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What is the acceleration of the the object during the first 4 seconds?

AVprozaik [17]

Answer:

Velocity (m/s) over time (s) graph

We could write out our average acceleration as:

a = Δv/ Δta=Δv/Δta, equals, Δ, v, slash, Δ, t

a = (15 m/s - 0 m/s) / 0.2 seconds

a = 15 m/s / 0.2 seconds

a = 75 m/s / second

Explanation:

What this formula is telling us is that if we know the acceleration of an object, and the ... we can plug in our acceleration of 12.5 m/s2 for a, and 4 seconds for t.

Velocity (m/s) over time (s) graph

We could write out our average acceleration as:

a = Δv/ Δta=Δv/Δta, equals, Δ, v, slash, Δ, t

a = (15 m/s - 0 m/s) / 0.2 seconds

a = 15 m/s / 0.2 seconds

a = 75 m/s / second

6 0

3 years ago

Can someone make a list of 10 simple things people can do to maintain fitness ? Please I literally need help.

ivanzaharov [21]

Answer:

1. drink lots of water

2. eat healthy breakfast

3. plan your meals

4. set a deadline

5. exercise at least 30 minutes a day even 10 minutes 3 times a day is good

6. eat fruits and vegetables

7. eat whole grains instead of white

8. get a good night's sleep

9. stay motivated

10. take time for mental health

6 0

3 years ago

Read 2 more answers

Ladybug walks 10 cm forward and 5 cm backwards in 20 seconds what is the average speed of ladybug what is the average velocity

lbvjy [14]

Answer:

Average speed = 0.0075 m/s

Average velocity = 0.0025 m/s along forward direction

Explanation:

Speed is the ratio of distance and time and velocity is the ratio of displacement and time.

Distance traveled = 10 + 5 = 15 cm = 0.15 m

Displacement = 10 - 5 = 5 cm = 0.05 m

Time = 20 seconds

$\texttt{Average speed = }\frac{\texttt{Distance}}{\texttt{Time}}\\\\\texttt{Average speed = }\frac{0.15}{20}=7.5\times 10^{-3}m/s\\\\\texttt{Average velocity = }\frac{\texttt{Displacement}}{\texttt{Time}}\\\\\texttt{Average velocity = }\frac{0.05}{20}=2.5\times 10^{-3}m/s$

Average speed = 0.0075 m/s

Average velocity = 0.0025 m/s along forward direction

3 0

3 years ago

Read 2 more answers

A spring with a constant k=400n/m shoots a 1. 00kg ball up a frictionless incline after being compressed 0. 150m. what is the ma

Zielflug [23.3K]

The maximum height reached by the ball is 0.46m.

To find the answer, we have to know about the potential energy of a spring mass system.

<h3>How to find the maximum height reached by the ball?</h3>

It is given that,

$k=400N/m\\m=1kg\\x=0.150m\\h=?\\$

We have to find the maximum height reached by the ball.
Thus, we have the expression for potential energy of a spring mass system and that of gravitational field as,

$U=\frac{1}{2}kx^2 \\U=mgh$