What gas is found at the atmosphere that causes globalisation

The heavyweight boxing champion of the world punches a sheet of paper in midair, bringing it from rest up to a speed of 26.5 m/s

enyata [817]

corrected question:The heavyweight boxing champion of the world punches a sheet of paper in midair, bringing it from rest up to a speed of 26.5 m/s in 0.044 s . The mass of the paper is 0.003 kg. Part A Find the force of the punch on the paper

Answer:

Force=1.8N

Explanation:

Newtons third law states that in every action there is equal and opposite reaction.

The force of the punch will be the force that moves the paper by a speed of 26.5m/s.

$F=ma$

$F=m\frac{v-u}{t}$

m=0.003kg , v=26.5m/s u=0(the paper is punched from rest) t=0.044s

$F=0.003*\frac{26.5-0.044}{0.044}$

F=1.8N

4 0

4 years ago

The best way to ward of ailments later in life is to exercise at least 30 min per day.

hichkok12 [17]

True, physical excercise helps

5 0

4 years ago

A tank, shaped like a cone has height 12 meter and base radius 1 meter. It is placed so that the circular part is upward. It is

Temka [501]

Answer:

376966.991 Joules

Explanation:

Given that :

the height = 12 m

Let assume the tank have a thickness = dh

The radius of the tank by using the concept of similar triangle is :

$\dfrac{1}{r} = \dfrac{12}{h}$

$r = \dfrac{h}{12}$

The area of the tank = $\mathbf{\pi r^2}$

The area of the tank = $\mathbf{\pi( \dfrac{h}{12})^2}$

The area of the tank = $\mathbf{ \dfrac{\pi}{144}h^2}$

The volume of the tank is = area × thickness

= $\mathbf{ \dfrac{\pi}{144}h^2 \ dh}$

Weight of the element = $\rho_ g * volume$

where;

$\rho_g$ = density of water ; which is given as 10000 N/m³

So;

Weight of the element = $\mathbf{ 10000 *\dfrac{\pi}{144}h^2 \ dh}$

Weight of the element = $\mathbf{69.44 \ \pi \ h^2 \ dh}$

However; the work required to pump this water = weight × height rise

where the height rise = 12 - h

the work required to pump this water = $\mathbf{69.44 \ \pi \ h^2 \ dh}$ (12 - h)

the work required to pump this water = $\mathbf{69.44 \pi (12h^2-h^3)dh}$

We can determine the total workdone by integrating the work required to pump this water

SO;

Workdone = $\mathbf{\int\limits^{12}_0 {69.44 \pi(12h^2-h^3)} dh}$

= $\mathbf{ 69.44 \pi \int\limits^{12}_0 {(12h^2-h^3)} dh}$

= $\mathbf{ 69.44 \pi[ \frac{12h^3}{3}- \frac{h^4}{4}]^{12}}_0} }$

= $\mathbf{69.44 \pi [ \frac{12^4}{3}-\frac{12^4}{4}]}$

= $\mathbf{69.44 \pi*12^4 [ \frac{4-3}{12}]}$

= $\mathbf{69.44 \pi*12^4 *\frac{1}{12}}$

= 376966.991 Joules

6 0

3 years ago

The mean distance of an asteroid from the Sun is 2.98 times that of Earth from the Sun. From Kepler's law of periods, calculate

lutik1710 [3]

Answer:

The asteroid requires 5.14 years to make one revolution around the Sun.

Explanation:

Kepler's third law establishes that the square of the period of a planet will be proportional to the cube of the semi-major axis of its orbit:

$T^{2} = a^{3}$ (1)

Where T is the period of revolution and a is the semi-major axis.

In the other hand, the distance between the Earth and the Sun has a value of $1.50x10^{8} Km$ . That value can be known as well as an astronomical unit (1AU).