Let car A's starting position be the origin, so that its position at time <em>t</em> is
A: <em>x</em> = (40 m/s) <em>t</em>
and car B has position at time <em>t</em> of
B: <em>x</em> = 100 m - (60 m/s) <em>t</em>
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They meet when their positions are equal:
(40 m/s) <em>t</em> = 100 m - (60 m/s) <em>t</em>
(100 m/s) <em>t</em> = 100 m
<em>t</em> = (100 m) / (100 m/s) = 1 s
so the cars meet 1 second after they start moving.
They are 100 m apart when the difference in their positions is equal to 100 m:
(40 m/s) <em>t</em> - (100 m - (60 m/s) <em>t</em>) = 100 m
(subtract car B's position from car A's position because we take car A's direction to be positive)
(100 m/s) <em>t</em> = 200 m
<em>t</em> = (200 m) / (100 m/s) = 2 s
so the cars are 100 m apart after 2 seconds.
Answer: B
Explanation:
Given that an object of mass 2 kg starts from rest and is allowed to slide down a frictionless incline so that its height changes by 20 m.
The parameters given from the question are:
Mass M = 2kg
Height h = 20m
Let g = 9.8m/s^2
At the bottom of the incline plane, the object will experience maximum kinetic energy.
From conservative of energy, maximum K.K.E = maximum P.E
Maximum P.E = mgh
Maximum P.E = 2 × 9.8 × 20 = 392 J
But
K.E = 1/2mv^2
Substitute the values of energy and mass into the formula
392 = 1/2 × 2 × V^2
V^2 = 392
V = sqrt( 392 )
V = 19.8 m/s
V = 20 m/s approximately
Answer:
3.5m/s^2
Explanation:
From Newton's second Law of Motion
F = ma
Where F is the applied force, m is the mass of the object and a is the acceleration.
F = 350 N
Mass = 100kg
350N = 100×a
a = 350/100
a = 3.5m/s^2
The acceleration of the object will be 3.5m/s^2
Answer:
λ = 102.78 nm
This radiation is in the UV range,
Explanation:
Bohr's atomic model for the hydrogen atom states that the energy is
E = - 13.606 / n²
where 13.606 eV is the ground state energy and n is an integer
an atom transition is the jump of an electron from an initial state to a final state of lesser emergy
ΔE = 13.606 (1 /
- 1 / n_{i}^{2})
the so-called Lyman series occurs when the final state nf = 1, so the second line occurs when ni = 3, let's calculate the energy of the emitted photon
DE = 13.606 (1/1 - 1/3²)
DE = 12.094 eV
let's reduce the energy to the SI system
DE = 12.094 eV (1.6 10⁻¹⁹ J / 1 ev) = 10.35 10⁻¹⁹ J
let's find the wavelength is this energy, let's use Planck's equation to find the frequency
E = h f
f = E / h
f = 19.35 10⁻¹⁹ / 6.63 10⁻³⁴
f = 2.9186 10¹⁵ Hz
now we can look up the wavelength
c = λ f
λ = c / f
λ = 3 10⁸ / 2.9186 10¹⁵
λ = 1.0278 10⁻⁷ m
let's reduce to nm
λ = 102.78 nm
This radiation is in the UV range, which occurs for wavelengths less than 400 nm.