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TEA [102]
4 years ago
5

I start going 10 m/sIf I slow down at 2 m/s/s how far will I go before I stop

Physics
1 answer:
klasskru [66]4 years ago
5 0

Answer:

25 m

Explanation:

Given:

v₀ = 10 m/s

v = 0 m/s

a = -2 m/s²

Find: Δx

v² = v₀² + 2aΔx

(0 m/s)² = (10 m/s)² + 2 (-2 m/s²) Δx

Δx = 25 m

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Tina is driving her sports car down US1, traveling 27m/s. She sees her friend Rita up ahead, waiting to cross the street at the
mrs_skeptik [129]

Answer:

434 Hz

Explanation:

According to the Doppler effect, when a source of a wave is moving towards an observer at rest, then the observer will observe an apparent frequency which is higher than the original frequency of the source.

In this situation, Tina is driving towards Rita. Tina is the source of the sound wave (the horn), while RIta is the observer. Since the original frequency of the sound is 400 Hz, Rita will hear a sound with a frequency higher than this value.

The only choice which is higher than 400 Hz is 434 Hz, so this is the frequency that Rita will hear.

3 0
3 years ago
Two technicians are discussing fuse testing. Technician A says that a test light should light on both test points of the fuse if
Murljashka [212]

Answer:

C. Both technicians A and B

Explanation:

Both technicians are absolutely correct because a functional test light is meant to light on both test point if the fuse is working fine which implies that, if the test light doesn't light on both sides then there must be a fault with the fuse. So, both technicians A and B are very correct.

4 0
3 years ago
A child blows a leaf from rest straight up in the air. The leaf has a constant upward acceleration of magnitude 1.0\,\dfrac{\tex
Novosadov [1.4K]

Answer:

Explanation:

Given

Acceleration a = 1.0m/s²

Displacement S = 1.0m

Required

Time t taken by the leaf to displace

Using the equation of motion

S = ut+1/2at²

Substitute

1.0 = 0+1/2(1)t²

1 = t²/2

Cross multiply

t² = 2

t = ±√2

t = 1.41secs

It takes the leaf to 1.41s to displace by 1m upward

6 0
3 years ago
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
3 years ago
A railroad car of mass 3.45 ✕ 104 kg moving at 2.60 m/s collides and couples with two coupled railroad cars, each of the same ma
Alex

Answer:

a) 1.67 m/s

b) 23kJ

Explanation:

We need to apply the linear momentum conservation formula, that states:

m1*v_{o1}+m2*v_{o2}=m1*v_{f1}+m2*v_{f2}

in this case:

3.45*10^4kg*2.60m/s+2*3.45*10^4kg*1.20m/s=3*m1*v_{f}\\v_f=1.67m/s

the initital kinetic energy is:

K_i=\frac{1}{2}*3.45*10^4kg*(2.60m/s)^2+2(\frac{1}{2}*3.45*10^4kg*(1.20m/s)^2\\K_i=167kJ

and the final:

K_f=3*\frac{1}{2}*3.45*10^4kg*(1.67m/s)^2\\K_f=144kJ

The energy lost is given by:

E_l=|K_f-K_i|\\E_l=23kJ

8 0
3 years ago
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