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1 year ago

the car turned a corner going 10m/s if it took two seconds to turn what is its acceleration is this acceleration

Physics

1 answer:

exis [7]1 year ago

6 0

The acceleration of this car is equal to 5 $m/s^2$ .

<u>Given the following data:</u>

Initial velocity = 0 m/s (assuming it's starting from rest).

Final velocity = 10 m/s.

Time = 2 seconds.

To determine the acceleration of this car:

<h3>How to calculate acceleration.</h3>

In Science, the acceleration of an object is calculated by subtracting the initial velocity from its final velocity and dividing by the time.

Mathematically, acceleration is given by this formula:

$a = \frac{V\;-\;U}{t}$

<u>Where:</u>

V is the final velocity.

U is the initial velocity.

is the time measured in seconds.

Substituting the given parameters into the formula, we have;

$a = \frac{10\;-\;0}{2}$

Acceleration, a = 5 $m/s^2$

Read more on acceleration here: brainly.com/question/24728358

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3 years ago

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[High Dive) above a pool of water. According to the announcer, the divers enter the water at a speed of 56 mi/h (25 m/s). (Air r

blsea [12.9K]

Answer:

(a) The announcer's claim is incorrect because the divers enter at a speed of 20.4 and not 25 m/s as announced

(b) it’s possible for a diver to enter the water with the velocity of 25 m/s if he has initial velocity of 14.4 m/s. The upward initial velocity can’t be physically attained

Explanation:

(a)

To find the final velocity $V_{f}$ for an object traveling distance h taking the initial vertical component of velocity as $V_{i}$ the kinematics equation is written as

$V_{f}^{2}=V_{i}^{2}+2ah$ where a is acceleration

Substituting g for a where g is gravitational force value taken as 9.81

$V_{f}^{2}=V_{i}^{2}+2gh$

Since the initial velocity is zero, we can solve for final velocity by substituting figures, note that 70 ft is 21.3 m for h

$V_{f}=\sqrt {(2gh)}= V_{f}=\sqrt {(2*9.81*21.3)}$ = 20.44275

Therefore, the divers enter with a speed of 20.4 m/s

The announcer's claim is incorrect because the divers enter at a speed of 20.4 and not 25 m/s as announced

(b)

The divers can enter water with a velocity of 25 m/s only if they have some initial velocity. Using the kinematic equation

$V_{f}^{2}=V_{i}^{2}+2gh$

Since we have final velocity of 25 m/s

$V_{i}^{2}=2gh-V_{f}^{2}$

$V_{i}=\sqrt{(V_{f}^{2}-2gh)}$

$V_{i}=\sqrt{(25^{2}-2*9.81*21.3)}$ = 14.390761 m/s

Therefore, it’s possible for a diver to enter the water with the velocity of 25 m/5 if he has initial velocity of 14.4 m/s

In conclusion, the upward initial velocity can’t be physically attained

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Answer:

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