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4 years ago

a mass spring system makes 50 complete osillations in 10 seconds. what is the period and frequency of the oscillations

Physics

1 answer:

Oxana [17]4 years ago

5 0

Answer: Period = 0.2 seconds; frequency = 5Hz

Explanation:

Number of oscillations = 50

Time required = 10 seconds

Period (T) = ?

Frequency of the oscillations (F) = ?

A) Recall that frequency is the number of oscillations that the mass spring system completes in one second.

i.e Frequency = (Number of oscillations / time taken)

F = 50/10 = 5Hz

B) Period, T is inversely proportional to frequency. i.e Period = 1/Frequency

T = 1/5Hz

T = 0.2 seconds

Thus, the the period and frequency of the oscillations are 0.2 seconds and 5Hz respectively.

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A photon is absorbed by an electron that is in the n = 3 state of a hydrogen atom, causing the hydrogen atom to become ionized.

Wittaler [7]

Answer:

$\lambda=3.99*10^{-7}m$

Explanation:

According to the law of conservation of energy, the energy of the absorbed photon must be equal to the binding energy of the electron plus the energy of the released electron:

$E_p=E_b+E_r\\\frac{hc}{\lambda}=\frac{13.6eV}{n^2}+\frac{m_ev^2}{2}$

1 eV is equal to $1.6*10^{-19}J$ , so:

$13.6eV*\frac{1.6*10^{-19}J}{1eV}=2.18*10^{-18}J$

Solving for $\lambda$ and replacing the given values:

$\lambda=\frac{hc}{\frac{2.18*10^{-18}J}{n^2}+\frac{m_ev^2}{2}}\\\lambda=\frac{6.63*10^{-134}J\cdot s(3*10^8\frac{m}{s})}{\frac{2.18*10^{-18}J}{3^2}+\frac{(9.11*10^{-31}kg)(7.5*10^5\frac{m}{s})^2}{2}}\\\lambda=3.99*10^{-7}m$

4 0

3 years ago

A car transports its passengers between 3 buildings. It moves from the first building to the second building, 4.76km away, in a

BigorU [14]

Answer:

1) a = 6.14 km

2) b = 4.69 km

Explanation:

Let the first building be A, second building be B and third building be C.

Now, bearing of A = 4.76 km in a direction 37° north of east

Bearing of B = 69° west of north

Bearing of C = 28° east of south

Thus if this 3 points form a triangle, we will have the following angles;

Angle at point A = 28 + (90 - 37) = 81°

Angle at point B = 28 + (90 - 69) = 49°

Angle at point C = 180 - (81 + 49) = 50°

Now, the distance between second and third building is "a" which is represented by BC in the triangle attached while the given distance of 4.76 represents side AB. Thus;

Using sine rule, we can find "a".

a/sin 81 = 4.76/sin 50

a = 6.14 km

B) distance between first and third building is AB in the triangle depicted by "b".

Similar to the first problem, we will use sine rule again.

b/sin49 = 4.76/sin 50

b = 4.69 km