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1 year ago

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Finally, you are ready to answer the main question. Cheetahs, the fastest of the great cats, can reach 50.0 miles/hour miles/hou

r in 2.22 s s starting from rest. Assuming that they have constant acceleration throughout that time, find their acceleration in meters per second squared.

2 answers:

dolphi86 [110]1 year ago

7 0

Answer:

a = 10.07m/s^2

Their acceleration in meters per second squared is 10.07m/s^2

Explanation:

Acceleration is the change in velocity per unit time

a = ∆v/t

Given;

∆v = 50.0miles/hour - 0

∆v = 50.0miles/hours × 1609.344 metres/mile × 1/3600 seconds/hour

∆v = 22.352m/s

t = 2.22 s

So,

Acceleration a = ∆v/t = 22.352m/s ÷ 2.22s

a = 10.07m/s^2

Their acceleration in meters per second squared is 10.07m/s^2

SCORPION-xisa [38]1 year ago

4 0

<h2>Answer:</h2>

10.1m/s²

<h2>Explanation:</h2>

<em>Using one of the equations of motion as follows;</em>

v = u + at ----------------------(i)

where;

v = final velocity of the body (Cheetahs) = 50.0 miles/hour

u = initial velocity of the body = 0 (since they start running from rest)

a = acceleration/deceleration

t = time taken for the motion = 2.22s

<em>First convert the final velocity (v) from miles/hour to m/s</em>

Remember that;

1 mile = 1609.34m

1 hour = 60 x 60s = 3600s

Therefore;

50miles/hour = $\frac{50miles}{1 hour}$ = $\frac{50*1609.34m}{3600s}$ = 22.35m/s

=> v = 22.35m/s

<em>Substituting the values of v, u and t into equation (i) gives;</em>

=> 22.35 = 0 + a (2.22)

=> 22.35 = 2.22a

=> a = $\frac{22.35}{2.22}$ = 10.07m/ $s^{2}$

=> a = 10.1m/s² (to 1 decimal place)

Therefore the acceleration in m/s² is 10.1

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1 year ago

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IRISSAK [1]

Answer:

Explanation:

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= k Q / r²

k is a constant , Q is charge and r is distance

= 9 x 10⁹ x 5 x 10⁻⁶ / .5²

= 180 x 10³ N /C

In vector form

E₁ = 180 x 10³ j

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= 9 x 10⁹ x 3 x 10⁻⁶ /.5² + .8²

= 30.33 x 10³ N / C

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Tan θ = .5 / √.5² + .8²

= .5 / .94

θ = 28°

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= 26.78 x 10³ i - 14.24 x 10³ j

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E = E₁ + E₂

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magnitude

= √(26.78² + 165.76² ) x 10³ N /C

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