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Fantom [35]
3 years ago
8

A student at the top of a building of height h throws one ball upward with a speed of vi and then throws a second ball downward

with the same initial speed vi. Just before it reaches the ground, is the final speed of the ball thrown upward __________ compared with the final speed of the ball thrown downward
Physics
1 answer:
Agata [3.3K]3 years ago
5 0

Answer:

The velocity of the one thrown up will be the same as the second one

Explanation:

They will fall and hit the ground at the same time although they have the same velocity because object one although has double height it has initial velocity of zero

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A skier starts from rest down a slope 500.0 M long, the skier accelerates at a constant rate of 2.00 m/s/s, what's the velocity
nevsk [136]
We can use the kinematic equation
(v_f)^2 = (v_i)^2 + 2*a*d
where Vf is what we are looking for
Vi is 0 since we start from rest
a is acceleration
and d is the distance

we get
(Vf)^2 = (0)^2 + 2*(2)*(500)
(Vf)^2 = 2000
Vf = about 44.721
or 44.7 m/s   [if you are rounding this by significant figures]
8 0
3 years ago
A star is moving toward the earth at 250 km/s. What can you say about the wavelength of the h-alpha absorption line that you wou
Juli2301 [7.4K]

The wavelength of the h-alpha absorption line that you would detect in the star's spectrum will be 54.67 *  10^{-2} nm

If the object is moving towards or away from us with some radial velocity, shifts can be observed in the location of the absorption

since, material is moving towards us  a shift to shorter wavelength will be observed

shift = rest wavelength * ( radial velocity / speed of light )

        = 656 * 10^{-9} * (250 * 10^{3}  / 3 * 10^{8} )

        = 54.67 *  10^{-2} nm

The wavelength of the h-alpha absorption line that you would detect in the star's spectrum will be 54.67 *  10^{-2} nm

learn more about radial velocity:

brainly.com/question/22409504?referrer=searchResults

#SPJ4

6 0
2 years ago
An insect is able to walk across the surface of a pond without sinking because of the
fiasKO [112]
<span>C. surface tension of the water.</span>
3 0
3 years ago
Read 2 more answers
A key lime pie in a 10.00 inch diameter plate is placed upon a rotating tray. Then, the tray is rotated such that the rim of the
juin [17]

Answer:

The angular distance in revolution is revolution = 3.439 \ revolution

The angular distance in radians  is \theta_{rad}= 21.6 \ radians

The angular distance in degrees  is   \theta =1238.04^o

The angular size is   Z = \frac{2}{9} \pi \ radians

Explanation:

From the question we are told that  

  The diameter is d = 10 \ inches

   The distance moved by the rim is D = 108 \ inches

Generally the circumference of the pie plate is mathematically represented as

               C = \pi d

Substituting the values  

          C = 10 *3.142

             = 31.42 \  inches

The number resolution carried out by the pie plate is evaluated as

           revolution  = \frac{D}{C}

Substituting value

            revolution  = \frac{108}{31.4}

                              revolution = 3.439 \ revolution

The angular  distance \theta_{rad} is mathematically evaluated as

                \theta_{rad} = \frac{D}{r}

             Where r is the radius which is mathematically evaluated as

                        r = \frac{d}{2} = \frac{10}{2}  = 5 \ inches

Substituting  this into the equation for angular distance

                      \theta_{rad} = \frac{108}{5}

                              \theta_{rad}= 21.6 \ radians

The angular distance traveled in degrees is

                   \theta =  3.439 *360

                      \theta =1238.04^o

When  the pie is cut into 9 equal parts

The angular size would be mathematically evaluated as

                         Z = \frac{2\pi}{9}

                           Z = \frac{2}{9} \pi \ radians

                             

7 0
4 years ago
A uranium ion and an iron ion are separated by a distance of =61.10 nm. The uranium atom is singly ionized; the iron atom is dou
scoray [572]

Answer:

Explanation:

Charge on uranium ion = charge of a single electron

= 1.6 x 10⁻¹⁹ C

charge on doubly ionised iron atom = charge of 2 electron

= 2 x 1.6 x 10⁻¹⁹ C = 3.2 x 10⁻¹⁹ C

Let the required distance from uranium ion be d .

force on electron at distance d from uranium ion

= 9 x 10⁹ x 1.6 x 10⁻¹⁹ / r²

force on electron at distance 61.10 x 10⁻⁹ - r from iron  ion

= 9 x 10⁹ x 3.2 x 10⁻¹⁹ / (61.10 x 10⁻⁹ - r )²

For equilibrium ,

9 x 10⁹ x 1.6 x 10⁻¹⁹ / r² = 9 x 10⁹ x 3.2 x 10⁻¹⁹ / (61.10 x 10⁻⁹ - r )²

2 d² = (61.10 x 10⁻⁹ - r )²

1.414 r = 61.10 x 10⁻⁹ - r

2.414 r = 61.10 x 10⁻⁹

r = 25.31 nm .

4 0
3 years ago
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