The answer is d that might help you
Answer:
1

2
The distance is
Explanation:
From the question we are told that
The maximum speed of the cheetah is 
The maximum of gazelle is 
The distance ahead is 
Let
denote the time which the cheetah catches the gazelle
Gnerally the equation representing the distance the cheetah needs to move in order to catch the gazelle is

=> 
=> 
=> 
Now at t = 7.5 s

=> 
=> 
=>
Hence the for the gazelle to escape the cheetah it must be 55.2 m
Answer:
Explanation:
Threshold frequency = 4.17 x 10¹⁴ Hz .
minimum energy required = hν where h is plank's constant and ν is frequency .
E = 6.6 x 10⁻³⁴ x 4.17 x 10¹⁴
= 27.52 x 10⁻²⁰ J .
wavelength of radiation falling = 245 x 10⁻⁹ m
Energy of this radiation = hc / λ
c is velocity of light and λ is wavelength of radiation .
= 6.6 x 10⁻³⁴ x 3 x 10⁸ / 245 x 10⁻⁹
= .08081 x 10⁻¹⁷ J
= 80.81 x 10⁻²⁰ J
kinetic energy of electrons ejected = energy of falling radiation - threshold energy
= 80.81 x 10⁻²⁰ - 27.52 x 10⁻²⁰
= 53.29 x 10⁻²⁰ J .
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