All categories

3 years ago

What is an example of a reservoir? 1. Beach 2. Cloud 3. Desert 4. Lake

Physics

2 answers:

otez555 [7]3 years ago

7 0

Answer: If you guys are in the future it is lake

Explanation: It is lake because Reservoirs include humans, animals, and the environment. The reservoir may or may not be the source from which an agent is transferred to a host. For example, the reservoir of Clostridium botulinum is soil, but the source of most botulism infections is improperly canned by cdc.org

miskamm [114]3 years ago

4 0

I think that the best answer is Lake.

A reservoir (/ˈrɛzərvwɑːr/; from French réservoir [ʁezɛʁvwaʁ]; lit. 'storehouse') is, most commonly, an enlarged natural or artificial lake, pond, or impoundment created using a dam or lock to store water.

Please give brainiest if correct, would rlly appreciate it :D

You might be interested in

What is meant by inertia???

kipiarov [429]

Continuing in an existing state. Resistance to change.

7 0

3 years ago

Read 2 more answers

Radar is used to determine distances to various objects by measuring the round-trip time for an echo from the object. (a) How fa

zloy xaker [14]

Answer:

150000000000 m

0.0000005 seconds

33.33 ns

Explanation:

Speed of electromagnetic waves through vacuum = $3\times 10^8\ m/s$

Echo time = 1000 seconds

Echo time is the time taken to reach the object and come back to the observer

Distance = Speed×Time

$Distance=3\times 10^8\times \frac{1000}{2}=150000000000\ m$

Venus is 150000000000 m away from Earth

Time = Distance / Speed

$Time=\frac{75}{3\times 10^8}=0.00000025\ seconds$

Echo time will be twice the time

$Echo\ time=2\times 0.00000025=0.0000005\ seconds$

The echo time will be 0.0000005 seconds

Difference in time = Difference in distance / Speed

$\Delta t=\frac{\Delta d}{v}\\\Rightarrow \Delta t=\frac{10}{3\times 10^8}\\\Rightarrow \Delta=33.33\ ns$

The accuracy by which I will be able to measure the echo time is 33.33 ns

5 0

3 years ago

Convert 1nanosecond in to its SI init

SOVA2 [1]

<em>Convert 1nanosecond in to its SI init</em>

<em>In SI units, nano is 1000th part of micro which in turn is 1000th part of mini which in turn is 1000th part of main unit. Now, when you affix nano to any unit, here in case, second, it means that you are referring to 1000th part of 1000th part of 1000th part of second or in short, 1000000000th(10^9) part of a second.</em>

8 0

3 years ago

How is property belonging to individuals protected under the Fifth Amendment?

kotegsom [21]

Answer:

By preventing the government from taking property without fair payment.

Explanation:

Answer 1

Please mark me as brilliant and thank you

7 0

3 years ago

Read 2 more answers

A stone is thrown horizontally at 60.0 m/sm/s from the top of a very tall cliff. Calculate its horizontal position and vertical

svp [43]

Answer:

X-Positions: Y-Positions

x(0) = 0 y(0) = 0

x(2) = 120 m y(2) = 19.6 m

x(4) = 240 m y(4) = 78.4 m

x(6) = 360 m y(6) = 176.4 m

x(8) = 480 m y(8) = 313 m

x(10) = 600m y (10) = 490 m

Explanation:

X-Positions

First, we choose to take the horizontal direction as our x-axis, and the positive x-axis as positive.
After being thrown, in the horizontal direction, no external influence acts on the stone, so it will continue in the same direction at the same initial speed of 60. 0 m/s
So, in order to know the horizontal position at any time t, we can apply the definition of average velocity, rearranging terms, as follows:

$x = v_{ox} * t = 60.0 m/s * t(s)$

It can be seen that after 2 s, the displacement will be 120 m, and each 2 seconds, as the speed is constant, the displacement will increase in the same 120 m each time.

Y-Positions

We choose to take the vertical direction as our y-axis, taking the downward direction as our positive axis.
As both axes are perpendicular each other, both movements are independent each other also, so, in the vertical direction, the stone starts from rest.
At any moment, it is subject to the acceleration of gravity, g.
As the acceleration is constant, we can find the vertical displacement (taking the height of the cliff as the initial reference level), using the following kinematic equation:

$y = \frac{1}{2} * g* t^{2} = \frac{1}{2} * 9.8 m/s2 * t(s)^{2}$

Replacing by the values of t, we get the following vertical positions, from the height of the cliff as y = 0:
y(2) = 2* 9.8 m/s2 = 19.6 m
y(4) = 8* 9.8 m/s2 = 78.4 m
y(6) = 18*9.8 m/s2 = 176.4 m
y(8) = 32*9.8 m/s2 = 313.6 m
y(10)= 50 * 9.8 m/s2 = 490.0 m

5 0

3 years ago