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3 years ago

13

Determine the value of a for which the maximum positive moment has the same magnitude as the maximum negative moment. ei is cons

tant.

1 answer:

horrorfan [7]3 years ago

4 0

Hey im Janae how are you i like math but not that much really it depends on the teacher all teachers are wack am i rt??????

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(a) Calculate the tension in a vertical strand of spiderweb if a spider of mass 5.00 ✕ 10-5 kg hangs motionless on it.

sertanlavr [38]

a)The tension in the vertical strand will be 0.00049 N.

b))The tension in the horizontal strand will be 0.001284 N.

c)The ratio of tension in the horizontal strand to the tension in the vertical strand will be 2.62.

<h3>What is tension force?</h3>

The tension force is described as the force transferred through a rope, string, or wire as it is pulled by opposing forces.

Given data;

Mass of spiderweb = 5.00 ✕ 10⁻⁵ kg

The tension in a horizontal strand of a spiderweb, Tₓ

The tension in a vertical strand of a spiderweb, Tₐ

The angle at which the strand sags, Θ = 10.0°

R is the ratio of tension in the horizontal strand to the tension in the vertical strand

a)

From the FBD,

The tension in the vertical strand is equal to the weight of the spider;

Tₐ = W

Tₐ = mg

Tₐ= 5 x 10⁻⁵ kg x 9.8 m / s²

Tₐ= 0.00049 N.

(b)

The super's total vertical forces are zero since it is not moving the value of the tension in the horizontal strand;

Tₓ = mg/( 2sinΘ)

Tₓ = 5 x 10⁻⁵ kg x 9.8 m /s² / ( 2 sin 11° )

Tₓ= 0.001284 N.

c)

R is the tension in the vertical strand divided by the tension in the horizontal strand.

R= Tₐ / Tₓ

R= 0.001284 N / 0.00049 N

R= 2.62

Hence, the tension in a vertical strand of a spiderweb, the tension in a horizontal strand of a spiderweb, and the ratio of tension in the horizontal strand to the tension in the vertical strand will be 0.00049 N, 0.001284 N, and 2.62 respectively.

To learn more about the tension force refer to the link;

brainly.com/question/2287912

#SPJ1

6 0

2 years ago

Find out the x-component and y-component of the following vectors:

Leno4ka [110]

Answer:

A = 15.0 cm; the vector is 30.00 north of west; = 15 cm N30W

Y - component = 7.5cm and X - component = -12.99 cm

B = 12.0 m; the vector is 55.00 north of east; = 12.0 m N55E

Y - component = 9.83 m and X - component = 6.88 m

C = 28.0 m/s; the vector is 40.00 south of west; = 28.0 m/s S40W

Y - component = - 17.998 m/s and X - component = - 21.448 m/s

D = 55.5 m/s2; the vector is 35.00 south of east; = 55.5 m/s² S35E

Y - component = - 31.835 m/s² and X - component = 45.466 m/s²

Explanation:

1. A = 15.0 cm; the vector is 30.00 north of west; = 15 cm N30W

Y - component = 15cm × sin30 = 7.5cm

X - component = -15cm × cos30 = -12.99 cm

2. B = 12.0 m; the vector is 55.00 north of east; = 12.0 m N55E

Y - component = 12.0 m × sin55 = 9.83 m

X - component =12.0 m × cos55 = 6.88 m

3. C = 28.0 m/s; the vector is 40.00 south of west; = 28.0 m/s S40W

Y - component = - 28.0 m/s × sin40 = - 17.998 m/s

X - component = - 28.0 m/s × cos40 = - 21.448 m/s

4. D = 55.5 m/s2; the vector is 35.00 south of east; = 55.5 m/s² S35E

Y - component = - 55.5 m/s² × sin35 = - 31.835 m/s²

X - component = 55.5 m/s² × cos35 = 45.466 m/s²

6 0

3 years ago

Read 2 more answers

In an insulated container, liquid water is mixed with ice. What can you conclude about the phases present in the container when

Ann [662]

Answer:

d. There is no way of knowing the phase composition without more information.

Explanation:

There is no way to ascertain the phases present in the container when equilibrium is established because we are not furnished with enough information.

The most important information needed to have a deeper understanding and provide solution is temperature.
Liquid water and ice can both exist at the same temperature.
So we cannot conclude based on the information at hand.

5 0

3 years ago

A 0.100 kg ball hangs from a spring of negligible mass. When the ball is hung on the spring and is at rest, the spring is stretc

anzhelika [568]

Answer:

a) 4.9 N/m

b) 1.4 m/s

c) 0.225 s

Explanation:

Hooke's law states that

F = k * Δx

Where

F: force applied to a spring

k: constant of the spring

Δx: elongation of the spring

The force applied in this case is the weight of the ball, this is

P = m * g = 0.1 kg * 9.81 m/s^2 = 0.981 N

Rearrainging Hooke's law:

k = F / Δx

k = 0.981 / 0.2 = 4.9 N/m

If the ball is pulled down the spring will acquire some potential energy, when it is released, the potential energy will be released as kinetic energy on the ball

$Ec = \frac{1}{2} * m * v^2$

Elastic potential energy is:

$U = \frac{1}{2} * k * \Delta x^2$

The energy gained from the 0.2m pull will be turned into kinetic energy

Ec = U

Therefore:

$\frac{1}{2} * m * v^2 = \frac{1}{2} * k * \Delta x^2$

Rearranging:

$v^2 = \frac{k}{m} * \Delta x^2$

$v = \Delta x * \sqrt{\frac{k}{m}}$

$v = 0.2 * \sqrt{\frac{4.9}{0.1}} = 1.4 m/s$

After being released the ball will oscillate at the natural frequency of the system, which is

$f = \frac{1}{2 * \pi} * \sqrt{\frac{k}{m}}$

And the period will be:

$T = 2 * \pi * \sqrt{\frac{m}{k}}$

The period in this case is:

$T = 2 * \pi * \sqrt{\frac{0.1}{4.9}} = 0.9 s$

The ball will move up and down taking T time to complete a cycle, the movement from the stretched position to the equilibrium position takes T/4 = 0.225 s

8 0

3 years ago

Lerok [7]

Answer:

1. acceleration value

2. 20 m

3a. when you throwing somethinb upward

3b. when a thing reach maximum height after throwing upward

4. (entire only) velocity = diaplacement/ time = 0/91 = 0

speed = distance/time = 200/91 = 2.2 m/s

5. (40-20)/4 = 5 m/s²

6. displacement using phytagoras get 50 m

velocity = displacement/time = 50/(6+4) = 5 m/s

7. 0, it turn back to the base

8. speed = distance/time = (240*2)/(4+6) = 48 km/h

9. s = (4+14)*3/2 =27 m

3 0

3 years ago