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An object’s speed is the distance it travels _____ the amount of time it takes.

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# Tag: speed

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Greg is in a bike race. at mile marker four (out of ten), his speed was measured at 13.5mph. which best describes the measured number

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True or False: Florida law prohibits you from increasing your speed until the vehicle that is passing you has completed the passing maneuver.

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Regularly Tuning up a computer can assist keeping it running at Peak speed. True or False?

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John goes for a run. From his house, he jogs north for exactly 5.0 min at an average speed of 8.0 km/h. He continues north at a speed of 12.0 km/h for the next 30.0 min. He then turns around and jogs south at a speed of 15.0 km/h for 15.0 min. Then he jogs south for another 20.0 min at 8.0 km/h. He walks the rest of the way home. Which is the correct plot of total distance as a function of time for John’s jog?

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**Answer:**

– 2.14 m/s^2

**Explanation:**

initial velocity , u = 245 m/s

final velocity, v = 230 m/s

time taken , t = 7 second

By use of first equation of motion,

**v = u + a t**

where, a be the acceleration.

Substitute the values of v, u and t

230 = 245 + a x 7

a = – 2.14 m/s^2

the acceleration is negative, it means the airplane is slowing.

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How long will it take Karen to drive 975 miles, if her average speed is 65 miles per hour? how to write a formula

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Swimming and baseball are also some hard sports especially swimming since most are your muscles are used during that activity.

Hope this helps 😉

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The correct answer is B.

You may be driving at the posted speed limit and still receive a ticket if the posted speed at that particular time when you’re driving.

It mostly occurs in : heavy rain or snowstorm, a hail, a hurricane or other natural disaster.

In such cases, it is advisable to reduce your speed and drive under ( or well under ) the posted or recommended speed. It is done as a measure to avoid accidents and unnecessary deaths.

Some other examples when you might receive a ticket for driving at the posted speed limit include: not slowing or accelerating for a passing ambulance or police vehicle and disobeying oral instructions given to you by traffic wardens.

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A projectile is able to orbit the Earth at what speed?

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Sasha sits on a horse on a carousel 3.5 m from the center of the circle. She makes a revolution once every 8.2 seconds. What is Sasha’s tangential speed?

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**Answer:**

0.68 m/s

**Explanation:**

The centripetal force that keeps the beetle moving in circle is given by:

where

m is the mass of the beetle

v is the tangential speed of the beetle

r is the distance of the beetle from the center of the record

In this problem, we know the force (F=0.070 N), the mass of the beetle (m=0.023 kg) and the distance from the center (r=0.15 m), therefore we can re-arrange the equation to find the tangential speed:

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To answer this problem, we will use the formula of

hypotenuse:

c^2 = a^2 + b^2

Where,

c = total distance traveled / displacement = 119 miles

a = distance traveled north = 5 hours * x mph = 5 x

b = distance traveled west = 4 hours * (x + 5) mph = 4 (x

+ 5)

Substituting to the equation:

119^2 = (5 x)^2 + [4(x + 5)]^2

14,161 = 25 x^2 + 16 (x + 5)^2

14,161 = 25 x^2 + 16 (x^2 + 10 x + 25)

14,161 = 25 x^2 + 16 x^2 + 160 x + 400

0 = 41 x^2 + 160 x – 13,761

x =[ – b ± sqrt (b^2 – 4ac)]/ 2a

x = [- 160 ± sqrt (160^2 – 4 * 41 * (– 13,761))] / 2 * 41

x = – 1.95 ± 18.42

x = -20.37 , 16.47

Since speed cannot be negative, therefore x = 16.47

Therefore the speed of the ship travelling west is:

**x +
5 = 21.47 mph (ANSWER)**

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v = 2*pi*R/T,

you have R in m and T is days, which multiplied by 86,400 s/day gives T in seconds.

Then v = 2*pi*3.84*10^8/(27.3*86,400) = 1,022.9 m/s ~ 1 km/s (about 3 times the speed of sound 🙂

For the Earth around the Sun, it would be v = 2*pi*149.5*10^9/(365*86,400)~ 29.8 km/s!

I know it’s not in the problem, but it’s interesting to know how fast the Earth moves around the Sun! And yet we do not feel it (that’s one of the reasons some ancient people thought crazy the Earth not being at the center, there would be such strong winds!)

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**Answer:**

Tangential speed is 0.68 m/s

**Explanation:**

It is given that,

Mass of the beetle, m = 0.023 kg

It is placed at a distance of 0.15 m from the center of record i.e. r = 0.15 m.

If it takes 0.070 n of force to keep the beetle moving in a circle on the record i.e. centripetal force acting on it is, F = 0.070 N

We have to find the tangential speed of the beetle. **The formula for centripetal force is given by :**

v is tangential speed

v = 0.675 m/s

or

**v = 0.68 m/s**

**Hence, the correct option for tangential speed is (A).**

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speed of train=x-30

A loaded moving truck is traveling 30 mph faster than a freight train. in the time it takes the train to travel 135 miles, the truck travels 225 miles. find the speed of the truck.

A loaded moving truck is traveling 30 mph faster than a freight train…

(x-30)=speed of train

the time it takes the train to travel 135 miles, the truck travels 225 miles…

135/(x-30)=225/x

the you multiply on both sides the x values and get…

135x=225x-6750

then you subtract 225x on both sides…

-90x=-6750

then you divide -90 on both sides…

x=75

the speed of the truck is 75mph

i hope this helped give me a thanks and a 5 star rating if it helped! c:

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**Answer:**

a) The time the police officer required to reach the motorist was 15 s.

b) The speed of the officer at the moment she overtakes the motorist is 30 m/s

c) The total distance traveled by the officer was 225 m.

**Explanation:**

The equations for the position and velocity of an object moving in a straight line are as follows:

x = x0 + v0 · t + 1/2 · a · t²

v = v0 + a · t

Where:

x = position at time t

x0 = initial position

v0 = initial velocity

t = time

a = acceleration

v = velocity at time t

a)When the officer reaches the motorist, the position of the motorist is the same as the position of the officer:

x motorist = x officer

Using the equation for the position:

x motirist = x0 + v · t (since a = 0).

x officer = x0 + v0 · t + 1/2 · a · t²

Let´s place our frame of reference at the point where the officer starts following the motorist so that x0 = 0 for both:

x motorist = x officer

x0 + v · t = x0 + v0 · t + 1/2 · a · t² (the officer starts form rest, then, v0 = 0)

v · t = 1/2 · a · t²

Solving for t:

2 v/a = t

t = 2 · 15.0 m/s/ 2.00 m/s² = 15 s

The time the police officer required to reach the motorist was 15 s.

b) Now, we can calculate the speed of the officer using the time calculated in a) and the equation for velocity:

v = v0 + a · t

v = 0 m/s + 2.00 m/s² · 15 s

v = 30 m/s

The speed of the officer at the moment she overtakes the motorist is 30 m/s

c) Using the equation for the position, we can find the traveled distance in 15 s:

x = x0 + v0 · t + 1/2 · a · t²

x = 1/2 · 2.00 m/s² · (15s)² = 225 m

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**Answer:**

a) The time the police officer required to reach the motorist was 15 s.

b) The speed of the officer at the moment she overtakes the motorist is 30 m/s

c) The total distance traveled by the officer was 225 m.

**Explanation:**

The equations for the position and velocity of an object moving in a straight line are as follows:

x = x0 + v0 · t + 1/2 · a · t²

v = v0 + a · t

Where:

x = position at time t

x0 = initial position

v0 = initial velocity

t = time

a = acceleration

v = velocity at time t

a)When the officer reaches the motorist, the position of the motorist is the same as the position of the officer:

x motorist = x officer

Using the equation for the position:

x motirist = x0 + v · t (since a = 0).

x officer = x0 + v0 · t + 1/2 · a · t²

Let´s place our frame of reference at the point where the officer starts following the motorist so that x0 = 0 for both:

x motorist = x officer

x0 + v · t = x0 + v0 · t + 1/2 · a · t² (the officer starts form rest, then, v0 = 0)

v · t = 1/2 · a · t²

Solving for t:

2 v/a = t

t = 2 · 15.0 m/s/ 2.00 m/s² = 15 s

The time the police officer required to reach the motorist was 15 s.

b) Now, we can calculate the speed of the officer using the time calculated in a) and the equation for velocity:

v = v0 + a · t

v = 0 m/s + 2.00 m/s² · 15 s

v = 30 m/s

The speed of the officer at the moment she overtakes the motorist is 30 m/s

c) Using the equation for the position, we can find the traveled distance in 15 s:

x = x0 + v0 · t + 1/2 · a · t²

x = 1/2 · 2.00 m/s² · (15s)² = 225 m

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**Answer:**

a) The time the police officer required to reach the motorist was 15 s.

b) The speed of the officer at the moment she overtakes the motorist is 30 m/s

c) The total distance traveled by the officer was 225 m.

**Explanation:**

The equations for the position and velocity of an object moving in a straight line are as follows:

x = x0 + v0 · t + 1/2 · a · t²

v = v0 + a · t

Where:

x = position at time t

x0 = initial position

v0 = initial velocity

t = time

a = acceleration

v = velocity at time t

a)When the officer reaches the motorist, the position of the motorist is the same as the position of the officer:

x motorist = x officer

Using the equation for the position:

x motirist = x0 + v · t (since a = 0).

x officer = x0 + v0 · t + 1/2 · a · t²

Let´s place our frame of reference at the point where the officer starts following the motorist so that x0 = 0 for both:

x motorist = x officer

x0 + v · t = x0 + v0 · t + 1/2 · a · t² (the officer starts form rest, then, v0 = 0)

v · t = 1/2 · a · t²

Solving for t:

2 v/a = t

t = 2 · 15.0 m/s/ 2.00 m/s² = 15 s

The time the police officer required to reach the motorist was 15 s.

b) Now, we can calculate the speed of the officer using the time calculated in a) and the equation for velocity:

v = v0 + a · t

v = 0 m/s + 2.00 m/s² · 15 s

v = 30 m/s

The speed of the officer at the moment she overtakes the motorist is 30 m/s

c) Using the equation for the position, we can find the traveled distance in 15 s:

x = x0 + v0 · t + 1/2 · a · t²

x = 1/2 · 2.00 m/s² · (15s)² = 225 m

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A man standing 9 feet from the base of a lamppost casts a shadow 6 feet long. if the man is 6 feet tall and walks away from the lamppost at a speed of 30 feet per minute, at what rate, in feet per minute, will his shadow lengthen?

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The functional unit of the nervous system is the nerve cell or the neuron. The neuron consists of a cell body and the axon. The cell body starts with the dendrites that receive the messages or the impulses from other neurons or from different sense organs or receptors. These impulses are then transmitted through the cell body. The cell body contains a nucleus and different organelles which help the nerve cell to perform its functions. The nerve impulse is then transmitted to the axon.

The axon is an extension from the cell body. There are some cells called Schwan cells that secrete a myelin sheath to insulate the axon from the surrounding medium. The insulated axons have more ability to conduct the impulses than non-insulated axons. The axon ends with the terminal arborizations. The terminal arborizations of a nerve cell connect to the dendrites of the next cell or to the afferent organ. The gaps between the dendrites and the terminal arborizations are called the synapses.

The nerve impulse is an electrochemical phenomenon i.e. an electrical phenomenon with a chemical nature. The membrane of the axon acts as a barrier between an outside positively charged medium and an inside negatively charged medium. This makes a potential difference of -70mV. This state is called the resting potential. When the membrane is subjected to a stimulus, the positive charges enter to inside and the negative charges exit to the outside. The potential difference now becomes +40mV. This state is called the depolarization state. The point of stimulation acts as a new stimulus for the next point and so on. The membrane sooner gains its permeability again and the positive charges return to the outside and the negative charges to inside. This state is called repolarization.

The nerve impulse reaches the synapse. There are some neurotransmitters that are excited by the nerve impulse coming and carry the message across the membrane. Some receptors receive theses neurotransmitters on the dendrites of the next neuron. These receptors act as a stimulus for the new cell.

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**The basis to respond this question are:**

**1) Perpedicular lines form a 90° angle between them.**

**2) The product of the slopes of two any perpendicular lines is – 1.**

So, from that basic knowledge you can analyze each option:

**a.Lines s and t have slopes that are opposite reciprocals.**

**TRUE. Tha comes the number 2 basic condition for the perpendicular lines.**

slope_1 * slope_2 = – 1 => slope_1 = – 1 / slope_2, which is what opposite reciprocals means.

b.Lines s and t have the same slope.

FALSE. We have already stated the the slopes are opposite reciprocals.

**c.The product of the slopes of s and t is equal to -1**

**TRUE: that is one of the basic statements that you need to know and handle.**

d.The lines have the same steepness.

FALSE: the slope is a measure of steepness, so they have different steepness.

e.The lines have different y intercepts.

FALSE: the y intercepts may be equal or different. For example y = x + 2 and y = -x + 2 are perpendicular and both have the same y intercept, 2.

f.The lines never intersect.

FALSE: perpendicular lines always intersept (in a 90° angle).

**g.The intersection of s and t forms right angle.**

**TRUE: right angle = 90°.**

h.If the slope of s is 6, the slope of t is -6

FALSE. – 6 is not the opposite reciprocal of 6. The opposite reciprocal of 6 is – 1/6.

**So, the right choices are a, c and g.**

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Activation energy is the difference in the energy level of the reactants and the peak in the potential energy diagram (the energy of the transition state).

For an endothermic reaction, the products will be closer in energy to the transition state than what the reactans will be; so, the activation energy of the reversed reaction is lower than the activation energy of the forward reaction.

Activation energy of reverse and forward reactions is related by:

Activation energy of reverse rxn = Activation energy of forward rxn – ΔH rxn

=> Activiation energy of reverse rxn = 102 kJ/mol – 55 kJ/mol = 47 kJ/mol

**Answer: 47 kJ/mol**

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**Answer:**

Option C – BD=76 cm

**Step-by-step explanation:**

**Given :** You are designing a diamond-shaped kite. you know that AD = 44.8 cm, DC = 72 cm, and AC = 84.8 cm.

**To find : **How long BD should it be?

**Solution : **

First we draw a rough diagram.

The given sides were AD = 44.8 cm, DC = 72 cm and AC = 84.8 cm.

According to properties of kite

**Two disjoint pairs of consecutive sides are congruent.**

**So,** AD=AB=44.8 cm

DC=BC=72 cm

**The diagonals are perpendicular.**

**S**o, AC ⊥ BD

Let O be the point where diagonal intersect let let the partition be x and y.

AC= AO+OC

AC= …….[1]

Perpendicular bisect the diagonal BD into equal parts let it be z.

BD=BO+OD

BD=z+z

**Applying Pythagorean theorem in ΔAOD**

where H=AD=44.8 ,P= AO=x , B=OD=z

………[2]

**Applying Pythagorean theorem in ΔCOD**

where H=DC=72 ,P= OC=y , B=OD=z

…………[3]

**Subtract [2] and [3]**

……….[4]

**Add equation [1] and [4], to get values of x and y**

**Substitute x in [1]**

**Substitute value of x in equation [2], to get z**

**We know, **BD=z+z

BD= 38.06+38.06

BD= 76.12

**Nearest to whole number **BD=76 cm

**Therefore, Option c – BD=76 cm is correct.**

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**Answer:**

a) The time the police officer required to reach the motorist was 15 s.

b) The speed of the officer at the moment she overtakes the motorist is 30 m/s

c) The total distance traveled by the officer was 225 m.

**Explanation:**

The equations for the position and velocity of an object moving in a straight line are as follows:

x = x0 + v0 · t + 1/2 · a · t²

v = v0 + a · t

Where:

x = position at time t

x0 = initial position

v0 = initial velocity

t = time

a = acceleration

v = velocity at time t

x motorist = x officer

Using the equation for the position:

x motirist = x0 + v · t (since a = 0).

x officer = x0 + v0 · t + 1/2 · a · t²

x motorist = x officer

x0 + v · t = x0 + v0 · t + 1/2 · a · t² (the officer starts form rest, then, v0 = 0)

v · t = 1/2 · a · t²

Solving for t:

2 v/a = t

t = 2 · 15.0 m/s/ 2.00 m/s² = 15 s

The time the police officer required to reach the motorist was 15 s.

v = v0 + a · t

v = 0 m/s + 2.00 m/s² · 15 s

v = 30 m/s

The speed of the officer at the moment she overtakes the motorist is 30 m/s

c) Using the equation for the position, we can find the traveled distance in 15 s:

x = x0 + v0 · t + 1/2 · a · t²

x = 1/2 · 2.00 m/s² · (15s)² = 225 m

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