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user100 [1]
3 years ago
12

Calculate the gravitational force of 2 masses of 131 kg and 145 kg if they are 5.2 m apart.

Physics
1 answer:
LekaFEV [45]3 years ago
7 0

A is the correct choice

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Which career professionals are part of the Architecture and Construction career cluster? Check all that apply.
oksian1 [2.3K]

Answer:

A,B,D,E,F

Explanation:

I took the test for yall.

3 0
3 years ago
A block of 250-mm length and 54 × 40-mm cross section is to support a centric compressive load P. The material to be used is a b
musickatia [10]

Answer:

P = 17.28*10⁶ N

Explanation:

Given

L = 250 mm = 0.25 m

a = 0.54 m

b = 0.40 m

E = 95 GPa = 95*10⁹ Pa

σmax = 80 MPa = 80*10⁶ Pa

ΔL = 0.12%*L = 0.0012*0.25 m = 3*10⁻⁴ m

We get A as follows:

A = a*b = (0.54 m)*(0.40 m) = 0.216 m²

then, we apply the formula

ΔL = P*L/(A*E)  ⇒ P = ΔL*A*E/L

⇒  P = (3*10⁻⁴ m)*(0.216 m²)*(95*10⁹ Pa)/(0.25 m)

⇒  P = 24624000  N = 24.624*10⁶ N

Now we can use the equation

σ = P/A

⇒  σ = (24624000  N)/(0.216 m²) = 114000000 Pa = 114 MPa > 80 MPa

So σ > σmax  we use σmax

⇒  P = σmax*A = (80*10⁶ Pa)*(0.216 m²) = 17280000 N = 17.28*10⁶ N

7 0
3 years ago
WRONG ANSWERS WILL BE REPORTED
denpristay [2]

Answer:

1 = C

2 = B

Explanation:

6 0
3 years ago
Suppose you design a new thermometer called the "x" thermometer. on the x scale, the boiling point of water is 130.0 ox and the
Hoochie [10]

You've told us:

-- 130°x  =  212°F

and

-- 10°x  =  32°F

Thank you.  Those are two points on a graph of °x vs °F .  With those, we can figure out the equation of the graph, and easily convert ANY temperature on one scale to the equivalent temperature on the other scale.

-- If our graph is going to have °x on the horizontal axis and °F on the vertical axis, then the two points we know are  (130, 212)  and  (10, 32) .

-- The slope of the line through these two points is

Slope = (32 - 212) / (10 - 130)

Slope = (-180) / (-120)

Slope = 1.5

So far, the equation of the graph is

F = 1.5 x + (F-intercept)

Plug one of the points into this equation.  I'll use the second point  (10, 32) just because the numbers are smaller:

32 = 1.5 (10) + F-intercept

32 = 15 + (F-intercept)

F-intercept = 17

So the equation of the conversion graph is

F = 1.5 x + 17

There you are !  Now you can plug ANY x temperature in there, and the F temperature jumps out at you.

The question is asking what temperature is the same on both scales. This seems tricky, but it's not too bad.  Whatever that temperature is, since it's the same on both scales, you can take the conversion equation, and write the same variable in BOTH places.

We can write [ x = 1.5x + 17 ], solve it for  x, and the solution will be the same temperature in  F  too.

or

We can write [ F = 1.5F + 17 ], solve it for  F, and the solution will be the same temperature in  x  too.

F = 1.5F + 17

Subtract  F  from each side:  0.5F + 17 = 0

Subtract 17 from each side:   0.5F = -17

Multiply each side by 2 :  F = -34

That should be the temperature that's the same number on both scales.

Let's check it out, using our handy-dandy conversion formula (the equation of our graph):

F = 1.5x + 17

Plug in -34 for  x:  

F = 1.5(-34) + 17

F = -51 + 17

<em>F = -34</em>

It works !  -34 on either scale converts to -34 on the other one too. If the temperature ever gets down to -34, and you take both thermometers outside, they'll both read the same number.

<em>yay !</em>

6 0
3 years ago
I'm not really sure how to go about creating the equation, can anyone help me?
AlexFokin [52]
The displacement vector (SI units) is
\vec{r} =At\hat{i}+A[t^{3}-6t^{2}]\hat{j}

The speed is a scalar quantity. Its magnitude is
v= \sqrt{A^{2}t^{2}+A^{2}(t^{3}-6t^{2})^{2}} \\ v=A \sqrt{t^{2}+t^{6}-12t^{5}+36t^{4}} \\ v=At \sqrt{t^{4}-12t^{3}+36t^{2}+1}

Answer: At√(t⁴ - 12t³ + 36t² + 1)
3 0
3 years ago
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