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4 years ago

If velocity is positive, which would most likely yield a negative acceleration?

Physics

1 answer:

VLD [36.1K]4 years ago

6 0

Sorry for inconvenience it says something is wrong so I took a screenshot of that

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1.- Del siguiente sistema de fuerzas en el cual la interacción de todas las tensiones hace que

Crazy boy [7]

Answer:

i can help you but go to the school Grade 4 like my pro

Explanation:

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4 years ago

Explain how the biosphere,hydrosphere,geosphere, and atmosphere are interconnected

erastova [34]

<span>gimme a second i a thinking.

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4 years ago

Professor Stefanovic is spinning a bucket of water by extending his arm and rotating his shoulder in class to show the effects o

anyanavicka [17]

Answer:

a ) 2.368 rad/s

b) 3.617 rad/s

Explanation:

the minimum angular velocity that Prof. Stefanovic needs to spin the bucket for the water not to fall out can be determined by applying force equation in a circular path

i.e

$F_{inward } = F_G + F_T$ ------ equation (1)

where;

$F_{inward} = m *a_c$

$F_{inward} = m*r* \omega^2$

Also

$F_G = m*g$

$F_T = 0$ since; that is the initial minimum angular velocity to keep the water in the bucket

Now; we can rewrite our equation as :

$mr \omega^2= mg + 0\\\omega^2 = \frac{m*g}{m*r}\\\omega^2 = \frac{g}{r}\\\omega = \sqrt{\frac{g}{r} \ \ } ------ equation \ \ \ {2}$

So; Given that:

The rope that is attached to the bucket is lm long and his arm is 75 cm long.

we have our radius r = 1 m + 75 cm

= ( 1 + 0.75 ) m

= 1.75 m

g = acceleration due to gravity = 9.81 m/s²

Replacing our values into equation (2) ; we have:

$\omega = \sqrt{ \frac{9.81}{1.75}}\\\omega = 2.368 \ rad/s$

b) if he detaches the rope and spins the bucket by holding it with his hand ; then the radius = 0.75 m

∴

$\omega = \sqrt{ \frac{9.81}{0.75}}\\\omega = 3.617 \ rad/s$

4 0

4 years ago

Total energy input is 1400 and my efficiency is 80% so what is the useful energy output in jules?

Ierofanga [76]

Answer

1400 * 0.8 = 1,120

the useful energy output is 1,120 joules

4 0

4 years ago

You hang a heavy ball with a mass of 94 kg from a platinum rod 3.2 m long by 2.2 mm by 3.1 mm. You measure the stretch of the ro

vlada-n [284]

Answer:

$1.67\times 10^{11} N/m^2$

Explanation:

Mass of the ball, $m=94 kg$

Force = weight of the ball

$F=mg = 94kg\times 9.81 m/s^2=922.14N$

For Young's modulus, substitute the values in the formula which is as follows:

$\frac{F}{A}=Y\frac{\Delta L}{L}\\\Rightarrow Y= \frac{FL}{A\Delta L}\\\Rightarrow Y =\frac{922.14 \times 3.2}{2.2\times 10^{-3}\times 3.1\times 10^{-3} \times 0.002588}=1.67\times 10^{11} N/m^2$

4 0

3 years ago