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lbvjy [14]
3 years ago
8

How are acceleration and speed related​

Physics
2 answers:
Gre4nikov [31]3 years ago
6 0

Answer:

Acceleration is the rate of change of velocity. Usually, acceleration means the speed is changing, but not always. When an object moves in a circular path at a constant speed, it is still accelerating, because the direction of its velocity is changing. Comment on robshowsides's post “Speed is the magnitude of velocity.

Explanation:

hope it helped tee hee

Serggg [28]3 years ago
4 0
Acceleration is the rate of change of velocity. Usually, acceleration means the speed is changing, but not always. When an object moves in a circular path at a constant speed, it is still accelerating, because the direction of its velocity is changing. Comment on robshowsides's post “Speed is the magnitude of velocity.
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An object buoyant force and wieght arent the smae thing
Alona [7]

I hope this is true or false



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3 years ago
A man hits a 50 grams golf ball such that it leaves the tee at an angle of 40o with the horizontal and strikes the ground at the
lukranit [14]

Answer:

Explanation:

Range of projectile R = 20 m

formula of range

R = u² sin2θ / g

u is initial velocity , θ is angle of projectile

putting the values

20 = u² sin2x 40 / 9.8

u² = 199

u = 14.10 m /s

At the initial point

vertical component of u

= u sin40 = 14.1 x sin 40

= 9.06 m/s

Horizontal component

= u cos 30

At the final point where the ball strikes the ground after falling , its speed remains the same as that in the beginning .

Horizontal component of velocity

u cos 30

Vertical component

= - u sin 30

= - 9.06 m /s

So its horizontal component remains unchanged .

change in vertical component = 9.06 - ( - 9.06 )

= 18.12 m /s

change in momentum

mass x change in velocity

= .050 x 18.12

= .906 N.s

Impulse = change in momentum

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3 0
4 years ago
Model a hydrogen atom as a three-dimensional potential well with Uo = 0 in the region 0 &lt; x a. 283 eV <br> b. 339 eV <br> c.
denis23 [38]

This question is incomplete, the complete question is;

Model a hydrogen atom as a three-dimensional potential well with U₀ = 0 in the region 0 < x < L, 0 < y < L and 0 < z < L, and infinite otherwise, with L = 1.0 × 10⁻¹⁰ m.

Which of the following is NOT one of the lowest three energy levels of an electron in this model?

a. 283 eV

b. 339 eV

c. 113   eV  

d. 226 eV        

Answer:

the lowest three energy are; 113 eV, 225 eV, and 339 eV.

Hence Option a) 283 eV is not among the three lowest energy

Explanation:

Given the data in the question;

Three dimension cube or particle in a cubic box

the energy value is given by;

E_{nx,ny,nz = ( n_x^2 + n_y^2 + n_z^2 ) × π²h"² / 2ml²

where h" = h/2π and h is Planck's constant ( 6.626 × 10⁻³⁴ m² kg / s )

m is mass of electron ( 9.1 × 10⁻³¹ kg )

l is length of side of box ( 1.0 × 10⁻¹⁰ m )

for ground level ( n_x = n_y = n_z = 1 )

so

( n_x^2 + n_y^2 + n_z^2 ) ×  π²h"² / 2ml²

since h" = h/2π

( n_x^2 + n_y^2 + n_z^2 ) × π²h² / (2π)²2ml²

so we substitute

E_{111 = ( 1² + 1² + 1² ) × [ π²( 6.626 × 10⁻³⁴ )² ] / [ (2π)² × 2 × 9.1 × 10⁻³¹ kg × ( 1.0 × 10⁻¹⁰)² ]

E_{111 = 3 × [ (4.333188779 × 10⁻⁶⁶) / ( 7.185072 × 10⁻⁴⁹ ) ]    

E_{111 = 3 × [ 6.03082165 × 10⁻¹⁸ ]

Now, we know that electric charge = 1.602 x 10⁻¹⁹

so

E_{111 = 3 × [ (6.03082165 × 10⁻¹⁸) / (1.602 x 10⁻¹⁹) ]

E_{111 = 3 × [ 37.645578 ]

E_{111 = 112.9 ≈ 113 eV

E_{211 = ( n_x^2 + n_y^2 + n_z^2 )  × π²h² / (2π)²2ml²

we substitute

E_{211 = ( 1² + 1² + 2² ) × [ 37.645578 ]

E_{211 = 6 × [ 37.645578 ]

E_{211 = 225.87 ≈ 226 eV

E_{221 = ( n_x^2 + n_y^2 + n_z^2 )  × π²h² / (2π)²2ml²

we substitute

E_{221 = ( 2² + 2² + 1² ) × [ 37.645578 ]

E_{211 = 9 × [ 37.645578 ]

E_{211 = 338.8 ≈ 339 eV

Therefore, the lowest three energy are; 113 eV, 225 eV, and 339 eV.

Hence Option a) 283 eV is not among the three lowest energy

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3 years ago
If the sun is a medium sized star, why does it look bigger than others?
nata0808 [166]

Answer:

The sun looks bigger than other stars because it is closer to the Earth, distance makes it look larger

5 0
3 years ago
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