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1 year ago

A 1.2kg ball rolls forward with an acceleration of 1.11 m/s. What is the net force on the ball

Physics

1 answer:

grandymaker [24]1 year ago

8 0

Answer:

1.332 N

Explanation:

Net Force = Mass x Acceleration
1.2 x 1.11 = 1.332 N

I'm so sorry if I'm wrong.

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The answer is B: 6; 4. Carbon has four electrons in its outermost shell, which are its valence electrons.

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Answer:

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Explanation:

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2 years ago

A new cat 6 cable run and key stone jack were installed for a workstation that was moved from the sales department to the shippi

Andrei [34K]

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Explanation:

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3 years ago

An electric turntable 0.730 mm in diameter is rotating about a fixed axis with an initial angular velocity of 0.240 rev/srev/s a

Zolol [24]

Answer:

a) $\omega = 0.421\,\frac{rev}{s}$ , b) $\Delta \theta = 0.066\,rev$ , c) $v = 0.966\,\frac{mm}{s}$ , d) $a = 3.293\,\frac{mm}{s^{2}}$

Explanation:

a) The angular velocity of the turntable after $0.200\,s$ .

$\omega = \omega_{o} + \alpha\cdot \Delta t$

$\omega = 0.240\,\frac{rev}{s} + (0.906\,\frac{rev}{s^{2}} )\cdot (0.2\,s)$

$\omega = 0.421\,\frac{rev}{s}$

b) The change in angular position is:

$\Delta \theta = \omega_{o}\cdot t + \frac{1}{2} \cdot \alpha \cdot t^{2}$

$\Delta \theta = (0.240\,\frac{rev}{s} )\cdot (0.2\,s) + \frac{1}{2}\cdot (0.906\,\frac{rev}{s^{2}} )\cdot (0.2\,s)^{2}$

$\Delta \theta = 0.066\,rev$

c) The tangential speed of a point on the rim of the turn-table:

$v = r\cdot \omega$

$v = (0.365\times 10^{-3}\,m)\cdot (0.421\,\frac{rev}{s} )\cdot (\frac{2\pi\,rad}{1\,rev} )$

$v = 9.655\times 10^{-4}\,\frac{m}{s}$

$v = 0.966\,\frac{mm}{s}$

d) The tangential and normal components of the acceleration of the turn-table:

$a_{t} = (0.365\times 10^{-3}\,m)\cdot (0.906\,\frac{rev}{s^{2}})\cdot (\frac{2\pi\,rad}{1\,rev} )$

$a_{t} = 2.078\times 10^{-3}\,\frac{m}{s^{2}}$

$a_{t} = 2.078\,\frac{mm}{s}$

$a_{n} = (0.365\times 10^{-3}\,m)\cdot \left[(0.421\,\frac{rev}{s} )\cdot (\frac{2\pi\,rad}{1\,rev} )\right]^{2}$

$a_{n} = 2.554\times 10^{-3}\,\frac{m}{s^{2}}$

$a_{n} = 2.554\,\frac{mm}{s^{2}}$

The magnitude of the resultant acceleration is:

$a = \sqrt{(2.078\,\frac{mm}{s} )^{2}+(2.554\,\frac{mm}{s} )^{2}}$

$a = 3.293\,\frac{mm}{s^{2}}$

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3 years ago

Find the ratio of the lengths of the two mathematical pendulums, if the ratio of periods is 1.5

juin [17]

Answer:

The ratio of lengths of the two mathematical pendulums is 9:4.

Explanation:

It is given that,

The ratio of periods of two pendulums is 1.5

Let the lengths be L₁ and L₂.

The time period of a simple pendulum is given by :

$T=2\pi \sqrt{\dfrac{l}{g}}$

$T^2=4\pi^2\dfrac{l}{g}\\\\l=\dfrac{T^2g}{4\pi^2}$

Where

l is length of the pendulum

$l\propto T^2$

$\dfrac{l_1}{l_2}=(\dfrac{T_1}{T_2})^2$ ....(1)

ATQ,

$\dfrac{T_1}{T_2}=1.5$

Put in equation (1)

$\dfrac{l_1}{l_2}=(1.5)^2\\\\=\dfrac{9}{4}$

So, the ratio of lengths of the two mathematical pendulums is 9:4.

3 0

2 years ago