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Sergio [31]
4 years ago
9

A 8.1-kg rock and a 8.4 × 10-4-kg pebble are held near the surface of the earth. (a) determine the magnitude of the gravitationa

l force exerted on each by the earth. (b) calculate the magnitude of the acceleration of each object when released.
Physics
1 answer:
Svetllana [295]4 years ago
6 0

a). Gravitational force exerted by the Earth = (mass) x (grav acceleration)

-- 8.1 kg rock:  (8.1 kg) x (9.81 m/s²)  =  79.46 Newtons

-- 8.4 x 10⁻⁴ kg pebble: (8.4 x 10⁻⁴ kg) x (9.81 m/s²)  =  0.00824 Newton

b). Acceleration following release.

Near the Earth's surface, and neglecting air resistance, all objects have the same acceleration when released.  If air resistance can be neglected, then a feather, the 0.00084 kg pebble, the neckline of an evening gown, a school-bus, and the battleship USS Midway, if dropped together, will fall together and hit the ground at the same time.  

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La frecuencia será la misma en los dos medios, y en el vacio, no varia.

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a

6 0
3 years ago
Which of the following is a characteristic of active transport? Check all answers that apply.
vlada-n [284]
B. It moves substances against a concentration gradient. It requires energy from the cell to.

Hope this helps :)
5 0
3 years ago
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Which type of wires (copper, aluminum, or string) are ferromagnetic metals?
sp2606 [1]

Answer:

Copper isn't ferromagnetic,

Aluminum isn't ferromagnetic,

String has a ferromagnetic property

So in conclusion:

The answer is strings :)

8 0
3 years ago
Give three examples that show the importance of gravity to humans
kow [346]
1-Without gravity you can't grow any plant or tree because dirt would fly away
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3 0
3 years ago
Calculate the sample standard deviation and sample variance for the following frequency distribution of hourly wages for a sampl
ollegr [7]
<h2>Answer:</h2>

(a) standard deviation = σ = 4.9996

(b) variance = σ² = 24.996

<h2>Explanation:</h2><h2 />

<em>Given frequency table (find attached as Table 1);</em>

<u></u>

(a) To find the sample standard deviation and sample variance, follow these steps;

<em>i. Calculate the mid-point c for each group by using the mid-point formula;</em>

c = (lower bound + upper bound) / 2

=> c = (6.51 + 8.50) / 2 = 7.505

=> c = (8.51 + 10.50) / 2 = 9.505

=> c = (10.51 + 12.50) / 2 = 11.505

=> c = (12.51 + 14.50) / 2 = 13.505

=> c = (14.51 + 16.50) / 2 = 15.505

<em>So the new table becomes (find attached as Table 2);</em>

<em>ii. Calculate the total number of samples (n) which is the sum of all the frequencies.</em>

n = 50+18+42+20+46

n = 176

<em>iii. Calculate the mean (M)</em>

This is done by first multiplying the midpoints by the corresponding frequencies and then dividing the result by the total number of samples (n).

M = [(7.505 x 50) + (9.505 x 18) + (11.505 x 42) + (13.505 x 20) + (15.505 x 46)] / 176

M = [375.25 + 171.09 + 483.21 + 270.1 + 713.23] / 176

M = [2012.88] / 176

M = 11.44

<em>iv. Find the variance (σ²);</em>

The variance is calculated using the following formula

σ² = [Σ(f x c²) - (n x M²)] / (n - 1)                ------------(i)

Where;

f = frequency of each boundary data point

<em>=>  Let's first calculate </em>Σ(f x c²).

This is done by finding the sum of the product of the frequency (f) of each boundary point and the square of their corresponding mid-points(c)

Σ(f x c²) = [(50 x 7.505²) + (18 x 9.505²) + (42 x 11.505²) + (20 x 13.505²) + (46 x 15.505²)]

Σ(f x c²) = [(2816.25125) + (1626.21045) + (5559.33105) + (3647.7005) + (11058.63115)]

Σ(f x c²) = 24708.1244

<em>=> Now calculate (n x M²)</em>

n x M² = 176 x 11.44²

n x M² = 23033.7536

<em>=> Now substitute these values into equation (i) to calculate the variance</em>

σ² = [Σ(f x c²) - (n x M²)] / (n - 1)

σ² = [24708.1244 - 23033.7536] / (176 - 1)

σ² = [4374.3708] / (175)

σ² = 24.996

Therefore, the variance is 24.996

<em>v. Find the standard deviation (σ)</em>

The standard deviation is the square root of the variance. i.e

σ = √σ²

σ = √24.996

σ = 4.9996

Therefore, the standard deviation is 4.9996

4 0
3 years ago
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