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Elena L [17]
3 years ago
15

What is NOT one of the three primary resources that families have to reach financial goals?

Physics
1 answer:
Elodia [21]3 years ago
8 0
What is NOT one of the three primary resources that families have to reach financial goals? It is c) education
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SOME HELP ME WITH MY QUIZ
Rufina [12.5K]

u, w, x, and z will light

7 0
3 years ago
A rectangular beam 200 mm deep and 300 mm wide is simply supported over a span of 8 m. What uniformly distributed load per metre
Alexandra [31]

The uniformly distributed load per meter the beam may carry will be 1.275 × 10 ⁶ N/mm.

<h3>What is bending stress?</h3>

When an object is subjected to a heavy load at a specific spot, it often experiences bending stress, which causes the object to bend and tire.

The given data in the problem is;  

Bending stress, σ = 120 N/mm2  

Moment of inertia, I = 8.5 × 106 mm⁴

Depth of beam, y = d/2 = 200/2 = 100 mm  

Length of beam, L = 8 m = 8000 mm  

Width of beam, W = 300 mm

The maximum bending moment of the beam with UDL;

\rm W = \frac{wL^2}{8}

From the bending equation;

\rm M = \frac{\sigma I}{y_{max}} \\\\ M = \frac{120 \ N / mm^2 \times 8.5 \times 10^6 }{100 \ mm } \\\\ M = 10.2 \times 10^6 \ N - mm

The maximum bending moment of the beam with UDL;

\rm M = \frac{wL^2}{8} \\\\  10.2 \times 10^6 = \frac{w \times 8^2}{8} \\\\ w = 1.275 \times 10^6 \  N/mm

Hence the uniformly distributed load per meter the beam may carry will be 1.275 × 10 ⁶ N/mm.

To learn more about the bending stress, refer;

brainly.com/question/24227487

#SPJ1

5 0
2 years ago
While a utility patent protects the way an invention is used and​ works, a​ ___ patent protects the way it looks?
AnnZ [28]
<span>While a utility patent protects the way an invention is used and​ works, a​ design patent protects the way it looks.
Patent is a grant from federal government conferring the rights to exclude others from making, using or selling an invention.
There are three types of patents: Utility, design and plant. Only the inventor or inventors together can apply for a patent.</span>
4 0
3 years ago
Vector A has a magnitude of 16 m and makes an angle of 44° with the positive x axis. Vector B also has a magnitude of 13 m and i
marshall27 [118]

Answer with explanation:

The given vectors in are reduced to their componednt form as shown

For vector A it can be written as

\overrightarrow{v}_{a}=16cos(44^{o})\widehat{i}+16sin(44^{o})\widehat{j}

Similarly vector B can be written as

\overrightarrow{v}_{b}=-13\widehat{i}

Hence The sum and difference is calculated as

\overrightarrow{v}_{a}+\overrightarrow{v}_{b}=16cos(44^{o})\widehat{i}+16sin(44^{o})\widehat{j}+(-13\widehat{i})\\\\\overrightarrow{v}_{a}+\overrightarrow{v}_{b}=(16cos(44^{o})-13)\widehat{i}+16sin(44^{o})\widehat{j}\\\\\therefore \overrightarrow{v}_{a}+\overrightarrow{v}_{b}=-1.49\widehat{i}+11.11\widehat{j}\\\\\therefore |\overrightarrow{v}_{a}+\overrightarrow{v}_{b}|=\sqrt{(-1.49)^{2}+11.11^{2}}=11.21m

The direction is given by

\theta =tan^{-1}\frac{r_{y}}{r_{x}}\\\\\theta =tan^{-1}\frac{11.11}{-1.49}=97.64^{o}with positive x axis.

Similarly

\overrightarrow{v}_{a}-\overrightarrow{v}_{b}=16cos(44^{o})\widehat{i}+16sin(44^{o})\widehat{j}-(-13\widehat{i})\\\\\overrightarrow{v}_{a}-\overrightarrow{v}_{b}=(16cos(44^{o})+13)\widehat{i}+16sin(44^{o})\widehat{j}\\\\\therefore \overrightarrow{v}_{a}-\overrightarrow{v}_{b}=24.51\widehat{i}+11.11\widehat{j}\\\\\therefore |\overrightarrow{v}_{a}-\overrightarrow{v}_{b}|=\sqrt{(24.51)^{2}+11.11^{2}}=26.91m

The direction is given by

\theta =tan^{-1}\frac{r_{y}}{r_{x}}\\\\\theta =tan^{-1}\frac{11.11}{24.51}=24.38^{o}with positive x axis.

3 0
3 years ago
Two curves on a highway have the same radii. However, one is unbanked and the other is banked at an angle θ. A car can safely tr
Leviafan [203]

Answer:

θ = 29.38°

Explanation:

The centripetal force is given by the formula;

F_c = F_n(sin θ) = mv²/r

Now, the vertical component of the normal force is; F_n(cos θ)

Now, this vertical component is also expressed as; F_n(cos θ) = mg

Thus, the slope is;

F_n(sin θ)/F_n(cos θ) = (mv²/r)/mg

tan θ = v²/rg

v² = rg(tan θ)

The initial speed will be gotten from the relation;

(v_o)² = μ_s(gr)

Plugging rg(tan θ) for (v_o)², we have;

μ_s(gr) = rg(tan θ)

rg will cancel out to give;

μ_s = (tan θ)

Thus, θ = tan^(-1) μ_s

μ_s is coefficient of static friction given as 0.563

θ = tan^(-1) 0.563

θ = 29.38°

7 0
3 years ago
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