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grandymaker [24]

2 years ago

10

How does inertia explain why it is difficult to stop a moving skateboard?

1 answer:

My name is Ann [436]2 years ago

6 0

Answer:

(Example Person) has to push hard to get the skateboard started, but once it begins moving, it takes much less effort to keep it rolling over the smooth, flat pavement. In fact, if (Example Person) tries to stop the rolling skateboard, it may take as much effort to stop it as it did to start it rolling in the first place.

Explanation:

Hope this helps :)

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Neil pogo sticks to his science class, but stops to pick up his backpack on his way. He travels 8 m east, then 4 m west. what di

Likurg_2 [28]

Answer:

8.9 m NorthEast

Explanation:

a² + b² = c²

8² + 4² = c²

64 + 16 = c²

80 = c²

c = √80

c ≈ 8.9 m NorthEast

8 0

2 years ago

A satellite of mass M = 270kg is in circular orbit around the Earth at an altitude equal to the earth's mean radius (6370 km). A

zubka84 [21]

To solve this problem we will apply the concepts related to Orbital Speed as a function of the universal gravitational constant, the mass of the planet and the orbital distance of the satellite. From finding the velocity it will be possible to calculate the period of the body and finally the gravitational force acting on the satellite.

PART A)

$V_{orbital} = \sqrt{\frac{GM_E}{R}}$

Here,

M = Mass of Earth

R = Distance from center to the satellite

Replacing with our values we have,

$V_{orbital} = \sqrt{\frac{(6.67*10^{-11})(5.972*10^{24})}{(6370*10^3)+(6370*10^3)}}$

$V_{orbital} = 5591.62m/s$

$V_{orbital} = 5.591*10^3m/s$

PART B) The period of satellite is given as,

$T = 2\pi \sqrt{\frac{r^3}{Gm_E}}$

$T = \frac{2\pi r}{V_{orbital}}$

$T = \frac{2\pi (2*6370*10^3)}{5.591*10^3}$

$T = 238.61min$

PART C) The gravitational force on the satellite is given by,

$F = ma$

$F = \frac{1}{4} mg$

$F = \frac{270*9.8}{4}$

$F = 661.5N$

5 0

3 years ago

A coin of mass m rests on a turntable a distance r from the axis of rotation. The turntable rotates with a frequency of f. What

Crank

Answer:

$\mu = \frac{r (2\pi f)^{2}}{g}$

Explanation:

$N$ = normal force acting on the coin

Normal force in the upward direction balances the weight of the coin, hence

$N = mg$

$f$ = frequency of rotation

Angular velocity of turntable is hence given as

$w = 2\pi f$

$r$ = distance from the axis of rotation

$\mu$ = minimum coefficient of static friction

static frictional force is given as

$f = \mu N\\f = \mu mg$

The static frictional force provides the necessary centripetal force , hence

Centripetal force = Static frictional force

$m r w^{2} = \mu mg\\r w^{2} = \mu g\\\\\mu = \frac{r w^{2}}{g} \\\mu = \frac{r (2\pi f)^{2}}{g}$

3 0

3 years ago

The electron has the positron as its antiparticle.<br><br> A True<br> B False

Sphinxa [80]

The answer would be true

8 0

2 years ago

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You are faced with situations every day where you have to make choices, and there is a deep reason that compels you to do so. So

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Why did you put this here?

6 0

3 years ago