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2 years ago

I will fan and give 2 medals!!!!!!

Physics

1 answer:

Maksim231197 [3]2 years ago

7 0

Corona

I. Rare gaseous envelope

II. Irregular pearly glow

III. Total solar eclipse

IV. Surrounds darkened moon

V. Caused by diffraction

Solar Prominence

I. loop shape

II. large bright gaseous

III. luminous hydrogen gas

IV. rises above chromospheres

V. anchored to surface (sun’s surface)

Solar Flare

I. Intense High-energy

II. Causes electromagnetic disruptions

III. In solar atmosphere

IV. Varies in brightness

V. Emits radiation

Sunspots

I. Surface dark spots

II. Strong magnetic fields

III. Solar magnetic storms

IV. Visible through telescope

V. Cooler than photospheres

I did this in class I got 100%

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The value of 1.0004 to the power 1 by 2 using Binomial approximation is

IgorLugansk [536]

Given:

The given value is $(1.0004)^{\frac{1}{2}}$ .

To find:

The value of the given expression by using the Binomial approximation.

Explanation:

We have,

$(1.0004)^{\frac{1}{2}}$

It can be written as:

$(1.0004)^{\frac{1}{2}}=(1+0.0004)^{\frac{1}{2}}$

$(1.0004)^{\frac{1}{2}}=1+\dfrac{1}{2}\times 0.0004$ $[\because (1+x)^n=1+nx]$

$(1.0004)^{\frac{1}{2}}=1+0.0002$

$(1.0004)^{\frac{1}{2}}=1.0002$

Therefore, the approximate value of the given expression is 1.0002.

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2 years ago

A 4.67-g bullet is moving horizontally with a velocity of +357 m/s, where the sign + indicates that it is moving to the right (s

Leni [432]

Answer:

(a)0.531m/s

(b)0.00169

Explanation:

We are given that

Mass of bullet, m=4.67 g= $4.67\times 10^{-3} kg$

1 kg =1000 g

Speed of bullet, v=357m/s

Mass of block 1, $m_1=1177g=1.177kg$

Mass of block 2, $m_2=1626 g=1.626 kg$

Velocity of block 1, $v_1=0.681m/s$

(a)

Let velocity of the second block after the bullet imbeds itself=v2

Using conservation of momentum

Initial momentum=Final momentum

$mv=m_1v_1+(m+m_2)v_2$

$4.67\times 10^{-3}\times 357+1.177(0)+1.626(0)=1.177\times 0.681+(4.67\times 10^{-3}+1.626)v_2$

$1.66719=0.801537+1.63067v_2$

$1.66719-0.801537=1.63067v_2$

$0.865653=1.63067v_2$

$v_2=\frac{0.865653}{1.63067}$

$v_2=0.531m/s$

Hence, the velocity of the second block after the bullet imbeds itself=0.531m/s

(b)Initial kinetic energy before collision

$K_i=\frac{1}{2}mv^2$

$k_i=\frac{1}{2}(4.67\times 10^{-3}\times (357)^2)$

$k_i=297.59 J$

Final kinetic energy after collision

$K_f=\frac{1}{2}m_1v^2_1+\frac{1}{2}(m+m_2)v^2_2$

$K_f=\frac{1}{2}(1.177)(0.681)^2+\frac{1}{2}(4.67\times 10^{-3}+1.626)(0.531)^2$

$K_f=0.5028 J$

Now, he ratio of the total kinetic energy after the collision to that before the collision

= $\frac{k_f}{k_i}=\frac{0.5028}{297.59}$

=0.00169

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2 years ago

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If the mass of the sun is 1x, at least one planet will fall into the habitable zone. if I place a planet in orbits 2, 6, and 75, and all planets will orbit the sun successfully.

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2 years ago

To practice Tactics Box 9.1 Calculating the Work Done by a Constant Force. Recall that the work W done by a constant force F⃗ at

insens350 [35]

Answer:

The vector magnitudes F and r are always postive, so the sign o W is determined entirely by the angle e between the force and the displacement.Submit Figure 1 off 1 part C

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