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3 years ago

1A, 3B, and 7A are examples of group ___ on the periodic table.

Physics

2 answers:

AysviL [449]3 years ago

7 0

Answer : The correct option is, Number

Explanation :

The periodic table is a type of table in which the element are systematically arranged on the basis of the atomic number.

The periodic table have the groups and periods. In the periodic table, there are 7 horizontal rows of elements are known as period and 18 vertical columns of elements are known as group.

Hence, 1A, 3B, and 7A are examples of group number on the periodic table.

german3 years ago

6 0

Answer:

Explanation:

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On august 10, 1972, a large meteorite skipped across the atmosphere above the western united states and western canada, much lik

RUDIKE [14]

(a)
The velocity of the meteorite just before hitting the ground is:
$v=20 km/s=20000 m/s$
The loss of energy of the meteorite corresponds to the kinetic energy the meteorite had just before hitting the ground, so:
$\Delta K = \frac{1}{2}mv^2= \frac{1}{2}(3.4 \cdot 10^6 kg)(20000 m/s)^2=6.8 \cdot 10^{14}J$

(b) 1 megaton of tnt is equal to $1 MT=4.2 \cdot 10^{15}J$
To find to how many megatons the meteorite energy loss $\Delta E$
corresponds, we can set the following proportion
$1 MT: 4.2 \cdot 10^{15}J=x: \Delta E$
And so we find
$x= \frac{\Delta E}{4.2 \cdot 10^{15}J} = \frac{6.8 \cdot 10^{14}J }{4.2 \cdot 10^{15}J} =0.162 MT$
So, 0.162 megatons.

(c) 1 Hiroshima bomb is equivalent to 13 kilotons (13 kT). The impact of the meteorite had an energy of $\Delta E=0.162 MT=162 kT$ . So, to find to how many hiroshima bombs it corresponds, we can set the following proportion:
$1:13 kT=x:162 kT$
And so we find
$x= \frac{162 kT}{13 kT}=12.46$
So, the energy released by the impact of the meteorite corresponds to the energy of 12.46 hiroshima bombs.

7 0

4 years ago

A rotating space station is said to create "artificial gravity" - a loosely-defined term used for an acceleration that would be

FrozenT [24]

Answer:

$\omega=0.31\frac{rad}{s}$

Explanation:

The artificial gravity generated by the rotating space station is the same centripetal acceleration due to the rotational motion of the station, which is given by:

$a_c=\frac{v^2}{r}(1)$

Here, r is the radius and v is the tangential speed, which is given by:

$v=\omega r(2)$

Here $\omega$ is the angular velocity, we replace (2) in (1):

$a_c=\frac{(\omega r)^2}{r}\\\\a_c=\omega^2r$

Recall that $r=\frac{d}{2}=\frac{200m}{2}=100m$ .

Solving for $\omega$ :

$\omega=\sqrt{\frac{a_c}{r}}\\\omega=\sqrt{\frac{9.8\frac{m}{s^2}}{100m}}\\\omega=0.31\frac{rad}{s}$

3 0

4 years ago

An athlete performing a long jump leaves the ground at a 32.9 ∘ angle and lands 7.78 m away.part a )what was the takeoff speed ?

MatroZZZ [7]

Answer:

Explanation:

Given

Inclination $\theta =32.9^{\circ}$

Distance of landing point $R=7.78\ m$

Considering athlete to be an Projectile

range of projectile is given by

$R=\frac{u^2\sin 2\theta }{g}$

where u=launch velocity

$7.78=\frac{u^2\sin (2\times 32.9)}{9.8}$

$u^2=83.58$

$u=\sqrt{83.58}$

$u=9.142\ m/s$

(b)If u is increased by 8% then

new velocity is $u'=9.87\ m/s$

$R'=\frac{u'^2\sin 2\theta }{g}$

$R'=9.07\ m$

3 0

4 years ago

Read 2 more answers

Q C Ganymede is the largest of Jupiter's moons. Consider a rocket on the surface of Ganymede, at the point farthest from the pla

guapka [62]

The answer is $v=15.6 \mathrm{~km} / \mathrm{s}$ .

<h3>What is kinetic energy?</h3>

A particle or an item that is in motion has a sort of energy called kinetic energy. An item accumulates kinetic energy when work, which involves the transfer of energy, is done on it by exerting a net force.
Kinetic energy comes in five forms: radiant, thermal, acoustic, electrical, and mechanical.
The energy of a body in motion, or kinetic energy (KE), is essentially the energy of all moving objects. Along with potential energy, which is the stored energy present in objects at rest, it is one of the two primary types of energy.
Explain that a moving object's mass and speed are two factors that impact the amount of kinetic energy it will possess.

Determine the escape speed for the rocket from the planet-satellite system:

The potential energy of the rocket due to Ganymede when it is on the surface of the Ganymede is,

$U_1=-\frac{G M_{\mathrm{G}} m}{R_G}$

The potential energy of the rocket due to Jupiter

when it is on the surface of the Ganymede is,

$U_2=-\frac{G M_{\mathrm{J}} m}{R}$

Here, R is a separation between Jupiter and Ganymede.

To escape from the surface of Ganymede potential energy of the rocket due to Jupiter and Ganymede is equal to the kinetic energy of the rocket.

$\begin{aligned}&\frac{1}{2} m v^2+U_1+U_2=0 \\&\frac{1}{2} m v^2=-U_1-U_2 \\&\frac{1}{2} m v^2=\frac{G M_{\mathrm{G}} m}{R_G}+\frac{G M_{\mathrm{J}} m}{R} \\&\frac{1}{2} v^2=\frac{G M_{\mathrm{G}}}{R_G}+\frac{G M_{\mathrm{J}}}{R}\end{aligned}$

$v^2=\frac{2 G M_{\mathrm{G}}}{R_G}+\frac{2 G M_{\mathrm{T}}}{R}$

$v=\sqrt{2 G\left(\frac{M_{\mathrm{G}}}{R_G}+\frac{M}{R}\right)}$

$v=\sqrt{2\left(6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^2 / \mathrm{kg}^2\right)\left(\frac{1.495 \times 10^{23} \mathrm{~kg}}{2.64 \times 10^6 \mathrm{~m}}+\frac{1.90 \times 10^{27} \mathrm{~kg}}{1.071 \times 10^9 \mathrm{~m}}\right)}$

$v=15.6 \times 10^3 \mathrm{~m} / \mathrm{s}$

$v=15.6 \mathrm{~km} / \mathrm{s}$

To learn more about kinetic energy, refer to:

brainly.com/question/25959744

#SPJ4

3 0

2 years ago

An observer moving toward Earth with a speed of 0.84 c notices that it takes 5.1 min for a person to fill her car with gas. Supp

stealth61 [152]

Answer:

t₂=6.35min

Explanation:

t₁ = first observed time (=5.1 min)

t₂ = unknown; this is the quantity we want to find

V₁ = observer's initial speed (=0.84c)

V₂ = observer's final speed (=0.90c)

Lorentz factors for V₁ and V₂:

γ₁ = 1/√(1−(V₁/c)²)

γ₂ = 1/√(1−(V₂/c)²)

The "proper time" (the time measured by the person filling her car) is:

t′ = t₁/γ₁

The proper time is stated to be the same for both observations, so we also have:

t′ = t₂/γ₂

Combine those two equations and solve for t₂

t₂ = t₁(γ₂/γ₁)

t₂= t₁√((1−(V₁/c)²)/(1−(V₂/c)²))

$t_{2}=5.1\sqrt{\frac{1-(0.84)^{2} }{1-(0.9)^{2} } }\\\\t_{2}=6.348min$

8 0

4 years ago