Plz help me

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.

saveliy_v [14]

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

6 0

3 years ago

What is the magnitude of the gravitational force of attraction to Jupiter exerts on IO

kotegsom [21]

The gravity force between Jupiter and Io will be 6.343 × 10²² N.

<h3>What is Newton's law of gravitation?</h3>

Newton's law of gravity states that each particle having mass in the universe attracts each other particle with a force known as the gravitational force.

Given data;

Mass of Jupiter, $\rm m_j = 1.9 \times 10^{27} \ kg$

Mass of moon of Jupiter, $\rm m_{i_0}= 8.9 \times 10^{22} \ kg$ ]

The gravitational constant is, $\rm G = 6.67 \times 10^{-11 } \ m^3 kg^{-1} s^{-2}$

Distance between Jupiter and Io, R = 421,700 km = 4,217,00,000 m

The gravitational force is proportional to the product of the masses of the two bodies and inversely proportional to the square of their distance.

The gravitational force is found as;

$\rm F = G \frac{ m_J m_{I_0}}{R^2} \\\\\ F = (6.67\times 10^{-11}) \frac{( (1.9\times 10^{27})\times (8.9\times 10^{22} )} { (421700000)^2}\\\\ F_g = 6.343 \times 10^{22} \ N$

Hence, the gravity force between Jupiter and Io will be 6.343 × 10²² N.

The complete question is

"Jupiter has a mass of 1.9 × 1027 kg, and its moon Io has a mass of 8.9 × 1022 kg. Their centers are separated by a distance of 421,700 km what is the force of gravity acting on Io? "

To learn more about Newton's law of gravitation, refer to the link.

brainly.com/question/9699135.

#SPJ1

8 0

2 years ago

Which phase of the Moon occurs when Earth, the Moon, and the Sun lie on the same line?

ss7ja [257]

The answer is B I think.

3 0

3 years ago

Danny is riding his bike at 12m/s he tries to pop a wheelie but he fails hits a curb flies through the air and comes to a comple

maxonik [38]

Answer:

a = -0.4m/s²

Explanation:

v_f = v_I + (a)(t)

a(t) = v_f-v_I

a = (v_f-v_I)/t

a = (0m/s-12m/s)/30s

a = -0.4m/s²

8 0

2 years ago

Challenge! A marshmallow is dropped from a 5-meter high pedestrian bridge and 0.83 seconds later, it lands right on the head of

WINSTONCH [101]

a₀). You know ...
       -- the object is dropped from 5 meters
   above the pavement;
         -- it falls for 0.83 second.

a₁). Without being told, you assume ...
   -- there is no air anyplace where the marshmallow travels,
   so it free-falls, with no air resistance;
   -- the event is happening on Earth,
where the acceleration of gravity is 9.81 m/s² .

b). You need to find how much LESS than 5 meters
   the marshmallow falls in 0.83 second.

c). You can use whatever equations you like.
   I'm going to use the equation for the distance an object falls in
   ' T ' seconds, in a place where the acceleration of gravity is ' G '.

d). To see how this all goes together for the solution, keep reading:

The distance that an object falls in ' T ' seconds
when it's dropped from rest is

   (1/2 G) x (T²) .

On Earth, ' G ' is roughly 9.81 m/s², so in 0.83 seconds,
such an object would fall

   (9.81 / 2) x (0.83)² = 3.38 meters .

It dropped from 5 meters above the pavement, but it
only fell 3.38 meters before something stopped it.
So it must have hit something that was

   (5.00 - 3.38) = 1.62 meters

above the pavement. That's where the head of the unsuspecting
person was as he innocently walked by and got clobbered.

7 0

3 years ago