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UkoKoshka [18]
3 years ago
11

How fast will it be traveling after it goes 87 m ?

Physics
1 answer:
swat323 years ago
6 0

To solve this we can use one of the main four kinematic equations, preferably we want something that we can use directly without solving for other values.

So in this problem we know that:

Inital velocity or Vi = 0 (started from rest)

The acceleration of the object = 7.5

The distance the object travels Δx = 87m

So, our best equation here with what we have would be:

Vf² = Vi² + 2aΔx

Now we now that Vi is 0 so:

Vf² = 2aΔx

We need final velocity so lets take the square root of Vf to isolate it:

Vf = √2aΔx

Now we plug in

Vf = √2(7.5)(87)

Vf = 36.124 m/s or 36.1 m/s

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A golfer gives a ball a maximum initial speed of 51.5 m/s. how far does it go
nata0808 [166]

<u>Answer:</u>

Golf ball will go a maximum of 270.36 meter.

<u>Explanation:</u>

  Projectile motion has two types of motion Horizontal and Vertical motion.

 Vertical motion:

          We have equation of motion, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.

          Considering upward vertical motion of projectile.

          In this case, Initial velocity = vertical component of velocity = u sin θ, acceleration = acceleration due to gravity = -g m/s^2 and final velocity = 0 m/s.

         0 = u sin θ - gt

          t = u sin θ/g

     Total time for vertical motion is two times time taken for upward vertical motion of projectile.

     So total travel time of projectile = 2u sin θ/g

Horizontal motion:

   We have equation of motion , s= ut+\frac{1}{2} at^2, s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

   In this case Initial velocity = horizontal component of velocity = u cos θ, acceleration = 0 m/s^2 and time taken = 2u sin θ /g

  So range of projectile,  R=ucos\theta*\frac{2u sin\theta}{g} = \frac{u^2sin2\theta}{g}

  Now in the given problem

     A golfer gives a ball a maximum initial speed of 51.5 m/s. how far does it go

     u = 51.5 m/s, for maximum range θ = 45⁰

   So maximum distance reached = \frac{51.5^2sin(2*45)}{9.81}=270.36 meter

So it will go a maximum of 270.36 meter.

5 0
4 years ago
The first few hundred million years of the solar system's history were the time of the __________, during which Earth suffered m
azamat

Answer:

heavy bombardment

Explanation:

5 0
3 years ago
Every 4 years we have a leap year on our calander to make up the extra.......
OverLord2011 [107]

The Earth takes very nearly (365 and 1/4) days to go around the sun.

If our calendar always had 365 days, then the year would end and re-start
too soon, and the beginning of Spring (and every other season) would
eventually drift into the months after March.

If our calendar always had 366 days, then the year would end and re-start
too late, and the beginning of Spring (and every other season) would
eventually drift into the months before March.

We can't make calendars with an extra quarter-day in each year.  But we
keep them lined up with the real year by saving up the quarters, and adding
one full day to the calendar every 4 years.

7 0
3 years ago
True or False. A projectile is an object that once set in motion, continues in motion by its own inertia.
bazaltina [42]
The answer is true.
5 0
3 years ago
Read 2 more answers
An iron anchor of density 7890.00 kg/m3 appears 299 N lighter in water than in air. (a) What is the volume of the anchor? (b) Ho
PtichkaEL [24]

Answer:

weigh is 2353.13 N

Explanation:

Given data

density = 7890.00 kg/m3

lighter =  299 N

to find out

the volume of the anchor and weigh in air

solution

from question we can say that

apparent weight = actual weight - buoyant force

we know weight = mg and buoyant force = water density × g

so volume of anchor is = actual weight - apparent weight / buoyant force

volume of anchor is = 299 / 1000 × 9.81

volume of anchor is = 0.0304791 m³

and

weight of anchor is mg

here mass m = density Fe g

density Fe = 7870 from table 14-1

so weight = 7870 × 0.0304791  × 9.81

weigh is 2353.13 N

7 0
4 years ago
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