Ask question

Ask question

All categories

2 years ago

12

A 210-N mass is suspended from a horizontal beam by two cables that make angles of 29 degrees and 53 degrees with the beam. Find

, using vectors, the tension in the cables.

1 answer:

Nonamiya [84]2 years ago

7 0

The tension in both cables are calculated as; T₁ = 127.63 N and T₂ = 185.48 N

<h3>How to find the Tension in Suspended Cables?</h3>

Resolving forces in the y direction gives us;

T₁ sin 29° + T₂ sin 53° = 210 ------(eq 1)

Resolving forces in the horizontal direction gives;

T₁ cos 29° = T₂ cos 53° ----(eq 2)

T₁ = T₂ cos 53°/cos 29°

Thus;

(T₂ cos 53°/cos 29°)sin 29° + T₂ sin 53° = 210

0.3336T₂ + 0.7986T₂ = 210

T₂ = 210/1.1322

T₂ = 185.48 N

Thus;

T₁ = 185.48 cos 53°/cos 29°

T₁ = 127.63 N

Read more about Tension in Cables at; brainly.com/question/11653585

#SPJ1

You might be interested in

Write the equation of a line that has a m= -5 and passes through the point (-1,4)

svetoff [14.1K]

Answer:

We have a slope and a point, this way, we do not need to find the slope using two points. We can find the line using the Slope Form to turn it into Slope-Intercept Form, also known as "y = mx + b". We can use the Slope Form which is "y - y1 = m(x - x1)". A coordinate is "(x,y)", so that means, "(-1,4)" means that x = -1 and y = 4. We can substitute y1 for 4 and x1 for -1, and m for -5 since m is the slope. Plugging that information into the Slope Form Formula we get "y - 4 = -5(x - (-1))". We know that a negative with a negative turn positive, so we can use this information to change the formula to "y - 4 = -5(x + 1)". Now we distribute and simplify, these steps are shown below.

y - 4 = -5(x + 1)

y - 4 = -5x -5

y - 4 + 4 = -5x - 5 + 4

y = -5x - 1

Now that we have simplified it, we can see that the slope is -5 and the -1 is the y-intercept. To check this is a linear equation is to graph it, or substitute three x-coordinate or y-coordinates and draw a line. In conclusion, the equations is y = -5x - 1.

Step-by-step explanation:

4 0

2 years ago

2/3x +4 = 2/5 rewrite the equation so it doesn’t have fractions. Do not use decimals.

erma4kov [3.2K]

Answer:

exfv

Step-by-step explanation:

6 0

3 years ago

Marcel made an error when he used the distributive property to solve the problem shown. Which sentence describes the error?

LenKa [72]

A.In Step 2, 90 should be multiplied by the quantity 10 – 12, not by 10.

correct:
90(10 - 12) = 90 · 10 - 90 · 12 or 90(10 - 12) = 90 · (-2)

4 0

3 years ago

Read 2 more answers

Simplify. Assume no variable is 0.

quester [9]

Step-by-step explanation:

Q16

$( {5x}^{3} {y}^{ - 5} )( {4xy}^{3} )$

First, we multiply the variable x together:

${x}^{3} \times x = {x}^{4}$

and remove brackets:

${5y}^{ - 5} \times {4x}^{4} \times {y}^{3}$

Now, we multiply the variable y:

${y}^{ - 5} \times {y}^{3} = {y}^{ - 2}$

Hence:

$5 \times 4 {x}^{4} \times {y}^{ - 2}$

Use the rule

${x}^{ - y} = \frac{1}{ {x}^{y} }$

Thus,

$5 \times 4 {x}^{4} \times \frac{1}{ {y}^{2} }$

Multiply:

$\frac{20 {x}^{4} }{ {y}^{2} }$

Q17

$( - 2 {b}^{3} c)(4 {b}^{2} {c}^{2} )$

$- 2 {b}^{3} c\times 4 {b}^{2} {c}^{2}$

Multiply the variable b together:

$- 2c \times 4 {b}^{5} {c}^{2}$

Multiply the variable c together:

$- 2 \times 4 {b}^{5} {c}^{3}$

Multiply.

$- 8 {b}^{5} {c}^{3}$

Q18:

$\frac{ {a}^{3} {n}^{7} }{a {n}^{4} }$

Divide the numerator and denominator by a:

$\frac{ {a}^{2} {n}^{7} }{ {n}^{4} }$

Apply the rule that

$\frac{ {x}^{y} }{ {x}^{z} } = {x}^{y - z}$

Hence, the equation will be

${a}^{2} {n}^{7 - 4}$

which is finally

${a}^{2} {n}^{3}$

Q19:

$\frac{ - {y}^{3} {z}^{5} }{ {y}^{2} {z}^{3} }$

Do the same rule said above:

$- \frac{ {z}^{5} {y}^{3 - 2} }{ {z}^{3} }$

$- \frac{y {z}^{5} }{ {z}^{3} }$

Do the rule again:

$- y {z}^{5 - 3}$

$- y {z}^{2}$

Q20:

$\frac{ - 7 {x}^{5} {y}^{5} {z}^{4} }{21 {x}^{7} {y}^{5} {z}^{2} }$

Cancel the common factor of 7:

$- \frac{ {x}^{5} {y}^{5} {z}^{4} }{3 {x}^{7} {y}^{5} {z}^{2} }$

$- \frac{ {y}^{5} {z}^{4} }{3 {y}^{5} {z}^{2} {x}^{7 - 5} }$

Thus,

$- \frac{ {y}^{5} {z}^{4} }{3 {y}^{5} {z}^{2} {x}^{2} }$

Cancel common factor of y^5

$- \frac{ {z}^{4} }{3 {z}^{2} {x}^{2} }$

$- \frac{ {z}^{4 - 2} }{3 {x}^{2} }$

$- \frac{ {z}^{2} }{3 {x}^{2} }$

Q21:

$\frac{9 {a}^{7} {b}^{5} {c}^{5} }{18 {a}^{5} {b}^{9} {c}^{3} }$

Cancel the common factor which is 9:

$\frac{{a}^{7} {b}^{5} {c}^{5} }{2 {a}^{5} {b}^{9} {c}^{3} }$

"Move" the variable a up using the rule:

$\frac{{a}^{7 - 5} {b}^{5} {c}^{5} }{2 {b}^{9} {c}^{3} }$

$\frac{{a}^{2} {b}^{5} {c}^{5} }{2 {b}^{9} {c}^{3} }$

"Move" the variable b down using the rule:

$\frac{{a}^{2} {c}^{5} }{2 {c}^{3} {b}^{9 - 5} }$

$\frac{{a}^{2} {c}^{5} }{2 {c}^{3} {b}^{4} }$

"Move" the variable c up using the rule

$\frac{{a}^{2} {c}^{5 - 3} }{2 {b}^{4} }$

$\frac{{a}^{2} {c}^{2} }{2 {b}^{4} }$

Q22:

$( {n}^{5} {)}^{4}$

Use the rule:

$({x}^{y} {)}^{z} = {x}^{yz}$

Hence,

${n}^{5 \times 4}$

${n}^{20}$

Q23:

$( {z}^{3} {)}^{6}$

Using the same rule above,

${z}^{3 \times 6}$

Therefore,

${z}^{18}$

5 0

3 years ago

Read 2 more answers

Joseph is making kebabs. He wants each kebab to have 3 pieces of meat and 4

tino4ka555 [31]

Answer:

Option D.

Step-by-step explanation:

From your comment, you said D was 30 pieces of meat and 40 pieces of vegetables. The starting ratio is 3:4. A true proportion is 3:4::30:40. 3 and 4 are both multiplied by 10 to get 30 and 40 therefore there would be no leftovers.

6 0

3 years ago