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Blizzard [7]
3 years ago
7

Write the formula for the compound created by the following two elements: selenium + gallium *

Physics
1 answer:
Elis [28]3 years ago
3 0

Explanation:

plz search in google

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It is claimed that if a lead bullet goes fast enough, it can melt completely when it comes to a halt suddenly, and all its kinet
TEA [102]

Answer:

v=346.05\ m.s^{-1}

Explanation:

Given:

initial temperature of the lead bullet, T_i=43^{\circ}C

latent heat of fusion of lead, L_f=2.32\times 10^4\ J.kg^{-1}

melting point of lead, T_m=327.3^{\circ}C

We have:

specific heat capacity of lead, c=129\ J.kg^{-1}.K^{-1}

<em>According to question the whole kinetic energy gets converted into heat which establishes the relation:</em>

\rm KE=(heat\ of\ rising\ the\ temperature\ from\ 43\ to\ 327.3\ degree\ C)+(heat\ of\ melting)

\frac{1}{2} m.v^2=m.c.\Delta T+m.L_f

\frac{1}{2} m.v^2=m(c.\Delta T+L_f)

\frac{v^2}{2} =129\times(327.3-43)+23200

v=346.05\ m.s^{-1}

3 0
3 years ago
Esercizio 1
timama [110]

Explanation:

can u please translate it to english language. I can't understand.

7 0
3 years ago
a. A rectangular pen is built with one side against a barn. 1900 m of fencing are used for the other three sides of the pen. Wha
Natalka [10]

Answer:

475 m , 950 m

Explanation:

Let l be the length of the side perpendicular to the barn.

1900-2l = length of the side parallel to the barn

Area A= l( 1900-2l)

A= 1900l-2l^2

now, the maximum value of l ( the equation being quadratic)

l_max= -b/2a

a= 2

b=1900

l_max= -1900/4= 475 m

then 1900-2l= 1900-2×(475) = 950 m

So, the dimensions that maximize area are

950 and 475

Now. A_max = -2( l_max)^2+1900×l_max

A_max=  -2(475)^2+1900×475

A_max= 451250 m^2

or, 475×950 = 451250 m^2

6 0
4 years ago
What are parasites? Give some example​
Inga [223]

Answer:

parasites are creatures the gain benefit off of other animals usually harming them eg:ticks on dogs

4 0
3 years ago
Zero, a hypothetical planet, has a mass of 4.5 x 1023 kg, a radius of 3.2 x 106 m, and no atmosphere. A 10 kg space probe is to
jok3333 [9.3K]

a.) K 2=K 1 +GmM( r 21− r 11)=2.2×10 7J

b.) ​K 2 +GmM( r 11− r 21)=6.9×10 7 J

Applying Law of  Energy conservation :

K 1+U 1

=K 2+U 2

⇒K 1− r 1GmM

=K 2− r 2 GmM

where M=5.0×10 23kg,r1

=> R=3.0×10 6m and m=10kg

(a) If K 1​

=5.0×10 7J and r 2

=4.0×10 6 m, then the above equation leads to

K 2=K 1 +GmM (r 21− r 11)=2.2×10 7J

(b) In this case, we require K 2

=0 and r2

=8.0×10 6m, and solve for K 1:K 1

​=K 2 +GmM (r 11− r 21)=6.9×10 7 J

Learn more about Kinetic energy on:

brainly.com/question/12337396

#SPJ4

7 0
2 years ago
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