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tamaranim1 [39]
3 years ago
14

30 POINTS

Physics
2 answers:
Serjik [45]3 years ago
8 0

Answer:

B

Explanation:

Travka [436]3 years ago
3 0

Answer:

to be honest I'm not sure

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6. What is ductility?​
gavmur [86]

Answer:

Explanation:

Best to look it up for better results. The measure of a metals ability to withstand tension or stress to it. Like when when metal is being pulled apart it would be the distance it's being pulled.

6 0
3 years ago
Read 2 more answers
a system of four particles moves along one dimension. the center of mass of the system is at rest, and the particles do not inte
Zinaida [17]

The velocity of the particle 4 at time, t = 2.83 s, is -14.1 m/s.

The given parameters;

m1 = 1.45 kg, v1(t) = (6.09m/s) + (0.299m/s^2) × t

m2 = 2.81 kg, v2(t) = (7.83m/s) + (0.357m/s^2) × t

m3 = 3.89 kg, v3(t) = (8.09m/s) + (0.405m/s^2) × t

m4 = 5.03kg

The velocity of the center mass of the particles is calculated as;

McmVcm = m1v1 + m2v2 +m3v3+m4v4

Vcm= m1v1 + m2v2 +m3v3 +m4v4/ Mcm

0 = m1v1 + m2v2 +m3v3 +m4v4/ Mcm

m1v1 + m2v2 +m3v3+m4v4 = 0

m4v4 = -(m1v1 + m2v2 +m3v3)

v4 =-(m1v1 + m2v2 +m3v3)/ m4

The velocity of particle 1 at time, t = 2.83 s;

vi = 6.09 + 0.299× 2.83

v1 = 6.94 m/s

The velocity of particle 3 at time, t = 2.83 s;

v2 = 7.83 + 0.357 × 2.83

v2 = 8.84 m/s

The velocity of particle 3 at time, t = 2.83 s;

v3 = 8.09 + 0.405 × 2.83

v3 = 9.24 m/s

The velocity of particle 3 at time, t = 2.83 s;

v4 = - (m1v1 + m2v2 + m3v3)/m4

v4 = -(1.45×6.94 + 2.81×8.84 + 3.89×9.24)/5.03

v4 = -14.4 m/s

Thus, the velocity of the particle 4 at time, t = 2.83 s, is -14.1 m/s.

Learn more about Velocity here:

brainly.com/question/18084516

#SPJ4

5 0
1 year ago
The geese fly 23 m/s to the south when they migrate for the winter. Identify if this example is speed or velocity.
kari74 [83]

Answer:

Its velocity.

7 0
3 years ago
Audio frequency range of a human ear is 20Hz - 20000 Hz. Express the range in terms of time period ? Answer needed urgently. Pl.
Feliz [49]
Using the term c in this case is a little confusing. It is more generic to use a general velocity, v. That way, in this case, we know to use the speed of sound. 

wavelength*frequency=v 

wavelength_20Hz = (345 m/s)/(1/20s) 

<span>wavelength_20kHz = (345 m/s)/(1/20000s)


I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
</span>
5 0
3 years ago
Yvette hangs a 2.4kg bird feeder in the middle of a rope tied between two trees, the feeder creates a
kicyunya [14]

Answer:

a) 0.049 m

b) Yes, increase

Explanation:

Draw a free body diagram.

In the y direction, there are three forces acting on the feeder.  Two vertical components of the tension forces in each rope pulling up, and weight force pulling down.

Apply Newton's second law to the feeder in the y direction.

∑F = ma

2Ty − mg = 0

Ty = mg/2

Let's say the distance the rope sags is d.  The trees are 4m apart, so the feeder is 2m horizontally from either tree.  Using Pythagorean theorem, we can find the length of the rope on either side:

L² = 2² + d²

L = √(4 + d²)

Using similar triangles, we can write a proportion using the forces and distances.

Ty / T = d / L

Substitute:

(mg/2) / T = d / √(4 + d²)

Solve for d:

Td = mg/2 √(4 + d²)

T² d² = (mg/2)² (4 + d²)

T² d² = (mg)² + (mg/2)² d²

(T² − (mg/2)²) d² = (mg)²

d² = (mg)² / (T² − (mg/2)²)

d = mg / √(T² − (mg/2)²)

Given m = 2.4 kg and T = 480 N:

d = (2.4) (9.8) / √(480² − (2.4×9.8/2)²)

d = 0.049 m

b) If a bird lands on a feeder, this will increase the tension in the rope to support the bird's weight.

8 0
3 years ago
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