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3 years ago

__________ is the most common type of stretching.

Physics

2 answers:

Verizon [17]3 years ago

6 0

Answer: static stretching

Explanation:

e.g rubberband

faust18 [17]3 years ago

4 0

Answer:

static stretching

Explanation:

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The current theory of the structure of the

Mariana [72]

Answers:

a) $2.82(10)^{21} kg$

b) $1410 J$

c) $36.62 m/s$

Explanation:

<h3>a) Mass of the continent</h3>

Density $\rho$ is defined as a relation between mass $m$ and volume $V$ :

$\rho=\frac{m}{V}$ (1)

Where:

$\rho=2720 kg/m^{3}$ is the average density of the continent

$m$ is the mass of the continent

$V$ is the volume of the continent, which can be estimated is we assume it as a a slab of rock 5300 km on a side and 37 km deep:

$V=(length)(width)(depth)=(5300 km)(5300 km)(37 km)=1,030,330,000 km^{3} \frac{(1000 m)^{3}}{1 km^{3}}=1.03933(10)^{18} m^{3}$

Finding the mass:

$m=\rho V$ (2)

$m=(2720 kg/m^{3})(1.03933(10)^{18} m^{3})$ (3)

$m=2.82(10)^{21} kg$ (4) This is the mass of the continent

<h3>b) Kinetic energy of the continent</h3>

Kinetic energy $K$ is given by the following equation:

$K=\frac{1}{2}mv^{2}$ (5)

Where:

$m=2.82(10)^{21} kg$ is the mass of the continent

$v=4.8 \frac{cm}{year} \frac{1 m}{100 cm} \frac{1 year}{365 days} \frac{1 day}{24 hours} \frac{1 hour}{3600 s}=1(10)^{-9} m/s$ is the velocity of the continent

$K=\frac{1}{2}(2.82(10)^{21} kg)(1(10)^{-9} m/s)^{2}$ (6)

$K=1410 J$ (7) This is the kinetic energy of the continent

<h3>c) Speed of the jogger</h3>

If we have a jogger with mass $m=77 kg$ and the same kinetic energy as that of the continent $1413 J$ , we can find its velocity by isolating $v$ from (5):

$v=\sqrt{\frac{2 K}{m}}$ (6)

$v=\sqrt{\frac{2 (1413 J)}{77 kg}}$

Finally:

$v=36.62 m/s$ This is the speed of the jogger

5 0

4 years ago

STATE THE HOOKE'S LAW

victus00 [196]

The Hooke's law is a principal of physics that states that the force needed to extend or compress a spring by some distance x scales linearly with respect to that distance.

6 0

4 years ago

Read 2 more answers

In a Hydrogen atom an electron rotates around a stationary proton in a circular orbit with an approximate radius of r =0.053nm.

leonid [27]

Answer:

(a): $F_e = 8.202\times 10^{-8}\ \rm N.$

(b): $F_g = 3.6125\times 10^{-47}\ \rm N.$

(c): $\dfrac{F_e}{F_g}=2.27\times 10^{39}.$

Explanation:

Given that an electron revolves around the hydrogen atom in a circular orbit of radius r = 0.053 nm = 0.053 $\times 10^{-9}$ m.

Part (a):

According to Coulomb's law, the magnitude of the electrostatic force of interaction between two charged particles of charges $q_1$ and $q_2$ respectively is given by

$F_e = \dfrac{k|q_1||q_2|}{r^2}$

where,

$k$ = Coulomb's constant = $9\times 10^9\ \rm Nm^2/C^2.$
$r$ = distance of separation between the charges.

For the given system,

The Hydrogen atom consists of a single proton, therefore, the charge on the Hydrogen atom, $q_1 = +1.6\times 10^{-19}\ C.$

The charge on the electron, $q_2 = -1.6\times 10^{-19}\ C.$

These two are separated by the distance, $r = 0.053\times 10^{-9}\ m.$

Thus, the magnitude of the electrostatic force of attraction between the electron and the proton is given by

$F_e = \dfrac{(9\times 10^9)\times |+1.6\times 10^{-19}|\times |-1.6\times 10^{-19}|}{(0.053\times 10^{-9})^2}=8.202\times 10^{-8}\ \rm N.$

Part (b):

The gravitational force of attraction between two objects of masses $m_1$ and $m_1$ respectively is given by

$F_g = \dfrac{Gm_1m_2}{r^2}.$

where,

$G$ = Universal Gravitational constant = $6.67\times 10^{-11}\ \rm Nm^2/kg^2.$
$r$ = distance of separation between the masses.

For the given system,

The mass of proton, $m_1 = 1.67\times 10^{-27}\ kg.$

The mass of the electron, $m_2 = 9.11\times 10^{-31}\ kg.$

Distance between the two, $r = 0.053\times 10^{-9}\ m.$

Thus, the magnitude of the gravitational force of attraction between the electron and the proton is given by

$F_g = \dfrac{(6.67\times 10^{-11})\times (1.67\times 10^{-27})\times (9.11\times 10^{-31})}{(0.053\times 10^{-9})^2}=3.6125\times 10^{-47}\ \rm N.$

The ratio $\dfrac{F_e}{F_g}$ :

$\dfrac{F_e}{F_g}=\dfrac{8.202\times 10^{-8}}{3.6125\times 10^{-47}}=2.27\times 10^{39}.$

6 0

3 years ago

In January 2006, astronomers reported the discovery of a planet comparable in size to the earth orbiting another star and having

Orlov [11]

Answer:

R = 5.28 103 km

Explanation:

The definition of density is

ρ = m / V

V = m /ρ

Where m is the mass and V the volume of the body

The volume of a sphere is

V = 4/3 π r³

Let's replace

4/3 π r³ = m / ρ

R =∛ ¾ m / ρ π

The mass of the planet is

M = 5.5 Me

R = ∛ ¾ 5.5 Me /ρ π

Let's reduce the density to SI units

ρ = 1.76 g / cm³ (1 kg / 10³ g) (10² cm / 1 m)³

ρ = 1.76 10³ kg / m³

Let's calculate

R = ∛ ¾ 5.5 5.97 10²⁴ / (1.76 10³ pi)

R = ∛ 0.14723 10²¹

R = 0.528 10⁷ m

R = 0.528 104 km

R = 5.28 103 km

8 0

3 years ago

1. El puente mas largo del mundo es el puente Akashi Kaikyo, en Japón. El puente mide 3910 m de largo y está construido con acer

bearhunter [10]

Answer:

It's 1.0000042 times longer in summer than in winter. It represents a 1.6 centimeters difference between seasons.

Explanation:

The linear coefficient of thermal expansion for steel is about $1.2*10^{-7}\°C^{-1}$ . From the equation of linear thermal expansion, we have:

$L_f=L_0(1+\alpha\Delta T)$

Taking the winter day as the initial, and the summer day as the final, we can take the relationship between them:

$L_{summer}=L_{winter}[1+(1.2*10^{-7}\°C^{-1})(30\°C+5\°C)]\\\\L_{summer}=(1.0000042)L_{winter}$

It means that the bridge is 1.0000042 times longer in summer than in winter. If we multiply it by the length of the bridge, we obtain that the difference is of about 1.6 centimeters between the two seasons.

8 0

4 years ago