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Gnesinka [82]
3 years ago
12

Precipitation is greater over what and less over what? A. Land, sea B. Sea, land

Physics
1 answer:
Ksju [112]3 years ago
3 0
I would say letter A to this because we need rain water for crops plants. Also we need it so that we can drink it. Its less great over sea because as it merge with the salt in the ocean it becomes less drinkable and less usable.
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Which of the following is responsible for the formation of sinkholes F. Eroding soil on Earths’s surface G. Ice that builds up o
monitta
Ans is <span>I. Flowing groundwater that dissolves rock , such as limestone</span>
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3 years ago
Which graph shows uniform motion.​
bagirrra123 [75]

Answer:

1. shows uniform motion

Explanation:

5 0
3 years ago
Read 2 more answers
A simple pendulum consists of a point mass suspended by a weightless, rigid wire in a uniform gravitation field. Check all that
9966 [12]

Answer:

f.The period is independent of the suspended mass.

Explanation:

The period of a pendulum is given by

T=2 \pi \sqrt{\frac{L}{g}}

where

L is the length of the pendulum

g is the acceleration due to gravity

From the formula, we see that:

1) the period of the pendulum depends only on its length, L, and it is proportional to the square root of the length

2) the period does not depend neither on the mass of the pendulum, nor on its amplitude of oscillation

So, the only correct statements are

f.The period is independent of the suspended mass.

Note: statement "e.The period is proportional to the length of the wire" is also wrong, because the period is NOT proportional to the length of the wire, but it is proportional to the square root of it.

3 0
3 years ago
You are standing in a building whose height (40m) you throw a ball downward at a angle of -30 at a speed of (10m/s) acceleration
jeyben [28]

Answer: 3.41 s

Explanation:

Assuming the question is to find the time t the ball is in air, we can use the following equation:

y=y_{o}+V_{o}sin \theta t-\frac{1}{2}gt^{2}

Where:

y=0m is the final height of the ball

y_{o}=40 m is the initial height of the ball

V_{o}=10 m/s is the initial velocity of the ball

t is the time the ball is in air

g=9.8 m/s^{2} is the acceleration due to gravity  

\theta=30\°

Then:

0 m=40 m+(10 m/s)(sin(30\°))t-\frac{1}{2}9.8 m/s^{2}t^{2}

0 m=40 m+5m/s t-4.9 m/s^{2}t^{2}

Multiplying both sides of the equation by -1 and rearranging:

4.9 m/s^{2}t^{2}-5m/s t-40 m=0

At this point we have a quadratic equation of the form at^{2}+bt+c=0, which can be solved with the following formula:

t=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}

Where:

a=4.9

b=-5

c=-40

Substituting the known values:

t=\frac{-(-5) \pm \sqrt{(-5)^{2}-4(4.9)(-40)}}{2(4.9)}

Solving the equation and choosing the positive result we have:

t=3.41 s  This is the time the ball is in air

5 0
3 years ago
Visible light with a wavelength of 480 nm appears Question 33 options:
AlladinOne [14]
It would appear as blue

5 0
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