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3 years ago

It takes 200 N to move a box 10 meters in 8 seconds. How much power is

Physics

1 answer:

svetoff [14.1K]3 years ago

8 0

Answer:

Explanation:

The equation of power is defined as Power = Workdone/Time Taken

And workdone = Force x Distance so using these equations we get they workdone is, 200x 10 = 2000Nm.

Dividing workdone with time will yield power, 2000 ÷ 8 = 250 Nm/s = 250W.

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a person when asked to speak up,increases her sound level from 30dB to 60dB.The amount of power per unit area increased by? a)30

aivan3 [116]

Answer:

d) 1000 times

Explanation:

As we know that difference of sound level is given as

$L_2 - L_1 = 10 Log \frac{I_2}{I_1}$

so here we need to find the ratio of two intensity

it is given as

$Log\frac{I_2}{I_1} = \frac{(L_2- L_1)}{10}$

$Log\frac{I_2}{I_1} = \frac{60 - 30}{10}$

$Log\frac{I_2}{I_1} = 3$

now we have

$\frac{I_2}{I_1} = 10^3$

so it is

d) 1000 times

4 0

2 years ago

Hi please help with this question! Need the workings.

andreyandreev [35.5K]

<span>Mass of the copper penny m = 2.6 g Atomic mass of copper = 63.55, Atomic number = 29, So the number of neutrons = Atomic mass - Atomic number = 63 - 29 = 34 a. Neutron mass = 34 x (2.6 / 63.55) = 1.4 grams Copper atoms per mole = 6.040 x 10^23 atoms/mol moles of copper = 2.6 / 63.06 = 0.04123 mol Total atoms in the copper = 6.040 x 10^23 atoms/mol x 0.04123 mol = 0.25 x 10^23 atoms Number of electrons in the copper = 29 per atom Mass of the electron = 9.085 x 10^-28 g b. Electron mass = 0.25 x 10^23 x 29 x 9.085 x 10^-28 = 65.86 x 10^-5 g</span>

6 0

3 years ago

If Angle "a" is 25°, Angle "b" is:

Leokris [45]

I would say B but I have no clue

6 0

3 years ago

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

Lyrx [107]

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

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

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,

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

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 $r2^{2}$

4 0

2 years ago

What is one way that early scientific practice differed from modern scientific practice?

mote1985 [20]

Early hypotheses were not based on observations.
Early hypotheses were not tested by experimentation.
Early hypotheses were formed from scientific questions.
Early hypotheses were influenced by creative thinking

6 0

2 years ago

Read 2 more answers