Answer:
The theory is supported by all the available observations and data.
Explanation: The scientific community will accept a theory when a sufficient body of evidence supports it. This includes experiments that refute other potential theories. Experiments should also be carried out that attempt to disprove the theory but cannot.
It should not matter who proposed the theory or who supports it, and instead should entirely be based on the quality and abundance of data supporting it.
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Answer:
initial velocity=12.31 m/s
Final speed= 16.234 m/s
Explanation:
Given Data
height=5.72 m
distance=13.30 m
To Find
Initial Speed=?
Solution
Use the following equation to determine the time of the stone is falling.
d = vi ×t ½ ×9.8 × t²
Where
d = 5.72m and vi = 0 m/s
so
5.72 = ½× 9.8 ×t²
t = √(5.72 ÷ 4.9)
t=1.08 seconds
To determine the initial horizontal velocity use the following equation.
d = v×t
13.30 = v ×1.08
v = 13.30 ÷ 1.08
v=12.31 m/s
To determine stone’s final vertical velocity use the following equation
vf = vi+9.8×t............vi=0 m/s
vf = 9.8×1.08
vf= 10.584 m/s
To determine stone’s final speed use the following equation
Final speed = √[Horizontal velocity²+Final vertical velocity²]
Final speed = √{(12.31 m/s)²+(10.584 m/s)²}
Final speed= 16.234 m/s
Answer:
This represents radiation in ultra-violet region .
Explanation:
Energy of the orbit where n = 3 is given as follows
= -1.511 eV
Energy of the orbit where n = 1 is given as follows
= 13.6 eV
Difference of [tex]E_3 and [tex]E_1 = - 1.511+ 13.6
= 12.089 eV.
The wavelength of light having this energy in nm is given by the expression as follows
Wavelength in nm = 1244 / energy in eV
= 1244 / 12.089
= 102.90 nm
This represents radiation in ultra-violet region .
Answer:
8.2 m/s²
Explanation:
m = mass of the block
μ = Coefficient of kinetic friction = 0.17
= Normal force on the block by the ramp
= kinetic frictional force
Force equation perpendicular to ramp surface is given as
Kinetic frictional force is given as
Force equation parallel to ramp surface is given as
m/s²
Answer:
A.) 4.81 seconds
B.) 44.6 m/s
Explanation:
He begins his dive by jumping up with a velocity of 5 (m/s).
Let us first calculate the maximum height reached by using third equation of motion
V^2 = U^2 - 2gH
At maximum height, V = 0
0 = 5^5 - 2 × 9.8H
19.6H = 25
H = 25 /19.6
H = 1.28 m
The time taken for the diver to reach the water from the maximum height can be calculated by using second equation of motion.
Where height h = 1.28 + 100 = 101.28 m
h = Ut + 1/2gt^2
As the diver drop from maximum height, U = 0
101.28 = 1/2 × 9.8 × t^2
4.9t^2 = 101.28
t^2 = 101.28/4.9
t^2 = 20.669
t = sqrt ( 20.669)
t = 4.55s
As the diver jumped up, the time taken to reach the maximum height will be
Time = 1.28 / 5 = 0.256
The time taken for him to hit the water below will be 0.256 + 4.55 = 4.81 seconds
B.) Velocity right before he hits the water will be
V^2 = U^2 + 2gH
But U = 0
V^2 = 2 × 9.8 × 101.28
V^2 = 1985.09
V = 44.6 m/s
