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Llana [10]
3 years ago
7

Problem 3.28 (a) In outer space, far from other objects, block 1 of mass 51 kg is at position <5, 15, 0> m, and block 2 of

mass 1550 kg is located at position <13, 15, 0> m. What is the (vector) gravitational force acting on block 2 due to block 1? It helps to make a sketch of the situation.
Physics
1 answer:
lbvjy [14]3 years ago
5 0

Answer: 8.242 × 10 exp -8 N

Explanation: F = G *m* M/r^2

r = √(13-5)^² + (51-15)^² + (0-0)²

r = 8; the distance between the masses

G =6.673 × 10 -¹¹ Nm²kg-²

F = gravitational force of attraction of m(51kg) on M {1550kg)

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What average net force is required to stop a 3.5 kg bowling ball initially travelling at a speed of 1.5 m/s over a distance of 0
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First the aceleration:

Vf² = Vo² - 2ad

a = (Vf² - Vo²) / 2d

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7 0
3 years ago
Need some help please
11Alexandr11 [23.1K]

Answer:

x^2\neq -\frac{5}{13}

First option

Explanation:

<u>Operations with functions </u>

Given two functions f, g, we can perform a number of operations with them including addition, subtraction, product, division, composition, and many others .

We have

f(x)=20x^3-7x^2+3x-7

g(x)=-13x^2-5

We are required to find

\displaystyle \frac{f}{g}(x)

We simply divide f by g as follows

\displaystyle \frac{f}{g}(x)=\frac{20x^3-7x^2+3x-7}{-13x^2-5}

We know rational functions may have problems if the denominator can be zero for some values of x. We must find out if there are such values and exclude them from the domain of the new-found function. We must ensure

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or equivalently

x^2\neq -\frac{5}{13}

Thus the first option is correct

Note: Since x^2 is always a positive number (for x real), our function does not really have any restriction in its domain

8 0
3 years ago
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