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3 years ago

Please help asap 100 POINTS and BRAINLIEST to best answer.

Physics

2 answers:

andrew11 [14]3 years ago

6 0

D. :)

lmk if im right or wrong

borishaifa [10]3 years ago

6 0

Answer:

D. 5ºC

Explanation:

Not B, because 1k= -457.87 fahrenheit. That means it is also not C. Since the icicles are melting, then that means it is not A either.

5ºC=about 41ºF

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A steel ball with mass m is suspended from the ceiling at the bottom end of a light, 17.0-m-long rope. The ball swings back and

iogann1982 [59]

Answer:

1. 18.25 m/s

2. 0 m/s

Explanation:

1.So the centripetal acceleration of the ball at this lowest point must be, taking gravity into account

$a_c = \frac{T - mg}{m} = \frac{3mg - mg}{m} = 2g$

The speed at this point would then be

$v^2 = a_c r = 2gr = 2*9.8*17 = 333.2$

$v = \sqrt{333.2} = 18.25 m/s$

2. Similarly, if T = mg, then the centripetal acceleration must be

$a_c = \frac{T - mg}{m} = \frac{mg - mg}{m} = 0$

As the ball has no centripetal acceleration, its speed must also be 0 as well.

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4 years ago

Whenever one object exerts a force on another object, the second object exerts a force of the same magnitude on the first object

Naddik [55]

Answer:

Explanation:

Newton third law of motion state that action and reaction are equal and opposite

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3 years ago

1-Autotrophic plants are also called

tresset_1 [31]

Answer:

1. Producers

2. cell (guards)

3. Rhizobium

4. Rafflesia Arnoldii

5. Xylem

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3 years ago

1 - Don't put the coat on the snowman - it will melt him 2 - I don't think the coat will make any difference 3 - I think it will

Vinil7 [7]

Explanation:

I don't think the coat will make any difference

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3 years ago

A heavy flywheel is accelerated (rotationally) by a motor that provides constant torque and therefore a constant angular acceler

ki77a [65]

Answer:

a) $t_1=\frac{w_1-w_o}{\alpha}=\frac{w_1}{\alpha}sec$

b) $\theta_1=\frac{w_1^2}{2\alpha}rad$

c) $t_2=\frac{\alpha t_1}{5\alpha}=\frac{t_1}{5}sec$

Explanation:

1) Basic concepts

Angular displacement is defined as the angle changed by an object. The units are rad/s.

Angular velocity is defined as the rate of change of angular displacement respect to the change of time, given by this formula:

$w=\frac{\Delat \theta}{\Delta t}$

Angular acceleration is the rate of change of the angular velocity respect to the time

$\alpha=\frac{dw}{dt}$

2) Part a

We can define some notation

$w_o=0\frac{rad}{s}$ ,represent the initial angular velocity of the wheel

$w_1=?\frac{rad}{s}$ , represent the final angular velocity of the wheel

$\alpha$ , represent the angular acceleration of the flywheel

$t_1$ time taken in order to reach the final angular velocity

So we can apply this formula from kinematics:

$w_1=w_o +\alpha t_1$

And solving for t1 we got:

$t_1=\frac{w_1-w_o}{\alpha}=\frac{w_1}{\alpha}sec$

3) Part b

We can use other formula from kinematics in order to find the angular displacement, on this case the following:

$\Delta \theta=wt+\frac{1}{2}\alpha t^2$

Replacing the values for our case we got:

$\Delta \theta=w_o t+\frac{1}{2}\alpha t_1^2$

And we can replace $t_1$ from the result for part a, like this:

$\theta_1-\theta_o=w_o t+\frac{1}{2}\alpha (\frac{w_1}{\alpha})^2$

Since $\theta_o=0$ and $w_o=0$ then we have:

$\theta_1=\frac{1}{2}\alpha \frac{w_1^2}{\alpha^2}$

And simplifying:

$\theta_1=\frac{w_1^2}{2\alpha}rad$

4) Part c

For this case we can assume that the angular acceleration in order to stop applied on the wheel is $\alpha_1 =-5\alpha \frac{rad}{s}$

We have an initial angular velocity $w_1$ , and since at the end stops we have that $w_2 =0$

Assuming that $t_2$ represent the time in order to stop the wheel, we cna use the following formula

$w_2 =w_1 +\alpha_1 t_2$

Since $w_2=0$ if we solve for $t_2$ we got

$t_2=\frac{0-w_1}{\alpha_1}=\frac{-w_1}{-5\alpha}$

And from part a) we can see that $w_1=\alpha t_1$ , and replacing into the last equation we got:

$t_2=\frac{\alpha t_1}{5\alpha}=\frac{t_1}{5}sec$

5 0

3 years ago