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valina [46]
4 years ago
12

A busy chipmunk runs back and forth along a straight line of acorns that has been set out between his burrow and a nearby tree.

At some instant the chipmunk moves with a velocity of -1.15 m/s. Then, 2.11 s later, it moves at a velocity of 1.63 m/s. What is the chipmunk's average acceleration during the 2.11-s time interval in m/s^2?
Physics
1 answer:
saveliy_v [14]4 years ago
6 0

Answer: a = 1.32m/s2

Therefore, the average acceleration is 1.32m/s2

Explanation:

Acceleration is the rate of change in the velocity per time

a = change in velocity/time

a = ∆v/t

average acceleration a = (v2 -v1)/t. ....1

Given;

Final velocity v2 = 1.63m/s

Initial velocity v1 = -1.15ms

time taken t = 2.11s

Substituting into eqn 1

a = [1.63 - (-1.15)]/2.11

a = (1.63+1.15)/2.11

a = 2.78/2.11

a = 1.32m/s2

Therefore, the average acceleration is 1.32m/s2

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1 hour = 3600 seconds.
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What is a plane mirror ? ​
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A plain mirrior is a mirrior with flat reflective surface.

hope it is helpful for you.

6 0
3 years ago
Which phrase best describes sir isaac newton's contributions to modern science and therefore to the industrial revolution?
astraxan [27]
Can you please give the phrases? 

But, I'll help what I can.

First, he was the first to discover gravity. He was not bonked by the head by an apple, rather he watched an apple fall from a tree before he decided to explore gravity further. 

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7 0
3 years ago
Please help ASAP guys !!! Thanks
andreev551 [17]

Answer:

(a) gravitational potential energy converted to kinetic energy

(b) chemical energy is converted to light or heat energy

(c) mechanical energy is converted into kinetic energy

7 0
2 years ago
Current passes through a solution of sodium chloride. In 1.00 second, 2.68×1016 Na+ ions arrive at the negative electrode and 3.
sladkih [1.3K]

Answer:

10.6 mA

Explanation:

t = time interval = 1.00 s

q = magnitude of charge on each ion = 1.6 x 10⁻¹⁹ C

n₁ = number of Na⁺ ions = 2.68 x 10¹⁶

q₁ = charge due to Na⁺ ions = n₁ q = (2.68 x 10¹⁶) (1.6 x 10⁻¹⁹) = 0.004288 C

n₂ = number of Cl⁻ ions = 3.92 x 10¹⁶

q₂ = charge due to Cl⁻ ions = n₂ q = (3.92 x 10¹⁶) (1.6 x 10⁻¹⁹) = 0.006272 C

i₁ = Current due to Na⁺ ions = \frac{q_{1}}{t} = \frac{0.004288}{1} = 0.004288 A

i₂ = Current due to Cl⁻ ions = \frac{q_{2}}{t} = \frac{0.006272}{1} = 0.006272 A

Current passing between the electrodes is given as

i = i₁ + i₂

i = 0.004288 + 0.006272

i = 0.01056 A

i = 10.6 x 10⁻³ A

i = 10.6 mA

8 0
3 years ago
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