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Tamiku [17]
3 years ago
12

What is true of the water cycle?

Physics
2 answers:
Zina [86]3 years ago
7 0

Answer:

D

Explanation:

A water cycle is constantly going on there is no definitive starting or stopping point

tresset_1 [31]3 years ago
3 0
D
Explanation: it’s d
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Which of the following is correct for speed?
Olin [163]

Answer:

Speed is a "scalar" quantity

(C) is the correct answer

An object could travel at 10 m/s to some point and then return to the origin at 10 m/s for an average speed of 10 m/s, however it's displacement over that time would be zero for a net velocity of zero.

4 0
2 years ago
In a wire with a 1.05 mm2 cross-sectional area, 7.93×1020 electrons flow past any point during 3.97 s. What is the current ????
34kurt

Answer:

The current in the wire is 31.96 A.

Explanation:

The current in the wire can be calculated as follows:

I = \frac{q}{t}

<u>Where</u>:

q: is the electric charge transferred through the surface

t: is the time      

The charge, q, is:

q = n*e

<u>Where</u>:

n: is the number of electrons = 7.93x10²⁰

e: is the electron's charge = 1.6x10⁻¹⁹ C

q = n*e = 7.93 \cdot 10^{20}*1.6 \cdot 10^{-19} C = 126.88 C

Hence, the current in the wire is:

I = \frac{126.88 C}{3.97 s} = 31.96 A

Therefore, the current in the wire is 31.96 A.

I hope it helps you!

3 0
3 years ago
Which chemical is released by brain cells along an activated reward pathway?
AfilCa [17]
The answer is Dopamine
3 0
3 years ago
Read 2 more answers
HeLp aSAp!!! If the speed and distance of an object are given, and I need to find the time, I will...
Helen [10]

Answer:

B - Divide Speed and Distance

Explanation:

Hope it helps!!

:D

5 0
3 years ago
Read 2 more answers
An object is dropped from a​ tower, 400 ft above the ground. The​ object's height above ground x seconds after the fall is ​s(x)
pav-90 [236]

1) 5 s

The vertical position of the object is given by

y(t) = h - \frac{1}{2}gt^2 =  400 - 16 t^2

where

h=400 ft represents the initial height

g = 32 ft/s^2 is the acceleration of gravity

t is the time

We want to find the time t at which the object reaches the ground, so the time t at which

y(t) = 0

By substituting this into the equation, we find

0 = 400 - 16t^2\\t=\sqrt{\frac{400}{16}}=5 s

2) 160 ft/s

The object is released from rest, so the initial velocity is zero

u = 0

The final vertical velocity can be found by using

v^2 - u^2 = 2ah

where

v is the final velocity

a = 32 ft/s^2 is the acceleration of gravity

h = 400 ft is the vertical distance covered

Solving for v, we find

v=\sqrt{u^2 +2ay}=\sqrt{2(32 ft/s)(400 ft)}=160 ft/s

8 0
3 years ago
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