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Airida [17]
3 years ago
7

A cheetah spotted a Gazelle. The cheetah runs at its top speed of 30 m/s for 15 seconds. Durning this time , a gazelle ,160 m fr

om the running cheetah notices the danger ahead. The gazelle accelerated at 4.5m/s^2 for 6.0s, then continued running at a constant speed.
A) how far does the cheetah travel during 15s duration
B) how fast does the gazelle reach after 6.0 s

C) how far does the gazelle travel during the first 6.0 s and last 9.0
D) who’s ahead at the end of 15secs
Physics
1 answer:
Xelga [282]3 years ago
5 0

Answer:

A) 450 m

B) 27 m/s

C) 81 m, 243 m

D) Gazelle

Explanation:

A)

Since, the Cheetah is running at constant speed. Therefore, we use the equation:

s₁ = v₁t

where,

s₁ = distance covered by Cheetah = ?

v₁ = speed of Cheetah = 30 m/s

t = time taken = 15 s

Therefore,

s₁ = (30 m/s)(15 s)

<u>s₁ = 450 m</u>

<u></u>

B)

For final speed of Gazelle at the end of 6 s acceleration time we use 1st equation of motion:

Vf = Vi + at

where,

Vf = Final Speed of Gazelle at the end of 6 s = ?

Vi = Initial Speed of Gazelle = 0 m/s (Since, Gazelle is initially at rest)

t = time taken = 6 s

a = acceleration = 4.5 m/s²

Therefore,

Vf = (0 m/s) + (4.5 m/s²)(6 s)

<u>Vf = 27 m/s</u>

C)

For the distance covered by Gazelle at the end of 6 s acceleration time we use 2nd equation of motion:

s₂ = Vi t + (0.5)at²

where,

Vi = Initial Speed of Gazelle = 0 m/s (Since, Gazelle is initially at rest)

t = time taken = 6 s

a = acceleration = 4.5 m/s²

Therefore,

s₂ = (0 m/s)(6 s) + (0.5)(4.5 m/s²)(6 s)²

<u>s₂ = 81 m</u>

<u></u>

Now, for the distance covered during the last 9 s at constant velocity Vf, we use equation:

s₃ = Vf t

where,

s₃ = distance covered by Gazelle in last 9 s = ?

t = time = 9 s

Therefore:

s₃ = (27 m/s)(9 s)

<u>s₃ = 243 m</u>

D)

We know that, at the end of 15 seconds:

Distance covered by Cheetah = s₁ = 450 m

Distance Covered by Gazelle = s₂ + s₃ = 81 m + 243 m = 324 m

If we take the initial position of Cheetah as origin. Then the positions of both Gazelle and Cheetah with respect to origin will be:

Position of Cheetah = 450 m ahead of origin

Position of Gazelle = 324 m + 160 m = 484 m (Since, Gazelle was initially 160 m ahead of Cheetah)

<u>Hence, it is clear that Gazelle is ahead at the end of 15 s.</u>

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=0.018 \mathrm{M} \times 0.200 \mathrm{~L}=0.004 \mathrm{~mol}

Find mass of Fe(II)

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  • Molarity is the concentration of a solution expressed as the number of moles of solute dissolved in each liter of solution.
  • concentration is the amount of a substance per defined space. Concentration usually is expressed in terms of mass per unit volume.

To know more about  MOLARITY    visit : brainly.com/question/8732513

#SPJ4

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