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natali 33 [55]
3 years ago
5

A circus acrobat is shot out of a cannon with an initial upward speed of 34 ft/s. if the acrobat leaves the cannon 4 ft above th

e ground, how long will it take him to reach a net that is 8 ft above the ground
Physics
1 answer:
jok3333 [9.3K]3 years ago
6 0
<h2>It will take 0.125 seconds to reach the net.</h2>

Explanation:

Initial speed, u = 34 ft/s = 10.36 m/s

Acceleration, a = -9.81 m/s²

Displacement, s = Final height - Initial height = 8 - 4 = 4 ft = 1.22 m

We have equation of motion, s = ut + 0.5 at²

Substituting

              s = ut + 0.5 at²

              1.22 = 10.36 x t + 0.5 x -9.81 x t²

              4.905t² - 10.36 t + 1.22 = 0

              t = 1.99 s     or    t = 0.125 seconds

Minimum time is 0.125 seconds.

It will take 0.125 seconds to reach the net.

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snow_lady [41]

Answer:

Their velocity after the collision is 1.2 m/s, to the right.

Explanation:

Given;

mass of the blue car, m₁ = 2 kg

initial velocity of the blue car, u₁ = 6 m/s

mass of the red car, m₂ = 3 kg

initial velocity of the red car, u₂ = 2 m/s

let the blue car moving to the right be in positive direction

also, let the red car moving to the left be in negative direction

Apply the principle of conservation of linear momentum for inelastic collision.

m₁u₁ - m₂u₂ = v(m₁ + m₂)

where;

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7 0
2 years ago
An electron in the n = 5 level of an h atom emits a photon of wavelength 1282.17 nm. to what energy level does the electron move
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This is an interesting (read tricky!) variation of Rydberg Eqn calculation.
Rydberg Eqn: 1/λ = R [1/n1^2 - 1/n2^2]
Where λ is the wavelength of the light; 1282.17 nm = 1282.17×10^-9 m
R is the Rydberg constant: R = 1.09737×10^7 m-1
n2 = 5 (emission)
Hence 1/(1282.17 ×10^-9) = 1.09737× 10^7 [1/n1^2 – 1/25^2]
Some rearranging and collecting up terms:
1 = (1282.17 ×10^-9) (1.09737× 10^7)[1/n2 -1/25]
1= 14.07[1/n^2 – 1/25]
1 =14.07/n^2 – (14.07/25)
14.07n^2 = 1 + 0.5628
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Plug in and solve for W:

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