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3 years ago

In the sum of 54.34 and 45.66, the number of significant figure for theanswer is

Physics

1 answer:

Natali5045456 [20]3 years ago

4 0

Answer:

Explanation:

54.34+45.66=100.00 (you have to use the .00 because when you add to numbers you keep the number of decimals)

so you get 100.00

all numbers that are not 0 are sig figs so 1 is a sig fig

If a number ends with a 0 after a decimal place that 0 is a sig fig

all numbers between two sig figs are sig figs so that would make all of the numbers sig figs

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The way that scientists ensure that data is reliable Clear Explanation

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3 years ago

Sawyer launches his 180 kg raft on the Mississippi River by pushing on it with a force of 75N. How long must Sawyer push on the

Daniel [21]

Answer: 4.8 s

Explanation:

We have the following data:

$m=180 kg$ the mass of the raft

$F=75 N$ the force applied by Sawyer

$V=2 m/s$ the raft's final speed

$V_{o}=0 m/s$ the raft's initial speed (assuming it starts from rest)

We have to find the time $t$

Well, according to Newton's second law of motion we have:

$F=m.a$ (1)

Where $a$ is the acceleration, which can be expressed as:

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Substituting (2) in (1):

$F=m\frac{V-V_{o}}{t-t_{o}}$ (3)

Where $t_{o}=0$

Isolating $t$ from (3):

$t=\frac{m(V-V_{o})}{F}$ (4)

$t=\frac{180 kg(2 m/s-0 m/s)}{75 N}$

Finally:

$t=4.8 s$

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3 years ago

Two thin concentric spherical shells of radii r1 and r2 (r1 < r2) contain uniform surface charge densities V1 and V2, respect

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Answer:

Answer is explained in the explanation section below.

Explanation:

Solution:

We know that the Electric field inside the thin hollow shell is zero, if there is no charge inside it.

So,

a) 0 < r < r1 :

We know that the Electric field inside the thin hollow shell is zero, if there is no charge inside it.

Hence, E = 0 for r < r1

b) r1 < r < r2:

Electric field =?

Let, us consider the Gaussian Surface,

E x 4 $\pi$ $r^{2}$ = $\frac{Q1}{E_{0} }$

So,

Rearranging the above equation to get Electric field, we will get:

E = $\frac{Q1}{E_{0} . 4 \pi. r^{2} }$

Multiply and divide by $r1^{2}$

E = $\frac{Q1}{E_{0} . 4 \pi. r^{2} }$ x $\frac{r1^{2} }{r1^{2} }$

Rearranging the above equation, we will get Electric Field for r1 < r < r2:

E= (σ1 x $r1^{2}$ ) /( $E_{0}$ x $r^{2}$ )

c) r > r2 :

Electric Field = ?

E x 4 $\pi$ $r^{2}$ = $\frac{Q1 + Q2}{E_{0} }$

Rearranging the above equation for E:

E = $\frac{Q1+Q2}{E_{0} . 4 \pi. r^{2} }$

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As we know from above, that:

$\frac{Q1}{E_{0} . 4 \pi. r^{2} }$ = (σ1 x $r1^{2}$ ) /( $E_{0}$ x $r^{2}$ )

Then, Similarly,

$\frac{Q2}{E_{0} . 4 \pi. r^{2} }$ = (σ2 x $r2^{2}$ ) /( $E_{0}$ x $r^{2}$ )

So,

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Replacing the above equations to get E:

E = (σ1 x $r1^{2}$ ) /( $E_{0}$ x $r^{2}$ ) + (σ2 x $r2^{2}$ ) /( $E_{0}$ x $r^{2}$ )

Now, for

d) Under what conditions, E = 0, for r > r2?

For r > r2, E =0 if

σ1 x $r1^{2}$ = - σ2 x $r2^{2}$

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3 years ago

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Answer:

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3 years ago

In the sum of 54.34 and 45.66, the number of significant figure for theanswer is​

In the sum of 54.34 and 45.66, the number of significant figure for theanswer is