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k0ka [10]
3 years ago
9

You are traveling on the interstate highway at a speed of 65 mph. What is your speed in km/h? The conversion factor is: 1.0 mph=

1.6 km/h.
Physics
1 answer:
chubhunter [2.5K]3 years ago
6 0
You have to cross multiply. So set up the equation like 1.6/x = 1/65 then multiply x by 1 and 1.6 by 65. The answer is 104 km/h.
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Explanation:

Niels Bohr improved Rutherford's model. Using mathematical ideas, he showed that electrons occupy shells or energy levels around the nucleus. The Dalton model has changed over time because of the discovery of subatomic particles .

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From the lab iodine clock reaction: explain the reason this result by comparing what is involved in changing Fe2+ into Fe 3+ to
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A diamond with a mass of 45 g hangs motionless from a chain. what is the upward force of the chain on the diamond?
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The upward force of the chain on the diamond would be the tension in the chain, and this tension would have to support the weight of the 45g that hangs from the chain.

mass = 45 g = 45/1000 kg = 0.045kg

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<span>So the upward force is ≈ </span><span>0.45N. </span>
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Which of the following is not a reason fluorescent lamps are advantageous over incandescent lamps?
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4 0
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Read 2 more answers
Two kids are playing on a newly installed slide, which is 3 m long. John, whose mass is 30 kg, slides down into William (20 kg),
yuradex [85]

Answer:

v=3.564\ m.s^{-1}

\Delta v =2.16\ m.s^{-1}

Explanation:

Given:

  • mass of John, m_J=30\ kg
  • mass of William, m_W=30\ kg
  • length of slide, l=3\ m

(A)

height between John and William, h=1.8\ m

<u>Using the equation of motion:</u>

v_J^2=u_J^2+2 (g.sin\theta).l

where:

v_J = final velocity of John at the end of the slide

u_J = initial velocity of John at the top of the slide = 0

Now putting respective :

v_J^2=0^2+2\times (9.8\times \frac{1.8}{3})\times 3

v_J=5.94\ m.s^{-1}

<u>Now using the law of conservation of momentum at the bottom of the slide:</u>

<em>Sum of initial momentum of kids before & after collision must be equal.</em>

m_J.v_J+m_w.v_w=(m_J+m_w).v

where: v = velocity with which they move together after collision

30\times 5.94+0=(30+20)v

v=3.564\ m.s^{-1} is the velocity with which they leave the slide.

(B)

  • frictional force due to mud, f=105\ N

<u>Now we find the force along the slide due to the body weight:</u>

F=m_J.g.sin\theta

F=30\times 9.8\times \frac{1.8}{3}

F=176.4\ N

<em><u>Hence the net force along the slide:</u></em>

F_R=71.4\ N

<em>Now the acceleration of John:</em>

a_j=\frac{F_R}{m_J}

a_j=\frac{71.4}{30}

a_j=2.38\ m.s^{-2}

<u>Now the new velocity:</u>

v_J_n^2=u_J^2+2.(a_j).l

v_J_n^2=0^2+2\times 2.38\times 3

v_J_n=3.78\ m.s^{-1}

Hence the new velocity is slower by

\Delta v =(v_J-v_J_n)

\Delta v =5.94-3.78= 2.16\ m.s^{-1}

8 0
3 years ago
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