What are some methods humans have used to control flooding and what has been done to worsen the effects of flooding?

Both formal and informal

Tema [17]

Explanation:

is this a question????...

5 0

3 years ago

Which part of the central nervous system controls reflexes?

larisa86 [58]

Answer:

spinal cord becuse it does all of the reflexes

6 0

3 years ago

Read 2 more answers

The power of motor is 50 w the motor drags a 2 object horizontally at a constant speed of 10m/s along a rough floor . What is th

Ilia_Sergeevich [38]

Frictional force between the object and the floor=5 N

Explanation:

power= 50 W

velocity= 10 m/s

power= force * velocity

50=F * 10

F=50/10

F=5 N

Thus the force of friction= 5 N

5 0

3 years ago

An artillery shell is launched on a flat, horizontal field at an angle of α = 40.8° with respect to the horizontal and with an i

Lina20 [59]

Answer:

1317.4 m

Explanation:

We are given that

Angle= $\alpha=40.8^{\circ}$

Initial speed = $v_0=346m/s$

We have to find the horizontal distance covered by the shell after 5.03 s.

Horizontal component of initial speed= $v_{ox}=v_0cos\theta=346cos40.8=261.9m/s$

Vertical component of initial speed= $v_{oy}=346sin40.8=226.1m/s$

Time=t=5.03 s

Horizontal distance = $Horizontal\;velocity\times time$

Using the formula

Horizontal distance= $261.9\times 5.03$

Horizontal distance=1317.4 m

Hence, the horizontal distance covered by the shell=1317.4 m

8 0

3 years ago

A 6.0 g marble is fired vertically upward using a spring gun. The spring must be compressed 9.4 cm if the marble is to just reac

RoseWind [281]

Answer:

a) $\Delta U_{g} = 12.945\,J$ , b) $\Delta U_{k} = 12.945\,J$ , c) $k = 2930.059\,\frac{N}{m}$

Explanation:

a) The change in the gravitational potential energy of the marble-Earth system is:

$\Delta U_{g} = (0.06\,kg)\cdot \left(9.807\,\frac{m}{s^{2}}\right)\cdot (22\,m)$

$\Delta U_{g} = 12.945\,J$

b) The change in the elastic potential energy of the spring is equal to the change in the gravitational potential energy, then:

$\Delta U_{k} = 12.945\,J$

c) The spring constant of the gun is:

$\Delta U_{k} = \frac{1}{2} \cdot k \cdot x^{2}$

$k = \frac{2\cdot \Delta U_{k}}{x^{2}}$

$k = \frac{2\cdot (12.945\,J)}{(0.094\,m)^{2}}$

$k = 2930.059\,\frac{N}{m}$

4 0

3 years ago