1. A wave on a rope has a wavelink of 2.0m And a frequency of 2.0 Hz. What is the speed of the wave?

Assume that a lightning bolt can be modeled as a long, straight line of current. If 16 C of charge passes by a point in 1.50 x 1

Reptile [31]

Answer:

$B = 7.9012*10^{-5}T$

Explanation:

To solve the problem, the concepts related to the magnetic field and the current produced in a lightning bolt are necessary.

The current is defined by the load due to time, that is to say

$I= \frac{q}{t}$

Where,

$q= Charge$

$t = time$

So the current can be expressed as:

$I = \frac{16}{1.5*10^{-3}}$

$I = 10666.67A$

Once the current is found it is now possible to find the magnetic field, as this is given by the equation,

$B =\frac{\mu_0}{2\pi}\frac{I}{r}$

Where,

$\mu_0 =$ Permeability Constant

I= Current

r= radius

Replacing the values we have

$B=\frac{4\pi*10^{-7}}{2\pi}(\frac{10666.67}{27})$

$B = 7.9012*10^{-5}T$

7 0

3 years ago

Suppose you could fit 100 dimes, end to end, between your card with the pinhole and your dime-sized sunball. how many suns could

Naddika [18.5K]

Answer: 100 suns

Explanation:

We can solve this with the following relation:

$\frac{d}{x_{sunball-pinhole}}=\frac{D}{x_{sun-pinhole}}$

Where:

$d=17.91 mm =17.91(10)^{-3} m$ is the diameter of a dime

$D$ is the diameter of the Sun

$x_{sun-pinhole}=150,000,000 km=1.5(10)^{11} m$ is the distance between the Sun and the pinhole

$x_{sunball-pinhole}=100 d=1.791 m$ is the amount of dimes that fit in a distance between the sunball and the pinhole

Finding $D$ :

$D=\frac{d}{x_{sunball-pinhole}}x_{sun-pinhole}$

$D=\frac{17.91(10)^{-3} m}{1.791 m} 1.5(10)^{11} m$

$D=1.5(10)^{9} m$ This is roughly the diameter of the Sun

Now, the distance between the Earth and the Sun is one astronomical unit (1 AU), which is equal to:

$1 AU=149,597,870,700 m$

So, we have to divide this distance between $D$ in order to find how many suns could it fit in this distance:

$\frac{149,597,870,700 m}{1.5(10)^{9} m}=99.73 suns \approx 100 suns$

8 0

3 years ago

At standard temperature and pressure, carbon dioxide has a density of 1.98 kg/m3. What volume does 1.70 kg of carbon dioxide occ

nikitadnepr [17]

Answer:

<h2>volume= 0.85m^3</h2>

Explanation:

<em>The density of a substance is defined as the mass per unit volume of the substance, the unit is in kg/m^3 and it is represented by the greek letter rho</em>

Step one:

given data

we are told that the density of Co2= 1.98 kg/m3

and the mass of Co2 is= 1.70 kg

we know the relation between mass, volume and density is

$density=mass/volume$

make volume subject of formula we have

$volume=mass/density$

substitute we have

$volume=1.7/1.98\\\\volume= 0.85m^3$

8 0

3 years ago

The acceleration due to gravity at the surface of a planet depends on the planet's mass and size; therefore other planets will h

Cloud [144]

Chaff has also written about his book of Mormon that is not true to say this but it doesn't have the

5 0

3 years ago

How would the composition of an atom change if both the atomic number and mass number each increase by one? A) The atom would ha

weqwewe [10]

Choice-B is the correct one.

-- The atomic number is the number of protons in the nucleus.
-- Each proton in the nucleus is usually matched by one electron in the 'cloud'.
-- The addition of a proton OR a neutron increases the mass number by 1 .
-- Electrons have such small mass that they don't figure into the atomic mass at all.

8 0

3 years ago