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4 days ago

What is the average acceleration of a car that is initially at rest at a stoplight and then accelerates to 24 m/s in 9.4 s?

Physics

2 answers:

kicyunya [2.2K]4 days ago

8 0

serg [2.5K]4 days ago

5 0

B) a= 2.6 m/s² Explanation: In the case of the car's uniform acceleration, the following formula is used: vf= v₀+a*t (Formula 1) where: t is time in seconds (s), v₀ denotes initial speed in m/s, vf represents final speed in m/s, and a signifies acceleration in m/s². Given that v₀=0, vf = 24 m/s, and t= 9.4 s, we substitute these values into formula (1): 24 m/s = 0 + a*9.4s, leading to a = (24 m/s) / (9.4s), which calculates to a= 2.6 m/s².

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Yuliya22 [2420]

Answer:

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Explanation:

Based on this,

the heart's weight constitutes about 0.5% of total body mass.

Total human weight = 185 lbs

Let the entire body weight be represented as w and the heart's weight as $w_{h}$ .

We aim to determine the heart's weight for a human

Using the provided information

$w_{h}=0.5\times w$

Where, h = heart weight

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$w_{h}=\dfrac{0.5}{100}\times 185$

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1 month ago

A rotating beacon is located 2 miles out in the water. Let A be the point on the shore that is closest to the beacon. As the bea

serg [2593]

Answer:

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Explanation:

The beacon is rotating at an angular speed of

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so we have

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We know that

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At this point we have

$r = 2 miles = 2(1609 m)$

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So we can conclude with

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$v = 3369.2 m/s$

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19 days ago

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Keith_Richards [2256]

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7 days ago

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A uniformly charged spherical droplet of mercury has electric potential Vbig throughout the droplet. The droplet then breaks int

kicyunya [2264]

Answer:

$\frac{V_{big}}{V_{small}} = n^{2/3}$

Explanation:

Let the charge on the large droplet be denoted as Q.

When the radius of the droplet is R, the electric potential for the larger droplet can be expressed as:

$V_{big} = \frac{KQ}{R}$

If it splits into n identical droplets, let the charge of each be "q" and their radius be "r".

Applying volume conservation gives us:

$\frac{4}{3}\pi R^3 = n(\frac{4}{3}\pi r^3)$

$r = \frac{R}{n^{1/3}}$

Now, the potential for the smaller droplets is given as:

$V_{small} = \frac{kq}{r}$

$V_{small} = \frac{K(Q/n)}{\frac{R}{n^{1/3}}}$

$V_{small} = \frac{1}{n^{2/3}}\frac{KQ}{R}$

$\frac{V_{big}}{V_{small}} = n^{2/3}$

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17 days ago

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9 days ago

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