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Dovator [93]
3 years ago
10

(HELP ASAP)

Physics
2 answers:
Lana71 [14]3 years ago
6 0

Answer:

The main connective tissue is blood which is found in <em><u>blood vessels</u></em> and capillaries.

harkovskaia [24]3 years ago
4 0

Answer:

The answer is tendons, bones, ligaments, blood vessels

Explanation:

I just took the test on edgenuity

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A pool ball moving 1.83 m/s strikes an identical ball at rest. Afterward, the first ball moves 1.15 m/s at a 23.3° angle. What i
Oksi-84 [34.3K]

Answer:

v_{1fy} = - 0.4549 m / s

Explanation:

6 0
3 years ago
A tube is sealed at both ends and contains a 0.0100-m long portion of liquid. The length of the tube is large compared to 0.0100
Ahat [919]

Answer:

31.321 rad/s

Explanation:

L = Tube length

A = Area of tube

\rho = Density of fluid

v = Fluid velocity

m = Mass = \rho Al

Centripetal force is given by

F=\dfrac{mv^2}{L}\\ F=\dfrac{m(\omega L)^2}{L}\\ F=m\omega^2\\ F= 0.01A\rho\omega^2L

Pressure is given by

P=\dfrac{F}{A}=\rho gL\\\Rightarrow \dfrac{0.01A\rho\omega^2L}{A}=\rho gL\\\Rightarrow 0.01\omega^2=g\\\Rightarrow \omega^2=\dfrac{g}{0.01}\\\Rightarrow \omega=\sqrt{\dfrac{g}{0.01}}\\\Rightarrow \omega=\sqrt{\dfrac{9.81}{0.01}}\\\Rightarrow \omega=31.321\ rad/s

The angular speed of the tube is 31.321 rad/s

5 0
3 years ago
Firecrackers A and B are 600 m apart. You are standing exactly halfway between them. Your lab partner is 300 m on the other side
pishuonlain [190]

Answer:

See the explanation

Explanation:

Given:

Distance of Firecrackers A and B = 600 m

Event 1 = firecracker 1 explodes

Event 2 = firecracker 2 explodes

Distance of lab partner from cracker A = 300 m

You observe the explosions at the same time

to find:

does event 1 occur before, after, or at the same time as event 2?

Solution:

Since the lab partner is at 300 m distance from the firecracker A and Firecrackers A and B are 600 m apart

So the distance of fire cracker B from the lab partner is:

600 m  + 300 m = 900 m

It takes longer for the light from the more distant firecracker to reach so

Let T1 represents the time taken for light from firecracker A to reach lab partner

T1 = 300/c

It is 300 because lab partner is 300 m on other side of firecracker A

Let T2 represents the time taken for light from firecracker B to reach lab partner

T2 = 900/c

It is 900 because lab partner is 900 m on other side of firecracker B

T2 = T1

900 = 300

900 = 3(300)

T2 = 3(T1)

Hence lab partner observes the explosion of the firecracker A before the explosion of firecracker B.

Since event 1 = firecracker 1 explodes and event 2 = firecracker 2 explodes

So this concludes that lab partner sees event 1 occur first and lab partner is smart enough to correct for the travel time of light and conclude that the events occur at the same time.

8 0
3 years ago
Solve the inequality 2(n+3) – 4&lt;6. Then graph<br> the solution.
Aloiza [94]
The solution is 22 2(n+3)-4&6
6 0
3 years ago
Students who embrace an innate mindset might view their poor performance in math as inevitable or beyond their control
PilotLPTM [1.2K]

Answer:

True

Explanation:

4 0
4 years ago
Read 2 more answers
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