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4 years ago

"I want to run, to skip, and to jump" is an example of _____.

2 answers:

6 0

<span>The question is asking us to define the sentence ""I want to run, to skip, and to jump". This is a parallel construction, which means that the three parts ("to run", "to skip" and "to jump" are in the same kind of construction - to <verb>. It could be considered comma splice, but a comma splice would rather be the case if there were different subjects : for example in "I want to run, she wants to skip". </span>

timofeeve [1]4 years ago

3 0

Answer:

parallel construction

Explanation:

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A 5.0 g coin is placed 15 cm from the center of a turntable. The coin has static and kinetic coefficients of friction with the t

Alexandra [31]

Answer:

the coin does not slide off

Explanation:

mass (m) = 5 g = 0.005 kg

distance (r) = 15 cm = 0.15 m

static coefficient of friction (μs) = 0.8

kinetic coefficient of friction (μk) = 0.5

speed (f) = 60 rpm

acceleration due to gravity (g) = 9.8 m/s^{2}

lets first find the angular speed of the table

ω = 2πf

ω = 2 x π x 60 x $\frac{1}{60}$

ω = 6.3 s^{-1]

Now lets find the maximum static force between the coin and the table so we can get the maximum velocity the coin can handle without sliding

static force (Fs) = ma

static force (Fs) = μs x Fn = μs x m x g

Fs = 0.8 x 0.005 x 9.8 = 0.0392 N

Fs = ma

0.0392 = 0.005 x a

a = 7.84 m/s^{2}

$(Vmax)^{2}$ = a x r

$(Vmax)^{2}$ = 7.84 x 0.15

Vmax = 1.08 m/s

ωmax = $\frac{Vmax}{r}$

ωmax = $\frac{1.08}{0.15}$ = 7.2 s^{-1}

now that we have the maximum angular acceleration of the table, we can calculate its maximum speed in rpm

Fmax = $\frac{ωmax}{2π}$

Fmax = $\frac{7.2}{2 x π}$ = 68.7 rpm

since the table is rotating at a speed less than the maximum speed that the static friction can hold coin on the table with, the coin would not slide off.

4 0

4 years ago

A group of students are provided with three objects all of the same mass and radius. The objects include a solid cylinder, a thi

SOVA2 [1]

Answer:

Sphere, cylinder hoop

Explanation:

To analyze Which student is right it is best to propose the solution of the problem. Let's look for the speed of the center of mass. Let's use the concept of mechanical energy

In the highest part of the ramp

Em₀ = U = mg y

In the lowest part

Here the energy has part of translation and part of rotation

$E_{mf}$ = $K_{T}$ + $K_{R}$

$E_{mf}$ = ½ m $v_{cm}$ ² + ½ I w²

Where I is the moment of inertia of the body and w the angular velocity that relates to the velocity of the center of mass

$v_{cm}$ = w r

w = $v_{cm}$ / r

Let's replace

$E_{mf}$ = ½ I ( $v_{cm}$ / r)²

Energy is conserved

mg y = ½ m $v_{cm}$ ² + ½ I $v_{cm}$ ² / r2

½ (m + I / r²) $v_{cm}$ ² = m g y

½ (1 + I / m r²) $v_{cm}$ ² = g y

$v_{cm}$ = √ [2gy / (1 + I / mr²)]

This is the velocity of the center of mass of the bodies, as they all have the same radius with comparing this point is sufficient. Now let's use the speed definition

v = d / t

t = d / v

t = d / (√ [2gy / (1 + I / mr²)])

t = (d / √ 2gy) √(1 + I / m r²)

Therefore we see that time is proportional to the square root. All quantities are constant and the one that varies is the moment of inertia.

The moments of inertia of

Sphere is Is = 2/5 M r²

Cylinder Ic = ½ M r²

Hoop Ih = M r²

Let's replace each one and calculate the time

Sphere

ts = (d / √2gy) √ (1 + 2/5 Mr² / mr²)

ts = (d / √ 2gy) √ (1 +2/5) = (d / √ 2gy) √(1.4)

ts = (d / √ 2gy) 1.1

Cylinder

tc = (d / √2gy) √ (1 + 1/2 Mr² / Mr²)

tc = (d / √2gy) √ (1 + ½) = (d / √ 2gy) √ 1.5

tc = (d / √ 2gy) 1.2

Hoop

th = (d / √2gy) √ (1 + mr² / mr²)

th = (d / √2gy) √(1 + 1) = (d / √ 2gy) √ 2

th = (d / √ 2gy) 1.41

We have the results for the time the body that arrives the fastest is the sphere and the one that is the most hoop. Therefore the correct answer is

ts < tc < th

Sphere, cylinder hoop

5 0

3 years ago

If the density of iron is 7.87 g/cm3, what is the volume of the 12.00 g piece of iron? use proper significant figures.

Lelechka [254]

1.524777636594663 cm^3

7 0

3 years ago

1. A car moves 20 m/s east for 5 sec, 10m/s north for 10 sec and then 30m/s west for 5 sec. Calculate a) displacement

aalyn [17]

Try the solutions described in the attached picture, note the answers are marked with green colour.

3 0

3 years ago

Please answer both questions thanks Will mark as brainiest

ololo11 [35]

D.

gas>liq>solid
as density increases, speed decreases.

4 0

3 years ago