Complete Question
An airplane takes off a runway at a constant speed of 49 m/s at constant angle 30 to the horizontal.How high (in meters ) is the airplane above the ground 13 seconds after takeoff?
Answer:
The height is 
Explanation:
From the question we are told that
The speed at which the plane takes off is 
The angle at which it takes off is 
The time taken is 
The vertical distance traveled is mathematically represented as
Substituting values
Momentum formula - P=MV
P=MV
P= (3.5 kg)(6 m/s) - multiply
P= 21 kg x m/s to the east
Hope this helps
Answer:
<h2>
E = 2.8028*10⁻¹⁹ Joules</h2>
Explanation:
The minimum energy needed to eject electrons from a metal with a threshold frequency fo is expressed as E = hfo
h = planck's constant
fo = threshold frequency
Given the threshold frequency fo = 4.23×10¹⁴ s⁻¹
h = 6.626× 10⁻³⁴ m² kg / s
Substituting this value into the formula to get the energy E
E = 4.23×10¹⁴ * 6.626 × 10⁻³⁴
E = 28.028*10¹⁴⁻³⁴
E = 28.028*10⁻²⁰
E = 2.8028*10⁻¹⁹ Joules
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Answer:
makes a transition from a high to a lower energy state
Explanation:
The emission spectrum is the spectrum of electromagnetic radiation emitted due to the transition from a state of high energy to a state of lower energy made by an electron in the atoms of the chemical element or compound. The emitted photons have an energy equal to the difference in energy between the two states.
