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3 years ago

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1 answer:

sdas [7]3 years ago

5 0

- my best friend shouldn’t die because if she does then who is going to make me laugh? There would be nobody to laugh at my bad jokes and there won’t be nobody there for me when I need it the most. My best friend is needed in this world with out them I would be lost.

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Okay so I've have 10 Brainliest, through my whole time using brainly- I want people to get brainly and points so Im giving away

Nookie1986 [14]

answer:

thank youu for the points cute drawing by the way (✿◠‿◠)

when did you join brainly though?

have a wonderful dayyy !!

4 0

3 years ago

Read 2 more answers

A square-based shipping crate is being designed that must contain a volume of 16 ft3 . The material that is used for the base an

Vlada [557]

Answer:

Explanation:

Given

volume $V=16 ft^3$

Suppose base is square with side L

height of crate is h

Volume $V=L^2\times h$

$16=L^2\times h$

Cost of top and bottom area $c_1=3L^2$

Cost of Side area $c_2=4Lh\times 2=8Lh=8L\times \frac{16}{L^2}=\frac{128}{L}$

Total Cost $C=c_1+c_2$

Total Cost $C=3L^2+\frac{128}{L}$

Differentiate C w.r.t Length

$\frac{dC}{dL}=6L-\frac{128}{L^2}$

$L^3=\frac{128}{6}$

$L=2.75 ft$

$h=\frac{16}{2.75^2}=11.46 ft$

Dimensions are $L\times L\times h=2.75\times 2.75\times 11.46$

6 0

3 years ago

Suppose that a person riding on the top of a freight car shines a searchlight beam in the direction in which the train is travel

Snezhnost [94]

Answer:

same

Explanation:

Acc. to Einstien's postulate of special theory of

Relativity , Velocity of the light beam is same in all frames of references

(a) If the freight car is at rest

The frame we can assumed as Non - inertial frame of reference s

In the inertial frame of reference , velocity of the light beam has its own value as : 3 x 10^8 m/s

(b) If the freight car is moving , the frame we can assumed as Non -inertial frame of reference

In thus case also , The velocity of the light beam will also have the same value as ; 3 x 108 m/s

6 0

3 years ago

Two charged spheres are 20 cm apart and exert an attractive force of 8 x 10-9 n on each other. What will the force of attraction

Pavel [41]

Answer:

$3.2\cdot 10^{-8} N$

Explanation:

The inital electrostatic force between the two spheres is given by:

$F=k\frac{q_1 q_2}{r^2}$

where

$F=8\cdot 10^{-9} N$ is the initial force

k is the Coulomb's constant

q1 and q2 are the charges on the two spheres

r is the distance between the two spheres

The problem tells us that the two spheres are moved from a distance of r=20 cm to a distance of r'=10 cm. So we have

$r'=\frac{r}{2}$

Therefore, the new electrostatic force will be

$F'=k\frac{q_1 q_2}{(r')^2}=k\frac{q_1 q_2}{(r/2)^2}=4k\frac{q_1 q_2}{r^2}=4F$

So the force has increased by a factor 4. By using $F=8\cdot 10^{-9} N$ , we find

$F'=4(8\cdot 10^{-9} N)=3.2\cdot 10^{-8} N$

6 0

3 years ago

The names of the seven days of the week are derived from the names of the bodies of the solar system that are visible to the nak

rjkz [21]

Answer:

Baja naja kanaka kanaka man's kana kana

7 0

3 years ago