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3 years ago

Based on how stars are named, which star is probably brightest

2 answers:

5 0

The name of the brightest star is ursae majoris. it is a dwarf star and lie in the constellation of ursa major. that is why it has such similar name. around this star , we have three planets around this star. the distance of this star is around 46 light years. also it's mass is almost same as that of the sun. it rotates with a period of 24 hours

kvv77 [185]3 years ago

4 0

Sirius<span> is the brightest star known to earth . </span>

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How much mass of ice at 0℃ must melt when ∆Q=948000 J of heat energy is added to it?

butalik [34]

<h2>The ice melted is nearly 2.8 kg </h2>

Explanation:

The quantity of heat required to melt the ice can be found by the relation .

ΔQ = m x L

here L is used for the latent heat of fusion .

Its value is 334 J per gram for ice .

Thus m = $\frac{\Delta Q}{L}$ = $\frac{948000}{334}$ = 2836 gram

or = 2.8 kg approx

6 0

3 years ago

Boxers attempt to move with an opponent's punch when it is thrown. In other words, a boxer moves in the same direction as their

maxonik [38]

When the force of the opponent's punch is extended, with time, the effect of the blow or the force of the blow is reduced thereby reducing chances for a knockout punch from the opponent to the boxer.

5 0

4 years ago

Read 2 more answers

The masses of the Earth and Moon are 5.98×1024kg and 7.35×1022kg respectively, and their centers are separated by 3.84×108m.

bija089 [108]

Answer:

C) The Earth-Moon CM follows the orbit around the Sun. Earth and Moon rotate around their CM. The radius of rotation of Moon around the CM is much greater than radius of rotation of Earth around the CM.

Explanation:

At first we assume that both Earth and Moon can be treated as particles, the center of mass of the Earth-Moon system is obtained by using this formula:

$r_{CM} = \frac{r_{E}\cdot m_{E}+r_{M}\cdot m_{M}}{m_{E}+m_{M}}$ (Eq. 1)

Where:

$r_{E}$ - Location of the center of the Earth, measured in kilometers.

$r_{M}$ - Location of the Moon, measured in kilometers.

$r_{CM}$ - Location of the center of mass, measured in kilometers.

$m_{E}$ - Mass of the Earth, measured in kilograms.

$m_{M}$ - Mass of the Moon, measured in kilograms.

If we know that $r_{E} = 0\,km$ , $r_{M} = 3.84\times 10^{8}\,m$ , $m_{E} = 5.98\times 10^{24}\,kg$ and $m_{M} = 7.35\times 10^{22}\,kg$ , the location of the center of mass respect to the Earth is:

$r_{CM} = \frac{(0\,km)\cdot (5.98\times 10^{22}\,kg)+(3.84\times 10^{8}\,m)\cdot (7.35\times 10^{22}\,kg)}{5.98\times 10^{24}\,kg+7.35\times 10^{22}\,kg}$

$r_{CM} = 4.662\times 10^{6}\,m$

The Earth has a radius of $6.371\times 10^{6}$ meters, we notice that center of mass in located inside the Earth and the radius of rotation of the Earth around the center of mass is much greater than the radius of rotation of the Moon around the center of mass. That center of mass follows an orbit around the sun.

In consequence, correct answer is C.

4 0

3 years ago

Help! pls, look at the attachment

GarryVolchara [31]

Answer:

the bottom right.

Explanation:

electricity cannot jump. that is the only complete circut

3 0

3 years ago

Projectile <br> SHOW WORK<br> WILL MARK BRANLIEST <br> (Draw Picture and Label)

m_a_m_a [10]

a) The horizontal distance covered by the projectile is 600 m

b) The projectile reaches its maximum height after 3.00 s

c) The altitude of the highest point is 44.1 m

Explanation:

The motion of the projectile consists of two independent motions:

- A uniform motion (constant velocity) along the horizontal direction

- A uniformly accelerated motion, with constant acceleration (acceleration of gravity) in the downward direction

In this part A, we just need to analyze the horizontal motion. We know that:

The projectile travels horizontally with a constant velocity ov $v_x = 100 m/s$
The total time of flight of the projectile is $t=6.00 s$

Therefore, the horizontal distance covered by the projectile is given by

$x=v_x t$

And substituting, we find

$x=(100)(6.0) = 600 m$

For this part, we need to analyze the vertical motion of the projectile.

First, we want to find the initial vertical velocity. We can do it by using the equation for the vertical displacement:

$s=u_y t + \frac{1}{2}at^2$

where:

$u_y$ is the initial vertical velocity

$a=g=-9.8 m/s^2$ is the acceleration of gravity (negative because it is downward)

t is the time

s is the vertical displacement

We know that the total time of flight is t = 6.00 s: this means that when t=6, the projectile returns to its initial vertical position, so s = 0. Substituting and solving for $u_y$ , we get

$u_y = - \frac{1}{2}at=-\frac{1}{2}(-9.8)(6)=29.4 m/s$

The vertical velocity then as a function of t is given by

$v_y = u_y + at$

And at the maximum height, it becomes zero: $v_y = 0$ . Substituting and solving for t, we find the time at which the projectile reaches the maximum height: