdot m \cdot s^{-1}}[/tex] even after the dive. Since the momentum of the beaver is [tex]30\; {\rm kg \cdot m \cdot s^{-1}[/tex] after the dive, the momentum of the log should become [tex](-30)\; {\rm kg \cdot m \cdot s^{-1}}[/tex].

Since the mass of the log is [tex]50\; {\rm kg}[/tex], the new velocity of this log would be:

[tex]\begin{aligned}v &= \frac{p}{m} \\ &= \frac{(-30)\; {\rm kg \cdot m \cdot s^{-1}}}{50\; {\rm kg}} \\ &= (-0.6)\; {\rm m \cdot s^{-1}}\end{aligned}[/tex].

(The new velocity of the log is negative because the log would be moving away from the beaver.)

The speed of an object is the magnitude of velocity. For this log, a velocity of [tex](-0.6)\; {\rm m \cdot s^{-1}}[/tex] would correspond to a speed of [tex]|(-0.6)\; {\rm m \cdot s^{-1}| = 0.6\; {\rm m \cdot s^{-1}[/tex].

A race car makes one lap around a track of radius 180 m in 90.0 s. What is the centripetal acceleration in m/s2?
a
0.5
12.6
Ob
OC
Od
.87
2.
2

Answers

Answer:

0.9m/s^2 (yours is 0.87, so choose that)

Explanation:

formula for centripetal acceleration:

v^2/r

to find v, we know that f=1/90s, and r=180m.

v=(2pir)/T

v=(2pi(180))/90

v=12.6m/s

now plug into a=v^2/r

a=(12.6)^2/180

a=0.9m/s^2


A car has a kinetic energy of 41.6 kJ.
The speed of the car is 8.0m/s.
Calculate the mass of the car.

Answers

K=1/2 mv2

M=?

41.6kj convert to joules by multiplying by 1000 so it will be 41,600J because the unit of kinetic energy is in joules.

41,600=1/2(m)(8)

Arrange the equation it will be:

M= 41,600/4 = 10,400

Final answer is:
m= 10,400 kg

What is the half-life of an isotope if after 30 days you have 31.25 g remaining from a 250 g beginning sample size?

Answers

The half-life of the given isotope will be 10 days, if after 30 days only 31.25 grams are remaining from a sample of 250 grams of the sample size taken in the beginning.

What is Half-life of an element?

The Half-life is the time which is required for a quantity to reduce the content to half of the amount present as its initial value. The term is used in nuclear physics to describe how quickly an unstable atom undergo radioactive decay or how long does stable atoms survive. The term is also used generally to characterize any type of exponential decay.

The half-life of the isotope can be calculated by the formula:

FR = 0.5n

FR = Fraction Remaining = 31.25 g / 250.0 g = 0.1250

n = number of half lives elapsed = ?

0.125 = 0.5n

log 0.125 = n log 0.5

-0.9031 = -0.3010 n

n = 3.000 half lives have elapsed

3 half lives = 30 days

1 half live = 10 days

Therefore, the half-life of the isotope will be 10 days.

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A car travels 80 km at an avarage speed of 40 km/h. It travels the remaining distance in 3h. What is its dis placement if the average speed of the car is 30 km/h.? And explain

A) 100 km

B)120 km

C)150km

D)180km

Answers

When traveling with average speed 40 km/h, the car would cover a distance of 80 km in time t such that

40 km/h = (80 km) / t   ⇒   t = 2 h

so the total travel time is 2 h + 3 h = 5 h.

Average speed is defined as the total distance traveled divided by the time it took to cover that distance. So if the overall average speed was 30 km/h, then

30 km/h = (80 km + x) / (5 h)   ⇒   x = 70 km

where x is the distance traveled in the last 3 h of the trip.

Then the total displacement of the car is 80 km + 70 km = 150 km.

9. How are ecological footprints useful?

Answers


As ecological footprints provide an easily communicable way of measuring the ecological bottom‐line condition for sustainability, it is a useful tool for promoting a sustainable future. It is particularly useful for cities, as it is in cities where the battle for sustainability will be won or lost.

Answer:

This is what the Ecological Footprint does: It measures the biologically productive area needed to provide for everything that people demand from nature: fruits and vegetables, meat, fish, wood, cotton and other fibres, as well as absorption of carbon dioxide from fossil fuel burning and space for buildings and roads.

Why it's useful:

Ecological footprint (EF), measure of the demands made by a person or group of people on global natural resources. It has become one of the most widely used measures of humanity’s effect upon the environment and has been used to highlight both the apparent unsustainability of current practices and the inequalities in resource consumption between and within countries.

please help me with this​

Answers

Answer:

>400N is needed to balance that lever

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