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Monica [59]
3 years ago
10

Find the order of magnitude of your age in seconds. 1)10^3 s 2)10^6 s 3)10^9 s 4)10^12 s

Physics
1 answer:
My name is Ann [436]3 years ago
8 0

Answer:

A) 3,  B) 3,  C)  3

Explanation:

The order of magnitude is a very useful unit for when you do not have a precision in the measurement, the way to find it is to take the number to scientific notation and then change the whole part to 101 if it is greater than or equal to five or 10 or less than five

A) take my age to seconds

for example if I am 25 years old.

I take the years to seconds

       age = 20 years (365 day / year) (24 h / 1 day) (3600 s / 1h)

      age = 630720000 s

we carry scientific notation

      age = 6.3072 10⁸ s

as the integer greater than five we change it by 10¹

     

      age = 10⁹s

The correct answer is 3

B) The number of beats per minute approximately 80 beats in a month we have

    # beats = 80 beats /min (60min / 1h) (24 h / 1 day) (30 day / 1 month)

    # _beats = 3456000 beats / month

   #_beats = 3,456 10⁶ beats / month

as the integer is less than five we change it by 10⁰

  #_beats = 10⁶

the correct answer is 3

C) the number of heartbeats in life

suppose a life expectancy of 80 years

    #_beats_total = 80 pulse /min (60min / 1h) (24 h / 1day) (365 day / year) 80 year

    #_total_beats = 3 363 840 000

    #_beats_total = 3.36384 10⁹

as the integer is less than five we change it by 10⁰

    #_beats_total = 10⁹

the correct answer is 3

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A 70.0 kg ice hockey goalie, originally at rest, has a 0.110 kg hockey puck slapped at him at a velocity of 31.5 m/s. Suppose th
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Answer

given,

mass of the goalie(m₁) = 70 kg

mass of the puck (m₂)= 0.11 kg

velocity of the puck = 31.5 m/s

elastic collision

v_1=\dfrac{m_2-m_1}{m_1+m_2}v_1+\dfrac{2m_2}{m_1+m_2}v_2

v_{pf}=\dfrac{0.11-70}{0.11+70}31.5+\dfrac{2m_2}{m_1+m_2}\times (0)

v_{pf}=-31.4\ m/s

v'_2 = \dfrac{2m_1v_1}{m_1+m_2}-\dfrac{(m_2-m_1)v_2}{m_2+m_1}

v_{gf} = \dfrac{2\times 0.11\times 31.5}{0.11+70}-\dfrac{(0.11-70)\times 0}{m_1+m_2}

v_{gf} = \dfrac{2\times 0.11\times 31.5}{0.11+70}

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4 0
3 years ago
A ski gondola is connected to the top of a hill by a steel cable of length 620 m and diameter 1.5 cm. As the gondola comes to th
xz_007 [3.2K]

Answer:

(a) 89 m/s

(b) 11000 N

Explanation:

Note that answers are given to 2 significant figures which is what we have in the values in the question.

(a) Speed is given by the ratio of distance to time. In the question, the time given was the time it took the pulse to travel the length of the cable twice. Thus, the distance travelled is twice the length of the cable.

v=\dfrac{2\times 620 \text{ m}}{14\text{ s}} = \dfrac{1240\text{ m}}{14\text{ s}}=88.571428\ldots \text{ m/s}= 89\text{ m/s}

(b) The tension, T, is given by

v =\sqrt{\dfrac{T}{\mu}}

where v is the speed, T is the tension and \mu is the mass per unit length.

Hence,

T = \mu\cdot v^{2}

To determine \mu, we need to know the mass of the cable. We use the density formula:

\rho = \dfrac{m}{V}

where m is the mass and V is the volume.

m=\rho\cdot V

If the length is denoted by l, then

\mu = \dfrac{m}{l} = \dfrac{\rho\cdot V}{l}

T = \dfrac{\rho\cdot V}{l} v^{2}

The density of steel = 8050 kg/m3

The cable is approximately a cylinder with diameter 1.5 cm and length or height of 620 m. Its volume is

V = \pi \dfrac{d^{2}}{4} l

T = \dfrac{\rho\cdot\pi d^2 l}{4l}v^2 = \dfrac{\rho\cdot\pi d^2}{4}v^2

T = \dfrac{8050\times\pi\times0.015^2}{4} \times 88.57^2

T = 11159.4186\ldots \text{ N} = 11000 \text{ N}

4 0
4 years ago
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