All categories

2 years ago

In the following calculations, be sure to express the answer in standard scientific notation with the appropriate number of

Physics

1 answer:

Ratling [72]2 years ago

4 0

Answer:

-45,597.07

Explanation:

if not in scientific calculator and yung answer nung sa scientific sa comment na lang dinadownload ko ka eh

You might be interested in

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 are some of the physical benefits to be derived from aerobic

VladimirAG [237]

Some of the benefits are increased heart muscles, increase in blood flow, and reduced body fat

6 0

3 years ago

Energy from solar radiation may be ________ or taken in by a surface or an object.

tester [92]

Answer:

Absorbed

Explanation:

I hope this helps you

7 0

3 years ago

An ambulance with a siren emitting a whine at 1790 Hz overtakes and passes a cyclist pedaling a bike at 2.36 m/s. After being pa

Deffense [45]

Answer:

The speed of the ambulance is 4.30 m/s

Explanation:

Given:

Frequency of the ambulance, f = 1790 Hz

Frequency at the cyclist, f' = 1780 Hz

Speed of the cyclist, v₀ = 2.36 m/s

let the velocity of the ambulance be 'vₓ'

Now,

the Doppler effect is given as:

$f'=f\frac{v\pm v_o}{v\pm v_x}$

where, v is the speed of sound

since the ambulance is moving towards the cyclist. thus, the sign will be positive

thus,

$v_x=\frac{f}{f'}(v+v_o)-v$

on substituting the values, we get

$v_x=\frac{1790}{1780}(343+2.36)-343$

vₓ = 4.30 m/s

Hence, <u>the speed of the ambulance is 4.30 m/s</u>

6 0

2 years ago

A person's body is producing energy internally due to metabolic processes. If the body loses more energy than metabolic processe

Allushta [10]

To solve this problem it is necessary to apply the concepts related to the Stefan-Boltzman law that is responsible for calculating radioactive energy.

Mathematically this expression can be given as

$P = \sigma Ae\Delta T^4$

Where

A = Surface area of the Object

$\sigma =$ Stefan-Boltzmann constant

e = Emissivity

T = Temperature (Kelvin)

Our values are given as

$A = 1.36m^2$

$\Delta T^4 = T_2^4 -T_1^4 = 307^4-T_1^4$

$\sigma = 5.67*10^{-8} J/(s m^2 K^4)$

$P = 122J/s$

$e = 0.7$

Replacing at our equation and solving to find the temperature 1 we have,

$P = \sigma Ae\Delta T^4$

$P = \sigma Ae (T_2^4 -T_1^4)$

$122 = (5.67*10^{-8})(1.36)(0.7)(307^4-T_1^4)$

$T_1 = 285.272K = 12.122\°C$

Therefore the the temperature of the coldest room in which this person could stand and not experience a drop in body temperature is 12°C

8 0

3 years ago