1. Using the data from M7:<br>Which four countries are the worst in releasing CO2?

The 1.5kg ball is launched straight upward with an initial velocity of 7m/s. What is the maximum height it will reach?

Temka [501]

Answer:

h = 2.5 m

Explanation:

Given that,

Mass of a ball, m = 1.5 kg

Initial velocity of the ball, u = 7 m/s

We need to find the maximum height reached by the ball. Let it is be h. Using the conservation of energy to find it such that,

$mgh=\dfrac{1}{2}mu^2\\\\h=\dfrac{u^2}{2g}$

Put all the values,

$h=\dfrac{7^2}{2\times 9.8}\\\\=2.5\ m$

So, it will reach to a height of 2.5 m.

8 0

3 years ago

Represent 7468 N with SI units having an appropriate prefix. Express your answer to four significant figures and include the app

Musya8 [376]

Answer:

7.468 kN

Explanation:

Here the force is given in Newton

Some of the prefixes of the SI units are

kilo = 10³

Mega = 10⁶

Giga = 10⁹

Tera = 10¹²

The number is 7468.0

Here, the only solution where the number of significant figures is kilo. If any other prefix is chosen then the significant figures will increase.

1 kilonewton = 1000 Newton

$1\ Newton=\frac{1}{1000}\ kilonewton$

$\\\Rightarrow 7468\ Newton=\frac{7468}{1000}\ kilonewton\\ =7.468\ kilonewton$

So, 7468 N = 7.468 kN

7 0

3 years ago

Explain in details the concept of economic geography

vova2212 [387]

Answer:

In the words of Hartshorn and Alexander: “Economic Geography is the study of the spatial variation on the earth’s surface of activities related to producing, exchanging and consuming goods and services. Whenever possible the goal is to develop generalizations and theories to account for these spatial variations.”

Explanation:

3 0

3 years ago

What is the magnitude of the magnetic field at a point midway between them if the top one carries a current of 19.5 A and the bo

Phantasy [73]

Answer:

The magnetic field will be $\large{\dfrac{1.4 \times 10^{-4}}{d}} T$ , '2d' being the distance the wires.

Explanation:

From Biot-Savart's law, the magnetic field ( $\large{\overrightarrow{B}}$ ) at a distance ' $r$ ' due to a current carrying conductor carrying current ' $I$ ' is given by

$\large{\overrightarrow{B} = \dfrac{\mu_{0}I}{4 \pi}} \int \dfrac{\overrightarrow{dl} \times \hat{r}}{r^{2}}}$

where ' $\overrightarrow{dl}$ ' is an elemental length along the direction of the current flow through the conductor.

Using this law, the magnetic field due to straight current carrying conductor having current ' $I$ ', at a distance ' $d$ ' is given by

$\large{\overrightarrow{B}} = \dfrac{\mu_{0}I}{2 \pi d}$

According to the figure if ' $I_{t}$ ' be the current carried by the top wire, ' $I_{b}$ ' be the current carried by the bottom wire and ' $2d$ ' be the distance between them, then the direction of the magnetic field at 'P', which is midway between them, will be perpendicular towards the plane of the screen, shown by the $\bigotimes$ symbol and that due to the bottom wire at 'P' will be perpendicular away from the plane of the screen, shown by $\bigodot$ symbol.