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3 years ago

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Physical science

1 answer:

attashe74 [19]3 years ago

3 0

Answer:

0.56 km/s

Explanation:

We will define a single system of units for measurement, for this case meters per second [m/s]. That is, we must convert the rest of units such as centimeters per second and kilometers per second to meters per second.

$560[\frac{cm}{s}]*(\frac{1m}{100cm} )=5.6[m/s]\\0.56[\frac{km}{s}]*(\frac{1000m}{1km} )=560[m/s]$

Therefore the speed of 0.56 [km/s] is the greatest of all

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Why do we get dizzy when we spin?

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The body senses whether it is upright or lying down or whether it is moving or standing still through the vestibular system, which is in the upper portion of the inner ear.

3 0

3 years ago

Several springs are connected as illustrated below in (a). Knowing the individual springs stiffness k1 = 20 N/m, k2 = 30 N/m, k3

Hatshy [7]

Answer:

The equivalent stiffness of the string is 8.93 N/m.

Explanation:

Given that,

Spring stiffness is

$k_{1}=20\ N/m$

$k_{2}=30\ N/m$

$k_{3}=15\ N/m$

$k_{4}=20\ N/m$

$k_{5}=35\ N/m$

According to figure,

$k_{2}$ and $k_{3}$ is in series

We need to calculate the equivalent

Using formula for series

$\dfrac{1}{k}=\dfrac{1}{k_{2}}+\dfrac{1}{k_{3}}$

$k=\dfrac{k_{2}k_{3}}{k_{2}+k_{3}}$

Put the value into the formula

$k=\dfrac{30\times15}{30+15}$

$k=10\ N/m$

k and $k_{4}$ is in parallel

We need to calculate the k'

Using formula for parallel

$k'=k+k_{4}$

Put the value into the formula

$k'=10+20$

$k'=30\ N/m$

$k_{1}$ ,k' and $k_{5}$ is in series

We need to calculate the equivalent stiffness of the spring

Using formula for series

$k_{eq}=\dfrac{1}{k_{1}}+\dfrac{1}{k'}+\dfrac{1}{k_{5}}$

Put the value into the formula

$k_{eq}=\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{35}$

$k_{eq}=8.93\ N/m$

Hence, The equivalent stiffness of the string is 8.93 N/m.

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3 years ago

What do MRI and ultrasound have in common as diagnostic imaging techniques? Check all that apply.

Gennadij [26K]

Answer:

low risk for tissue damage

uses radio waves

the last three are not correct

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8 0

3 years ago

Read 2 more answers

A 300 g glass thermometer initially at 23 ◦C is put into 236 cm3 of hot water at 87 ◦C. Find the final temperature of the thermo

DIA [1.3K]

Answer:

$74^{\circ} C$

Explanation:

We are given that

Mass of glass, $m=300 g$

$T_1=23^{\circ}$

Volume, $V=236cm^3$

Mass of water= $density\times volume=1\times 236=236 g$

Density of water= $1g/cm^3$

Temperature of hot water, $T=87^{\circ}$

Specific heat of glass, $C_g=0.2cal/g^{\circ}C$

Specific heat of water, $C_w=1 cal/g^{\circ}C$

$Q_{glass}=m_gC_g(T_f-T_1)=300\times 0.2(T_f-23)$

$Q_{water}=m_wC_w(T_f-T)=236\times 1(T_f-87)$

$Q_{glass}+Q_{water}=0$

$300\times 0.2(T_f-23)+236\times 1(T_f-87)$

$60T_f-1380+236T_f-20532=0$

$296T_f=20532+1380=21912$

$T_f=\frac{21912}{296}=74^{\circ} C$

5 0

3 years ago

For thermal equilibrium at temperature Tan appropriate measure of energy is kT where k is Boltzmann's constant. Convert the foll

Schach [20]

Answer:

1 cm⁻¹ =1.44K 1 ev = 1.16 10⁴ K

Explanation:

The relationship between temperature and thermal energy is

E = K T

The relationship of the speed of light

c =λ f = f / ν 1/λ= ν

The Planck equation is

E = h f

Let's start the transformations

c = f λ = f / ν

f = c ν

E = h f

E = h c ν

E = KT

h c ν = K T

T = h c ν / K =( h c / K) ν

Let's replace the constants

h = 6.63 10⁻³⁴ J s

c = 3 10⁸ m / s

K = 1.38 10⁻²³ J / K

v = 1 cm-1 (100 cm / 1 m) = 10² m-1

T = (6.63 10⁻³⁴ 3. 10⁸ / 1.38 10⁻²³) 1 10²

A = h c / K = 1,441 10⁻²

T = 1.44K

ν = 103 cm⁻¹ = 103 10² m

T = (6.63 10⁻³⁴ 3. 10⁸ / 1.38 10⁻²³) 103 10²

T = 148K

1 Rydberg = 1.097 10 7 m

As we saw at the beginning the λ=1 / v

T = (h c / K) 1 /λ

T = 1,441 10⁻² 1 / 1,097 10⁷

T = 1.3 10⁻⁹ K

E = 1Ev (1.6 10⁻¹⁹ J /1 eV) = 1.6 10⁻¹⁹ J

E = KT

T = E/K

T = 1.6 10⁻¹⁹ /1.38 10⁻²³

T = 1.16 10⁴ K

3 0

3 years ago