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2 years ago

You are standing on roller skates if you try to throw a 20-pound bowling ball while standing on skates,

Physics

1 answer:

Elina [12.6K]2 years ago

5 0

Answer:

I'd probably drop it on my toes. That would happen.

Explanation:

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Find the length of a pendulum that oscillates with a frequency of 0.16 hz. the acceleration due to gravity is 9.81 m/s 2 . answe

Vsevolod [243]

The period of the pendulum is the reciprocal of the frequency:
$T= \frac{1}{f}= \frac{1}{0.16 Hz}=6.25 s$

The period of the pendulum is given by
$T=2 \pi \frac{L}{g}$
where L is the length of the pendulum, and g the acceleration of gravity. By re-arranging the formula and using the value of T we found before, we can calculate the length of the pendulum L:
$L=g \frac{T^2}{(2 \pi)^2}=(9.81 m/s^2) \frac{(6.25 s)^2}{(2 \pi)^2}=9.71 m$

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2 years ago

GIVING BRAINLIEST PLEASE HELP!!

OLga [1]

A. because I had this question yesterday.

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3 years ago

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geniusboy [140]

These anisotropies in the temperature map correspond to areas of varying density fluctuations in the early universe. Eventually, gravity would draw the high-density fluctuations into even denser and more pronounced ones.

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3 years ago

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Calculate the current passing through a conductor of resistance 4ohms. If a potential difference of 15V its ends______

statuscvo [17]

Explanation:

current = velocity/resistance

I = V/R

15/4

current = 3.75A

hope this helps...

6 0

2 years ago

A curve that has a radius of 90 m is banked at an angle of =10.8∘. If a 1100 kg car navigates the curve at 75 km/h without skidd

PilotLPTM [1.2K]

The minimum coefficient of static friction between the pavement and the tires is 0.69.

The given parameters;

radius of the curve, r = 90 m
angle of inclination, θ = 10.8⁰
speed of the car, v = 75 km/h = 20.83 m/s
mass of the car, m = 1100 kg

The normal force on the car is calculated as follows;

$F_n = mgcos(\theta)$

The frictional force between the car and the road is calculated as;

$F_k = \mu_k F_n\\\\F_k = \mu_k mgcos(\theta)$

The net force on the car is calculated as follows;

$mgsin(\theta) + \mu_s mgcos(\theta) = \frac{mv^2}{r} \\\\mg(sin\theta \ + \ \mu_s cos\theta)= \frac{mv^2}{r} \\\\g(sin\theta \ + \ \mu_s cos\theta)= \frac{v^2}{r}\\\\sin\theta \ + \ \mu_s cos\theta = \frac{v^2}{rg}\\\\\mu_s cos\theta = sin\theta \ + \ \frac{v^2}{rg}\\\\\mu_s = \frac{sin\theta}{cos \theta} + \frac{v^2}{cos (\theta)rg}\\\\\mu_s = tan(\theta) + \frac{v^2}{cos (\theta)rg}\\\\\mu_s = tan(10.8) + \frac{(20.83)^2}{cos(10.8) \times 90 \times 9.8} \\\\\mu_s = 0.19 + 0.5\\\\$

$\mu_s = 0.69$

Thus, the minimum coefficient of static friction between the pavement and the tires is 0.69.

Learn more here:brainly.com/question/15415163

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2 years ago