1.)
Velocity is in m/s, and acceleration is in m/s^2 like you said. Because of this, we can calculate this by dividing the speed by the time it took to get to that speed.
(20 meters/second) / 10 seconds = 2 meters/ second^2

2.)
Same thing with the first one.
(100 meters/second) / 4 seconds = 25 meters / seconds^2

7 0

2 years ago

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What type of bond results from the side‑on overlap of orbitals?

Serga [27]

Answer:

A pi bond

Explanation:

A pi bond is a type of covalent bond that results from the formation of a molecular orbital by the side-to-side overlap of atomic orbitals along a plane perpendicular to a line connecting the nuclei of the atoms.

4 0

2 years ago

The tension in cable da has a magnitude of tda=6.27 lb. find the cartesian components of tension tda, which is directed from d t

Oliga [24]

Complete Question

The Complete Question is attached below

We have that the Cartesian components of tension $T_{da}$ is

$T_{DA}=-4.433i-3.49j+2.735k$

From the Question we are told that

$M_{da}=6.27 lb\\\\w=9.50ft\\\\d=6.60ft\\\\h=4.50ft$

$\vec {DA}=-4.7i-3.7j+2.9k)ft$

$\vec {DB}=-1.9i-3.7j+1.9k)ft\\\\\vec {DC}=-1.9i+5.8j-1.6k)ft$

Generally the equation for $T_{DA}$ is mathematically given as

$T_{DA}=\phi_{DA}* M_{da}$

Where

$\phi_{DA}=\frac{-4.7i-3.7j+2.9k}{(-4.7)^2+(-3.7)^2+(2.9)^2}\\\\\phi_{DA}=\frac{-4.7i-3.7j+2.9k}{6.65}$

Therefore

$T_{DA}=\phi_{DA}* M_{da}$

$T_{DA}=\frac{-4.7i-3.7j+2.9k}{6.65}* 6.27$

$T_{DA}=-4.433i-3.49j+2.735k$

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5 0

3 years ago

Consider two circular metal wire loops each carrying the same current I as shown below. In what region(s) could the net magnetic

nadezda [96]

Answer:

you counteract the Chromozones with the numerzones which contacts the specimen in the brain cells which then has voltage in the hand that connects with the wire then wallah

Explanation:

7 0

3 years ago