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frutty [35]
3 years ago
10

two objects are in uniform circular motion at the same speed but at different radii. the ones with the __ radius has the largest

centripetal acceleration.
Physics
2 answers:
ira [324]3 years ago
8 0

The one with the smaller radius has the greater centripetal FORCE.

We can't say anything about the centripetal acceleration, because that also depends on their masses, which you haven't told us.

Cerrena [4.2K]3 years ago
5 0

Answer:the one with the smaller radius has the highest centripetal force

Explanation:

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Investigations provide large amounts of information about a wide range of variables.
Klio2033 [76]

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8 0
3 years ago
A projectile is launched horizontally from a height of 8.0 m. The projectile travels 6.5 m before hitting the ground.
Effectus [21]
Time it takes the projectile to hit the ground after being thrown up:

√h/1/2a

√8/(.5)(9.81)

√8/4.905

√1.630988787

= 1.277101714

= 1. 28

hope this helps :)
7 0
3 years ago
A uniform electric field has a magnitude 1.80 kV/m and points in the +x direction. (a) What is the electric potential difference
EastWind [94]

Answer:

a)ΔV = 6.48 KV

b)ΔU =18.79 mJ

Explanation:

Given that

E= 1.8 KV/m

a)

We know that

Electric potential difference  ΔV given as

ΔV = E .d

Here

E= 1.8 KV/m

d= 3.6 m

ΔV = E .d

ΔV = 1.8 x 3.6 KV

ΔV = 6.48 KV

b)

Given that

q=+2.90 µC

Change in electric potential energy ΔU given as

ΔU = q .ΔV

\Delta U=2.9\times 10^{-6}\times 6.48\times 10^3\ J

ΔU =18.79 mJ

8 0
3 years ago
By calculating its wavelength (in nm), show that the second line in the Lyman series is UV radiation.
Rashid [163]

Answer:

 λ = 102.78  nm

This radiation is in the UV range,

Explanation:

Bohr's atomic model for the hydrogen atom states that the energy is

           E = - 13.606 / n²

where 13.606 eV   is the ground state energy and n is an integer

an atom transition is the jump of an electron from an initial state to a final state of lesser emergy

            ΔE = 13.606 (1 / n_{f}^{2} - 1 / n_{i}^{2})

the so-called Lyman series occurs when the final state nf = 1, so the second line occurs when ni = 3, let's calculate the energy of the emitted photon

            DE = 13.606 (1/1 - 1/3²)

            DE = 12.094 eV

let's reduce the energy to the SI system

            DE = 12.094 eV (1.6 10⁻¹⁹ J / 1 ev) = 10.35 10⁻¹⁹ J

let's find the wavelength is this energy, let's use Planck's equation to find the frequency

            E = h f

             f = E / h

            f = 19.35 10⁻¹⁹ / 6.63 10⁻³⁴

            f = 2.9186 10¹⁵ Hz

now we can look up the wavelength

           c = λ f

           λ = c / f

           λ = 3 10⁸ / 2.9186 10¹⁵

           λ = 1.0278  10⁻⁷ m

let's reduce to nm

            λ = 102.78  nm

This radiation is in the UV range, which occurs for wavelengths less than 400 nm.

5 0
3 years ago
18. The orbits of planets being elliptical was one the planetary laws developed by
TEA [102]
It was a man named <span>Johannes Kepler. </span>
6 0
3 years ago
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