If the velocity of an object changes from 65 m/s to 98 m/s during a time interval of 12 s, what is the acceleration of the object?

Answers

Answer 1
Answer: a=acceleration
vf=final velocity
v₀=original velocity
t=time period

a=(vf-v₀)/t
a=(98 m/s-65 m/s)/12s=33 m/s / 12 s=2.75 m/s²

Answer: the acceleration of the object is 2.75 m/s²
Answer 2
Answer:
     Size of acceleration = (change in speed) / (time for the change)

Change in speed = (end speed) - (start speed)

                             =     (98 m/s)    -    (65 m/s)  =  33 m/s .

Time for the change = 12 s

Size of acceleration =  (33 m/s) / (12 s)

                                   =    (33/12) (m/s²)

                                   =        2.75 m/s² .

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Can anyone help me with this?

Answers

Answer:

The answer for this question is 146 neutrons

With his Square Deal, Theodore Roosevelt hoped toa. help pull the United States out of the Great Depression.
b. increase the influence and power of the United States among the nations of the world.
c. keep the wealthy and powerful from taking advantage of small business owners and the poor.
d. help business trusts remain competitive.

Answers

3/C seems right.4/D is off for the subject.1/A FDR was the Roosevelt during the Depression, not Teddy.2/B is wrong because Teddy was focused on the US

As you (increase, decrease) in altitude, air pressure decreases.

Answers

Answer:

as you increase in altitude, air pressure decreases

True of false A millimeter I'd larger than a kilometer. So, these prefixes are listed in order of increasing size going from left to right

Answers

False, a millimeter is smaller than a kilometer.

A 30cm diameter wheel rotates with a constant angular acceleration of 2.00 rad/s2. It starts from rest at t = 0, and a chalk line drawn to a point, P, on the rim of the wheel makes an angle of 37.1° with the horizontal at this time. At t = 2.00 s, find (a) the angular speed of the wheel (b) the linear velocity and tangential acceleration of P

Answers

Answer:

a) 4.00 rad/s

b) 0.6 m/s, 0.3 m/s²

Explanation:

Given:

ω₀ = 0 rad/s

α = 2.00 rad/s²

t = 2.00 s

Find:

a) ω

b) v, a

a) ω = αt + ω₀

ω = (2.00 rad/s²) (2.00 s) + (0 rad/s)

ω = 4.00 rad/s

b) v = ωr

v = (4.00 rad/s) (0.15 m)

v = 0.6 m/s

a = αr

a = (2.00 rad/s²) (0.15 m)

a = 0.3 m/s²

Name the physical quantity which changes contenously during uniform-circular motion.

Answers

The direction of the motion is constantly changing during
motion over any closed path, not only circular.