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KonstantinChe [14]

3 years ago

12

Various types of tree sloths share the honor of being the slowest moving mammals. An average tree sloth moves at a speed of 0.74

3 m/s. How long does it take a sloth moving at this speed to travel 22.30 m?

1 answer:

WARRIOR [948]3 years ago

3 0

22.30/0.743 = 30.013 seconds.

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16 to 19 year old male and females are how much more likely to be involved in a crash?

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I say around 40% - 60%

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(I just corrected the question. Sorry if it is still incorrect.)

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The national government would like to start a program that would give states money for flood relief. Which of the following stat

rewona [7]

Answer:

When a disaster is declared, the Federal government, led by the Federal Emergency Management Agency (FEMA), responds at the request of, and in support of, States, Tribes, Territories, and Insular Areas and local jurisdictions impacted by a disaster.

Explanation:

3 0

2 years ago

A wheel of radius R, mass M, and moment of inertia I is mounted on a frictionless, horizontal axles. A light cord wrapped around

Alex_Xolod [135]

Answer:

$\alpha =\frac{m*g*R}{I-m*R^2}$

$a = \frac{m*g*R^2}{I-m*R^2}$

$T=\frac{I*m*g}{I-m*R^2}$

Explanation:

By analyzing the torque on the wheel we get:

$T*R=I*\alpha$ Solving for T: $T=I/R*\alpha$

On the object:

$T-m*g = -m*a$ Replacing our previous value for T:

$I/R*\alpha-m*g = -m*a$

The relation between angular and linear acceleration is:

$a=\alpha*R$

So,

$I/R*\alpha-m*g = -m*\alpha*R$

Solving for α:

$\alpha =\frac{R*m*g}{I+m*R^2}$

The linear acceleration will be:

$a =\frac{R^2*m*g}{I+m*R^2}$

And finally, the tension will be:

$T =\frac{I*m*g}{I+m*R^2}$

These are the values of all the variables: α, a, T

8 0

3 years ago

A boy is whirling a stone around his head by means of a string. The string makes one complete revolution every second, and the t

slega [8]

Answer:

The tension increases to four times its original value.

Explanation:

v = Velocity

r = Radius

m = Mass of stone

The centripetal force is

$F_c=m\dfrac{v^2}{r}$

The tension will balance the centripetal force

$T=m\dfrac{v^2}{r}$

$\\\Rightarrow T\propto v^2$

$\dfrac{T_1}{T_2}=\dfrac{v_1^2}{v_2^2}\\\Rightarrow \dfrac{T_1}{T_2}=\dfrac{v_1}{2^2v_1^2}\\\Rightarrow \dfrac{T_1}{T_2}=\dfrac{1}{4}\\\Rightarrow T_2=4T_1$

The new tension will be 4 times the old tension

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3 years ago