All categories

3 years ago

Which of Earth's spheres have the greatest impact from the forest fire?

1 answer:

6 0

I would say the atmosphere because air and oxygen are essential for human life however it is debatable. You should check out this prezi and judge for yourself which sphere is effected the most from forest fires based on the descriptions on how a forest fire effects all four spheres.

https://prezi.com/m/50ogw-qsyoxc/the-four-spheres/

You might be interested in

a golf ball is hit with a velocity of 30.0 m/s at an angle of 25 degrees about the horizontal. how long is the ball in the air a

Pie

Ok so use trigonometry to work out the vertical component of velocity.

sin(25) =opp/hyp
rearrange to:
30*sin(25) which equals 12.67ms^-1

now use SUVAT to get the time of flight from the vertical component,

V=U+at

Where V is velocity, U is the initial velocity, a is acceleration due to gravity or g. and t is the time.

rearranges to t= (V+u)/a

plug in some numbers and do some maths and we get 2.583s

this is the total air time of the golf ball.

now we can use Pythagoras to get the horizontal component of velocity.

30^2-12.67^2= 739.29
sqrt739.29 = 27.19ms^-1

and finally speed = distance/time

so--- 27.19ms^-1*2.583s= 70.24m

The ball makes it to the green, and the air time is 2.58s

4 0

2 years ago

A 2.00 kg block hangs from a spring balance calibrated in Newtons that is attached to the ceiling of an elevator.(a) What does t

tresset_1 [31]

Answer:

Part a)

$Reading = 2.00 kg$

Part b)

$Reading = 2.00 kg$

Part c)

$Reading = 4.04 kg$

Part d)

from t = 0 to t = 4.9 s

so the reading of the scale will be same as that of weight of the block

Then its speed will reduce to zero in next 3.2 s

from t = 4.9 to t = 8.1 s

The reading of the scale will be less than the actual mass

Explanation:

Part a)

When elevator is ascending with constant speed then we will have

$F_{net} = 0$

$T - mg = 0$

$T = mg$

So it will read same as that of the mass

$Reading = 2.00 kg$

Part b)

When elevator is decending with constant speed then we will have

$F_{net} = 0$

$T - mg = 0$

$T = mg$

So it will read same as that of the mass

$Reading = 2.00 kg$

Part c)

When elevator is ascending with constant speed 39 m/s and acceleration 10 m/s/s then we will have

$F_{net} = ma$

$T - mg = ma$

$T = mg + ma$

Reading is given as

$Reading = \frac{mg + ma}{g}$

$Reading = 2.00\frac{9.81 + 10}{9.81}$

$Reading = 4.04 kg$

Part d)

Here the speed of the elevator is constant initially

from t = 0 to t = 4.9 s

so the reading of the scale will be same as that of weight of the block

Then its speed will reduce to zero in next 3.2 s

from t = 4.9 to t = 8.1 s

The reading of the scale will be less than the actual mass

3 0

2 years ago

Which of the following best describes what went wrong with the scientists’ study? an improper experimental procedure the lack of

Rasek [7]

B- lack of control.

the scientists did not have a proper control group.

3 0

3 years ago

Read 2 more answers

The core of a certain reflected reactor consist of a cylinder 10 ft high 10 ft in diameter The measured maximum-to-average flux

Westkost [7]

Answer:

The maximum power density in the reactor is 37.562 KW/L.

Explanation:

Given that,

Height = 10 ft = 3.048 m

Diameter = 10 ft = 3.048 m

Flux = 1.5

Power = 835 MW

We need to calculate the volume of cylinder

Using formula of volume

$V =\pi r^2 h$

Put the value into the formula

$V=\pi\times(1.524)^2\times 3.048$

$V= 22.23\m^3$

$V = 22.23\times10^{3}\ Liter$

We need to calculate the maximum power density in the reactor

Using formula of power density

$P=\dfrac{E}{V}$

Where, P = power density

E = energy

V = volume

Put the value into the formula

$P=\dfrac{835\times10^{6}}{22.23\times10^{3}}$

$P=37561.85 = 37.562\times KW/L$

Hence, The maximum power density in the reactor is 37.562 KW/L.

6 0

3 years ago

Water enters the constant 130-mm inside-diameter tubes of a boiler at 7 MPa and 65°C and leaves the tubes at 6 MPa and 450°C wit

snow_lady [41]

The inlet velocity is 1.4 m/s and inlet volume is 0.019 m³/s.

Explanation:

When water entering the tube of constant diameter flows through the tube, it exhibits continuity of mass in the hydrostatics. So the mass of water moving from the inlet to the outlet tend to be same, but the velocity may differ.

As per mass flow equality which states that the rate of flow of mass in the inlet is equal to the product of area of the tube with the velocity of the water and the density of the tube.

Since, the inlet volume flow is measured as the product of velocity with the area.

Inlet volume flow=Inlet velocity*Area*time

And the mass flow rate is

Mass flow rate in the inlet=density*area*inlet velocity*time

Mass flow rate in the outlet=density*area*outlet velocity*time

Since, the time and area is constant, the inlet and outlet will be same as

(Mass inlet)/(density*inlet velocity)=Area*Time

(Mass outlet)/(density*outlet velocity)=Area*Time

As the ratio of mass to density is termed as specific volume, then

(Specific volume inlet)/(Inlet velocity)=(Specific volume outlet)/(Outlet velocity)

Inlet velocity= (Specific volume inlet)/(Specific volume outlet)*Outlet velocity

As, the specific volume of water at inlet is 0.001017 m³/kg and at outlet is 0.05217 m³/kg and the outlet velocity is given as 72 m/s, the inlet velocity

$Inlet velocity = \frac{0.001017}{0.05217}*72 =1.4035 m/s$

So, the inlet velocity is 1.4035 m/s.

Then the inlet volume will be

$Inlet volume = inlet velocity*area of circle=\pi r^{2}*inlet velocity$

As the diameter of tube is 130 mm, then the radius is 65 mm and inlet velocity is 1.4 m/s

$Inlet volume = 1.4*3.14*65*65*10^{-6} =0.019 \frac{m^{3} }{s}$

So, the inlet volume is 0.019 m³/s.

Thus, the inlet velocity is 1.4 m/s and inlet volume is 0.019 m³/s.

4 0

2 years ago