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3 years ago

8

Darryl finds a bottle of what looks like clear water, with dirt settled at the bottom. When he shakes the bottle, the water gets

dirty. Which term best describes the matter in the bottle?

1 answer:

Alexus [3.1K]3 years ago

7 0

Was there any choice of answers so i can help or something u have to figure out

You might be interested in

Compare and contrast instantaneous and average speed.

Harlamova29_29 [7]

Answer:

instantaneous velocity is a velocity covered at an instant while average velocity is the change in distance/ the change in time taken

6 0

2 years ago

A merry-go-round of radius 2 m is rotating at one revolution every 5 s. A

galben [10]

Answer:

a) The angular speed of the child is approximately 1.257 rad/s

b) The angular speed of the teenager is approximately 1.257 rad/s

c) The tangential speed of the child is approximately 1.257 m/s

d) For the child, r = 2 m

The tangential speed of the teenager is approximately 2.513 m/s

Explanation:

The revolutions per minute, r.p.m. of the merry-go-round = 1 revolution/(5 s)

The radius of the merry-go-round = 2 m

The location of the child = 1 m from the axis

The location of the teenager = 2 m from the axis

1 revolution = 2·π radians

Therefore, we have;

The angular speed, ω = (Angle turned)/(Time elapsed) = (2·π radians)/(5 s)

∴ The angular speed of the merry-go-round, ω = 2·π/5 radians/second

a) The angular speed of the child = The angular speed of the merry-go-round = 2·π/5 radians/second ≈ 1.257 rad/s

b) The angular speed of the teenager = The angular speed of the merry-go-round = 2·π/5 radians/second ≈ 1.257 rad/s

c) The tangential speed, v = r × The angular speed, ω

Where;

r = The radius of rotation of the object

For the child, r = 1 m

The tangential speed of the child = 1 m × 2·π/5 radians/second = 2·π/5 m/s ≈ 1.257 m/s

d) For the child, r = 2 m

The tangential speed of the teenager = 2 m × 2·π/5 radians/second = 4·π/5 m/s ≈ 2.513 m/s

8 0

2 years ago

What are three elements that are in the same group

cluponka [151]

Iodine, chloride, and bromide

3 0

2 years ago

The y-component of the force F which a person exerts on the handle of the box wrench is known to be 86 lb. Determine the x-compo

emmainna [20.7K]

Answer:

x-component of force is 38.18 lb where as magnitude of Force is 93.16

Explanation:

Fy of the force F exerted on the handle of the box wrench = 86 lb

Considering the triangle in Fig 1

magnitude of perpendicular = P = 12

magnitude of base = B = 5

using Pythagoras theorem

$H= \sqrt{P^{2} + B^{2}}$

$H= \sqrt{12^{2} + 5^{2}}$

$H=13\\\implies cos \theta = \frac{5}{13} \\\implies sin \theta = \frac{12}{13}\\$

y-component of force is given given as:

$86 = Fsin\theta\\F = 86 (\frac{13}{12})\\F = 93.16 lb\\F_{x} =F cos \theta\\F_{x} = 93.16 (\frac{5}{13})\\F_{x} =38.18 lb.$

5 0

3 years ago

A car initially traveling at 24 m/s slams on the brakes and moves forward 196 m before coming to a complete halt. What was the m

Marrrta [24]

Answer:

$-1.47 m/s^2$

Explanation:

We can use the following SUVAT equation to solve the problem:

$v^2 - u^2 = 2ad$

where

v = 0 is the final velocity of the car

u = 24 m/s is the initial velocity

a is the acceleration

d = 196 m is the displacement of the car before coming to a stop

Solving the equation for a, we find the acceleration:

$a=\frac{v^2-u^2}{2d}=\frac{0-(24)^2}{2(196)}=-1.47 m/s^2$

4 0

3 years ago