How many mL beakers would be when seprating salt,sand,iron fillings,and beads

Physics

1 answer:

-BARSIC- [3]3 years ago

7 0

Explanation:

Mixtures consist of two or more substances chemically combined. Mixtures cannot be separated by physical means. Mixtures consist of only one kind of atom. ***Mixtures have two or more substances .a 50-mL beaker with the same label.

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An atom containing 16 protons, 31 neutrons, and 16 electrons would have an atomic mass of

Eddi Din [679]

47 amu is the correct answer

7 0

3 years ago

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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;

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