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3 years ago

There are_ periods included in the periodic table. Answer here SUBMIT s

Physics

1 answer:

Assoli18 [71]3 years ago

4 0

Answer:

Explanation

There are currently seven complete periods in the periodic table, comprising the 118 known elements. Any new elements will be placed into an eighth period; see extended periodic table.

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You plan to take a trip to the moon. Since you do not have a traditional spaceship with rockets, you will need to leave the eart

hichkok12 [17]

Answer:

$v = 3.5 \times 10^5 m/s$

Explanation:

At some distance from the Earth the force of attraction due to moon is balanced by the force due to Moon

so we will have

$\frac{GM_em}{r^2} = \frac{GM_m}{(d-r)^2}$

now we have

$\frac{d - r}{r} = \sqrt{\frac{M_m}{M_e}}$

$\frac{3.844\times 10^8 - r}{r} = \sqrt{\frac{7.36 \times 10^{22}}{5.9742\times 10^{24}}}$

so we will have

$r = 3.46 \times 10^8 m$

Now by energy conservation

$-\frac{GM_e}{R_e} - \frac{GM_m}{d - (R_e + R_m)} + \frac{1}{2}v^2 = -\frac{GM_e}{r} - \frac{GM_m}{d - r}$

$-6.26 \times 10^{8} - 13046 + \frac{1}{2}v^2 = -1.15 \times 10^6 - 1.28 \times 10^5$

$v = 3.5 \times 10^5 m/s$

7 0

3 years ago

Ft-Fg= ma solve for a

densk [106]

The answer is a= Ft-Fg/m

8 0

3 years ago

Read 2 more answers

How much heat h1 is transferred to the skin by 25.0 g of steam onto the skin? the latent heat of vaporization for steam is l=2.2

elena-s [515]

The heat transferred by the steam to the skin is given by
$Q=m L_v$
where
m is the mass of the steam
$L_v$ is the latent heat of vaporization.

In our problem, the mass of the steam is (converting into kg)
$m=25.0 g=0.025 kg$
while the latent heat of vaporization of the steam is
$L_v = 2.256 \cdot 10^6 J/kg$
Substituting into the previous formula, we find the heat transferred to the skin:
$Q=m L_v = (0.025 kg)(2.256 \cdot 10^6 J/kg)=56400 J = 2.56 \cdot 10^4 J$

8 0

3 years ago

Light travels through a vacuum at a speed of 3 x 10 m/s. What is the speed of

Andrew [12]

Answer:

v = c / n (n = 1 for air)

v = c / 1.33 = 3 * 10E8 m/s / 1.33 = 2.25 * 10E8 m/s

3 0

3 years ago

Two wires of the same material and having the same volume, are fixed

Setler79 [48]

Answer:

48 kg

Explanation:

Given that the two wires are of same material, so their value of young's modulus will be same

Assuming that the wires are cylindrical in shape

As radius of the first wire is half that of the second wire and therefore the area of cross-section of the first wire will be one-fourth of the second wire( ∵ wire is cylindrical, the cross-sectional part will be circle and the area of the circle = π × r² )

As the volume is same for both wires

∴ π × ( $r_{1}$ )² × $l_{1}$ = π × ( $r_{2}$ )² × $l_{2}$

Here

$r_{1}$ is the radius of the first wire

$r_{2}$ is the radius of the second wire

$l_{1}$ is the length of the first wire

$l_{2}$ is the length of the second wire

⇒ π × (( $r_{2}$ )² ÷ 4) × $l_{1}$ = π × ( $r_{2}$ )² × $l_{2}$ (∵ radius of first wire is half that of the second wire)

By cancelling the same terms on both sides

we get

$l_{1}$ = 4 × $l_{2}$

⇒ Length of first wire will be four times of the length of second wire

<h3>Strain is defined as the elongation per unit length</h3>

Strain in first wire = ΔL ÷ $l_{1}$ = ΔL ÷ (4 × $l_{2}$ )

where ΔL is the elongation of the wire which in this case is same in both wires

Strain in second wire = ΔL ÷ $l_{2}$

∴ Strain in second wire is four times of strain in first wire

<h3>Stress = F ÷ A</h3>

where F is the force perpendicular to the cross-sectional area

A is the area of cross-section

Force in first wire = $m_{1}$ × g

where $m_{1}$ is the mass hanged to the first wire

g is the acceleration due to gravity

Force in second wire = $m_{2}$ × g

where $m_{2}$ is the mass hanged to the second wire

g is the acceleration due to gravity

Let $A_{1}$ be the cross-sectional area of first wire

$A_{2}$ be the cross-sectional area of second wire

$A_{2}$ = 4 × $A_{1}$ (∵ cross=sectional area of the wire = π × (radius of the wire)² )

Stress in first wire = ( $m_{1}$ × g) ÷ ( $A_{1}$ )

Stress in second wire = ( $m_{2}$ × g) ÷ ( $A_{2}$ ) = ( $m_{2}$ × g) ÷ (4 × $A_{1}$ )

<h3>Young's modulus is defined as Stress per unit strain</h3>

As Young's modulus is same for both wires, Stress per unit strain must be same for both wires

Stress per unit strain of first wire = (( $m_{1}$ × g) ÷ ( $A_{1}$ )) ÷ (ΔL ÷ (4 × $l_{2}$ ))

Stress per unit strain of second wire = (( $m_{2}$ × g) ÷ (4 × $A_{1}$ )) ÷ (ΔL ÷ $l_{2}$ )

By equating them we get

$m_{2}$ = 16 × $m_{1}$

⇒ $m_{2}$ = 16 × 3 = 48 kg

∴ $m_{2}$ = 48 kg

5 0

3 years ago