Answer:
Wavelength λ = 7.31 × 10^-37 m
Explanation:
From De Broglie's equation;
λ = h/mv
Where;
λ = wavelength in meters
h = plank's constant = 6.626×10^-34 m^2 kg/s
m = mass in kg
v = velocity in m/s
Given;
v = 24 mi/h
Converting to m/s
v = 24mi/h × 0.447 m/s ×1/(mi/h)
v = 10.73m/s
m = 84.5kg
Substituting the values into the equation;
λ = (6.626×10^-34 m^2 kg/s)/(84.5kg × 10.73m/s)
λ = 7.31 × 10^-37 m
Answer:
Mass and thus force depends on the reference frame chosen
Explanation:
This can be explained as Newton's law of gravity provides action which are instantaneous at a distance and involves the evaluation of all the quantities at present time or at the instant they occur.
If the body undergoes a change in its mass distribution there will be an immediate change in its gravitational force without any lag.
Now, if we talk about special relativity, it would be absurd to say that an information can travel faster than light. The effect is in synchronization with the cause in one reference frame where the effect occurs after the cause for some observer in some other reference frame.
In order to observe Newton's law of gravity all the observer's in different reference frames must observe the same phenomena which could only be possible if time were absolute and in special relativity, time is not absolute.
Therefore, Newton's law of gravity was inconsistent with the Einstein's Special Relativity.
1. circle graph
2. Bar graph
3. line graph
hope this helps
Answer:
1.36 x 10^-3 cm
Explanation:
Area = 50 ft^2 = 46451.5 cm^2
mass = 6 oz = 170.097 g
density = 2.70 g/cm^3
Let t be the thickness of foil in cm.
mass = volume x density
mass = area x thickness x density
170.097 = 46451.5 x t x 2.70
t = 1.36 x 10^-3 cm
Thus, the thickness of aluminium foil is 1.36 x 10^-3 cm.
A voltmeter<span> its </span>instrument<span> used for </span>measuring<span> electrical potential difference between two points in an electric circuit. </span>An ammeter<span> is a </span>measuring device<span> used to</span>measure<span> the electric current in a circuit.
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