Acceleration = (change in speed) / (time for the change)
change in speed = (speed at the end) - (speed at the beginning)
change in speed = (37 km/hr) - (89 km/hr) = -52 km/hr
Acceleration = (-52 km/hr) / (6 sec)
Acceleration = (-26/3) km/(hr·sec)
Units: (1/hr·sec) · (hr/3600 sec) = 1 / 3600 sec²
(-26/3) km/(hr·sec) = (-26/3) km/(3600 sec²)
= -26,000/(3 · 3600) m/s²
<em>Acceleration = -2.41 m/s²</em>
The answer is C, as there is not increase or decrease in speed during that time frame.
Answer:
E3 = 3.03 10⁻¹⁶ kJ, E4 = 4.09 10⁻¹⁶ kJ and E5 = 4.58 10⁻¹⁶ kJ
Explanation:
They give us some spectral lines of the Balmer series, let's take the opportunity to place the values in SI units
n = 3 λ = 656.3 nm = 656.3 10⁻⁹ m
n = 4 λ = 486.1 nm = 486.1 10⁻⁹ m
n = 5 λ=434.0 nm = 434.0 10⁻⁹ m
Let's use the Planck equation
E = h f
The speed of light equation
c = λ f
replace
E = h c /λ
Where h is the Planck constant that is worth 6.63 10⁻³⁴ J s and c is the speed of light that is worth 3 10⁸ m / s
Let's calculate the energies
E = 6.63 10⁻³⁴ 3 10⁸ / λ
E = 19.89 10⁻²⁶ /λ
n = 3
E3 = 19.89 10⁻²⁶ / 656.3 10⁻⁹
E3 = 3.03 10⁻¹⁹ J
1 kJ = 10³ J
E3 = 3.03 10⁻¹⁶ kJ
n = 4
E4 = 19.89 10⁻²⁶ /486.1 10⁻⁹
E4 = 4.09 10⁻¹⁹ J
E4 = 4.09 10⁻¹⁶ kJ
n = 5
E5 = 19.89 10⁻²⁶ /434.0 10⁻⁹
E5 = 4.58 10⁻¹⁹ J
E5 = 4.58 10⁻¹⁶ kJ
Answer:
The acceleration that earth experiences due to gravitational pull is = 9.81 
.
Explanation:
The acceleration that earth experiences due to gravitational pull is called the acceleration due to gravity. its value is 9.81 and its unit is .
When the object move upwards than in that case the earth gravitational force pulls down the body.
The formula of force due to gravity on the body is given as
F = mg
where g = acceleration due to gravity.
Due to this acceleration the body falls upon the surface of the earth.
