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3 years ago

Hey girls and guys.. I AM TAKING MY FINAL RN IN CLASS!! I NEED HELP AND WUICKLY

Physics

1 answer:

nordsb [41]3 years ago

7 0

Same as the speed as continental drift

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The angular velocity of a process control motor is (13−12t2) rad/s, where t is in seconds. Part A At what time does the motor re

mihalych1998 [28]

Answer:

Explanation:

Given

$\omega =13-\frac{1}{2}\cdot t^2$

Motor reverse its direction when \omega =0

$13-0.5t^2=0$

$26=t^2$

$t=\sqrt{26}=5.099\approx 5.1 s$

(b)

$\frac{\mathrm{d} \theta }{\mathrm{d} t}=\omega$

$\int d\theta =\int_{0}^{5.1}\omega dt$

$\int d\theta =\int_{0}^{t}(13-.05t^2)dt$

$\theta =(13t-0.1667\times t^3)_0^{5.1}$

$\theta =44.192^{\circ}$

4 0

3 years ago

An astronaut has a mass of 50.0 kg on earth. what is her mass on the moon, where gravity is 1 6 that on earth?

never [62]

I believe the correct gravity on the moon is 1/6 of Earth. Take note there is a difference between 1 6 and 1/6.

HOWEVER, we should realize that the trick here is that the question asks about the MASS of the astronaut and not his weight. Mass is an inherent property of an object, it is unaffected by external factors such as gravity. What will change as the astronaut moves from Earth to the moon is his weight, which has the formula: weight = mass times gravity.

<span>Therefore if he has a mass of 50 kg on Earth, then he will also have a mass of 50 kg on moon.</span>

6 0

3 years ago

How would you write the number 6,500,000,000 in scientific notation?

photoshop1234 [79]

Scientific form = 6.5 x 109.

8 0

3 years ago

A 100 N force is applied to a box at an angle of 60° to the

Anarel [89]

Answer: lift force = 100sin60 = 86.6 N

pull force = 100sin60 = 50.0 N

Explanation:

8 0

2 years ago

a race car accelerates uniformly from 18.5 m/s to 46.1m/s in 2.47 seconds detrrmine the acceleration of the car and distance tra

LenaWriter [7]

Because acceleration is constant, the acceleration of the car at any time is the same as its average acceleration over the duration. So

$a=\dfrac{\Delta v}{\Delta t}=\dfrac{46.1\,\frac{\mathrm m}{\mathrm s}-18.5\,\frac{\mathrm m}{\mathrm s}}{2.47\,\mathrm s}=11.2\,\dfrac{\mathrm m}{\mathrm s^2}$

Now, we have that

${v_f}^2-{v_0}^2=2a\Delta x$

so we end up with a distance traveled of

$\left(46.1\,\dfrac{\mathrm m}{\mathrm s}\right)^2-\left(18.5\,\dfrac{\mathrm m}{\mathrm s}\right)^2=2\left(11.2\,\dfrac{\mathrm m}{\mathrm s^2}\right)\Delta x$

$\implies\Delta x=79.6\,\mathrm m$

6 0

3 years ago