Please need help with this

Physics

2 answers:

Stella [2.4K]3 years ago

4 0

Answer:

D. The objects particles move faster

ozzi3 years ago

3 0

Answer:

D. The objects particles move faster

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As a mercury atom absorbs a photon of

Allisa [31]

The energy absorbed by photon is 1.24 eV.
This is the perfect answer.

8 0

3 years ago

The rocket is fired vertically and tracked by the radar station shown. When θ reaches 66°, other corresponding measurements give

Flauer [41]

Answer:

velocity = 1527.52 ft/s

Acceleration = 80.13 ft/s²

Explanation:

We are given;

Radius of rotation; r = 32,700 ft

Radial acceleration; a_r = r¨ = 85 ft/s²

Angular velocity; ω = θ˙˙ = 0.019 rad/s

Also, angle θ reaches 66°

So, velocity of the rocket for the given position will be;

v = rθ˙˙/cos θ

so, v = 32700 × 0.019/ cos 66

v = 1527.52 ft/s

Acceleration is given by the formula ;

a = a_r/sinθ

For the given position,

a_r = r¨ - r(θ˙˙)²

Thus,

a = (r¨ - r(θ˙˙)²)/sinθ

Plugging in the relevant values, we obtain;

a = (85 - 32700(0.019)²)/sin66

a = (85 - 11.8047)/0.9135

a = 80.13 ft/s²

4 0

3 years ago

A 3kg object has an initial velocity (6i - 2j) m/s (a ) what is its kinetic energy at this time? (b) Find total work done on the

guapka [62]

Answer:

$K.E = \frac{1}{2} m {v}^{2} \\ {v}^{2}_i = {v}^{2} _x + {v}^{2} _y \\ = {(6)}^{2} + {( - 2)}^{2} = 36 + 4 = 40m. {s}^{ - 1} \\ K.E_i = \frac{1}{2} (3) (40) = 60J \\ \\ {v}^{2}_f = {v}^{2} _x + {v}^{2} _y \\ = {(8)}^{2} + {(4)}^{2} = 80m. {s}^{ - 1} \\ K.E_i = \frac{1}{2} (3) (80) = 120J \\ W_{net}=K.E_f-K.E_i \\ = 120J - 60J \\ = 60J$