All categories

2 years ago

If an object has an initial velocity of 2.0 m/s and its rate of change in velocity is 3.0 m/s2

Physics

1 answer:

erica [24]2 years ago

6 0

Answer:

14m/s2

Explanation:

initial velocity=2m/s

final velocity =?

time=4s

acceleration=3.0m/s

v=u+at

v=(2)+(3*4)

v=14m/s

by substituting the values in the equation we get the value 14m/s.

You might be interested in

There are 13 boys are 4 girls in the jazz band find the ratio of the number of boys in the band to the total number of students

Nikitich [7]

13:17 b/c 13 + 4 =17...

7 0

2 years ago

Which of the following refers to stored energy that an object has due to its position?

mezya [45]

Explanation:

Energy stored in an object due to its position is Potential Energy. Energy that a moving object has due to its motion is Kinetic Energy.

5 0

2 years ago

The Sun has more gravity than the other planets in our solar system because the Sun is

IRISSAK [1]

B the mass of the solar object determines whether the gravity of the object unless the solar object has Sonic property's such as a neutron star which can be the size of Pluto but have the mass of 900 solar masses

5 0

2 years ago

Read 2 more answers

wo slightly out of tune instruments will create beats when the same note is played. This is a pattern of louder and softer sound

Lostsunrise [7]

This interaction is known as constructive interference. It's a result of linear superposition.

5 0

2 years ago

Of the following transitions in the Bohr hydrogen atom, the ________ transition results in the absorption of the highest-energy

8_murik_8 [283]

Answer:

The n = 2 → n = 3 transition results in the absorption of the highest-energy photon.

Explanation:

$E_n=-13.6\times \frac{Z^2}{n^2}ev$

Formula used for the radius of the $n^{th}$ orbit will be,

where,

$E_n$ = energy of $n^{th}$ orbit

n = number of orbit

Z = atomic number

Here: Z = 1 (hydrogen atom)

Energy of the first orbit in H atom .

$E_1=-13.6\times \frac{Z^2}{1^2} eV=-13.6 eV$

Energy of the second orbit in H atom .

$E_2=-13.6\times \frac{Z^2}{(2)^2} eV=-3.40 eV$

Energy of the third orbit in H atom .

$E_3=-13.6\times \frac{Z^2}{(9)^2} eV=-1.51 eV$

Energy of the fifth orbit in H atom .

$E_5=-13.6\times \frac{Z^2}{(2)^2} eV=-0.544 eV$

Energy of the sixth orbit in H atom .

$E_6=-13.6\times \frac{Z^2}{(2)^2} eV=-0.378 eV$

Energy of the seventh orbit in H atom .

$E_7=-13.6\times \frac{Z^2}{(2)^2} eV=-278 eV$

During an absorption of energy electron jumps from lower state to higher state.So, absorption will take place in :

1) n = 2 → n = 3

2) n= 5 → n = 6

Energy absorbed when: n = 2 → n = 3

$E=E_3-E_2$

$E=(-1.51 eV) -(-3.40 eV)=1.89 eV$

Energy absorbed when: n = 5 → n = 6

$E'=E_6-E_5$

$E'=(-0.378 eV)-(-0.544 eV) =0.166 eV$

1.89 eV > 0.166 eV

E> E'

So,the n = 2 → n = 3 transition results in the absorption of the highest-energy photon.

4 0

2 years ago