Which substance(s) will sink if placed in a fluid that has a density of 1.19 g/ml? (List all)

James is planning a science fair project on sound waves. He places an alarm inside a jar which he can remove the

ss7ja [257]

Answer:

D

Explanation:

this is simple, because when air is removed, means that there is no particles in the jar so vacuum is achieved, and when you can't hear a sound means that the sound couldn't travel through the vacuum. which means that sound cannot travel through vacuum,

as a result, sound requires a medium ( air) travel from one point to another.

hope it helps, if not please report it so that someone else gets to try it

7 0

3 years ago

The drill used by most dentists today is powered by a small air-turbine that can operate at angular speeds of 350000 rpm. These

alexdok [17]

Answer:

θ = 6.3 *10³ revolutions

Explanation:

Angular acceleration of the drill

We apply the equations of circular motion uniformly accelerated

ωf= ω₀ + α*t Formula (1)

Where:

α : Angular acceleration (rad/s²)

ω₀ : Initial angular speed ( rad/s)

ωf : Final angular speed ( rad

t : time interval (s)

Data

ω₀ = 0

ωf = 350000 rpm = 350000 rev/min

1 rev = 2π rad

1 min= 60 s

ωf = 350000 rev/min =350000*(2π rad/60 s)

ωf = 36651.9 rad/s

t = 2.2 s

We replace data in the formula (2) :

ωf= ω₀ + α*t

36651.9 = 0 + α* (2.2)

α = 36651.9 / (2.2)

α = 17000 rad/s²

Revolutions made by the drill

We apply the equations of circular motion uniformly accelerated

ωf²= ω₀ ²+ 2α*θ Formula (2)

Where:

θ : Angle that the body has rotated in a given time interval (rad)

We replace data in the formula (2):

(ωf)²= ω₀²+ 2α*θ

(36651.9)²= (0)²+ 2( 17000 )*θ

θ = (36651.9)²/ (34000 )

θ = 39510.64 rad = 39510.64 rad* (1 rev/2πrad)

θ = 6288.31 revolutions

θ = 6.3 *10³ revolutions

3 0

3 years ago

A student applies a force of 245 N to move a bicycle 12.0 meters. How much work is done on the bicycle

atroni [7]

Answer:

Explanation:

The work done by an object can be found by using the formula

workdone = force × distance

From the question we have

workdone = 245 × 12

We have the final answer as

Hope this helps you

8 0

3 years ago

When reading the printout from a laser printer, you are actually looking at an array of tiny dots.

solniwko [45]

Answer:

The value is $y = 3.097 * 10^{-5} \ m$

Explanation:

From the question we are told that

The diameter of the pupil is $d_p = 4.2 \ mm = 4.2 *10^{-3} \ m$

The distance of the page from the eye $d = 29 \ cm = 0.29 \ m$

The wavelength is $\lambda = 500 \ nm = 500 *10^{-9} \ m$

The refractive index is $n_r = 1.36$

Generally the minimum separation of adjacent dots that can be resolved is mathematically represented as

$y = [ \frac{1.22 * \lambda }{d_p * n_r } ]* d$

$y = [ \frac{1.22 * 500 *10^{-9} }{4.2 *10^{-3} * 1.36} ]* 0.29$

$y = 3.097 * 10^{-5} \ m$

7 0

3 years ago

A girl weighing 50 kgf wears sandals of pencil heel of area of cross section 1 cm^2, stands on the floor.An elephant weighing 20

Klio2033 [76]

Answer:

$\boxed{{\boxed{\blue{ 12.5}}}}$

Explanation:

Given, for girl : Weight or force;

$\rm \: F_1 = 50 \: kgf$

Area of both heels;

$\rm \: A_1 = \; 2 ×1 \; cm^2 = 2 \: cm^2$

$\rm \: Pressure \: P_1 = \cfrac{F_1}{ A_1 } = \dfrac{50 \: kgf}{2 \: cm {}^{2} } = 25 \: kgf \: cm {}^{ - 1}$

For elephant, Weight = Force $\rm F_2$ = 2000 kg•f

Area of 4 feet;

$\rm \: A_2 = \; 4 \times 250 \; cm^2 = 1000 \: cm^2$

$\rm \: Pressure \: P_2 = {F_2}/{A_2} \; = \cfrac{2 \cancel{0 00 }\: kgf}{1 \cancel{000} \: cm^2} = 2 \: kgf \: { \:cm}^{- 1}$

Now;

$\rm = \dfrac{Pressure \: Exerted \: by \: the \: Girl}{Pressure \: exerted \: by \: the \: elephant}$

$= \rm \: P_1/P_2$

$\implies \rm\cfrac{25 \: kgf \: \: cm {}^{ - 2} }{2 \: kgf \: cm {}^{ - 2} } = \rm\cfrac{25 \: \cancel{kgf \: \: cm {}^{ - 2}} }{2 \: \cancel{ kgf \: cm {}^{ - 2}} } = \boxed{12.5}$

Thus, the girl's pointed heel sandals exert 12.5 times more pressure P than the pressure P exerted by the elephant.

I aspire this helps!

3 0

2 years ago