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german
2 years ago
15

Can someone help me with these questions please?

Physics
1 answer:
Alenkinab [10]2 years ago
8 0

Answer:

pang anong grade po yan kasi grade 4 palang po

kasi ako

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Using complete sentences and your own words describe some of the ways humans use water.
Lera25 [3.4K]

Answer:

Humans use water for many different things. We use water to stay hydrated. Our bodies need water to live. Back in the day people used water for transportation and trading. This was a way to become wealthy and exchange goods and ideas from one place to another. We also use water to clean ourselves off. If we don't we can become sick with illnesses that can harm our bodies.

Explanation:

6 0
3 years ago
4 properties of light
Neko [114]
-reflection and refraction of light
-dispersion of light
-absorption of light
-polarization of light
3 0
3 years ago
If you could choose one job or career related to environmental science what would it be
tatuchka [14]

Answer:

I would love to be a gardener :)

Explanation:

Plants are kool!!!

4 0
3 years ago
How much work output can a 150hp snowmobile do in 15 seconds if it is 18 % efficiency?
Sav [38]

The work output is 3.02\cdot 10^5 J

Explanation:

The power of the snowmobile is

P=150 hp

Keeping in mind that

1 hp = 746 W

We can convert it into Watts:

P=(150)(746)=111,900 W

The energy used by the snowmobile can be found as follows:

E=Pt

where

P = 111,900 W is the power used

t = 15 s is the time interval

Substituting,

E=(111,900)(15)=1.68\cdot 10^6 J

However, the snowmobile is 18% efficient: this means that only 18% of this energy is converted into useful work. Therefore, the work output is

W=0.18E=(0.18)(1.68\cdot 10^6)=3.02\cdot 10^5 J

Learn more about power and work:

brainly.com/question/7956557

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

3 0
3 years ago
A charge partides round a 1 m radius circular particle accelerator at nearly the speed of light. Find : (a) The period (b) The c
Scrat [10]

Explanation:

It is given that,

Radius of circular particle accelerator, r = 1 m

The distance covered by the particle is equal to the circumference of the circular path, d = 2πr

d = 2π × 1 m

(a) The speed of satellite is given by total distance divided by total time taken as :

speed=\dfrac{distance}{time}

Let t is the period of the particle.

t=\dfrac{d}{s}

d = distance covered

s = speed of particle

It is given that the charged particle is moving nearly with the speed of light

t=\dfrac{d}{c}

t=\dfrac{2\pi\times 1\ m}{3\times 10^8\ m/s}

t=2.09\times 10^{-8}\ s

(b) On the circular path, the centripetal acceleration is given by :

a=\dfrac{c^2}{r}

a=\dfrac{(3\times 10^8\ m/s)^2}{1\ m}

a=9\times 10^{16}\ m/s^2

Hence, this is the required solution.

8 0
3 years ago
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