Which of the following best explains the importance of water to humans?a.Water is essential for sustaining human life.
b.Water is important for shipping and trade.
c.Water is a valuable, profitable commodity.
d.All of the above
Answers
Answer;
-All of the above
- Water is essential for sustaining human life.
- Water is important for shipping and trade.
- Water is a valuable, profitable commodity.
Explanation;
-Water is one of the most important substances on earth. All plants and animals must have water to survive. If there was no water there would be no life on earth. Apart from drinking it to survive, people have many other uses for water such as cooking, washing clothes, cleaning, recreation, etc.
-Water is also essential for the healthy growth of farm crops and farm stock and is used in the manufacture of many products. Water is also important in our bodies; It facilitates digestion of food and absorption of nutrients through their dissolving in water, transports nutrients, chemicals and heat throughout the body. Water is our primary source of food.
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7.5 m. Determine the force of friction
causing the puck to slow.
Answers
F=ma
m=170g=0,17kg
Acceleration using kinematic equations:
vk²=vp²+2aΔd
vp-initial velocity=25m/s
vk-ending velocity=10m/s
Δd-distance
a=
a=
a=
a=-35
F=0,17kg*(-35m/s²)=-5,95N
a
A metallic liquid substance
b
A very cold substance
c
A spherical solid metal made up of Nickel, Iron and small amount of sulfur, silicon and oxygen.
d
A flatten but thick structure similar to a granite counter top
Answers
Answer:
c
Explanation:
A spherical solid metal made up of Nickel, Iron and small amount of sulfur, silicon and oxygen.
you are welcome
Answers
potental energy(PE) =0
kinetic energy(KE) = 0.5mv^2 = 0.5(1)(30^2) = 450
at the highest point the ball does not moved, v=0
potential energy at its maximum
kinetic energy = 0
Energy is conserved then
total energy before = after
PE1 + KE1 = PE2 + KE2
0 + 450 = PE2 + 0
conservation of energy is fun fact
g=10m/s^2 (assuming)
u=30m/s
height(h)=u^2/2g
=900/20=45m
P.E.=mgh
=1×10×45=450 joules
Answers
work is distance * force so 15*100=1500
and to find time you know power = diastance * force / time
so 25=15*100/t
25=1500/t
25/1500=t
.016=time
Final answer:
The work done is 1500 Joules and the time required to do the work is 60 seconds.
Explanation:
The subject at hand is related to concepts in physics, specifically work and power.
To find the work done, we use the formula Work = Force x Distance. Substituting the given values:
Work = 100 Newtons x 15 meters = 1500 Joules
To find the time required to do the work, we use the formula Power = Work / Time. Substituting the given values and rearranging the formula to solve for Time gives us:
Time = Work / Power = 1500 Joules / 25 Watts = 60 Seconds
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(1 Point)
4 m/s2
2.5 m/s2
1.5 m/s2
1 m/s2
Answers
Answer:
1.5
Explanation:
resultant force=8-5=3N
using f=ma
a=f/m
=3/2
=1.5
Answers
Answer:
The average angular acceleration of the Earth, α = 6.152 X 10⁻²⁰ rad/s²
Explanation:
Given data,
The period of 365 rotation of Earth in 2006, T₁ = 365 days, 0.840 sec
= 3.1536 x 10⁷ +0.840
= 31536000.84 s
The period of 365 rotation of Earth in 2006, T₀ = 365 days
= 31536000 s
Therefore, time period of one rotation on 2006, Tₐ = 31536000.84/365
= 86400.0023 s
The time period of rotation is given by the formula,
Tₐ = 2π /ωₐ
ωₐ = 2π / Tₐ
Substituting the values,
ωₐ = 2π / 365.046306
= 7.272205023 x 10⁻⁵ rad/s
Therefore, the time period of one rotation on 1906, Tₓ = 31536000/365
= 86400 s
Time period of rotation,
Tₓ = 2π /ωₓ
ωₓ = 2π / T
= 2π /86400
= 7.272205217 x 10⁻⁵ rad/s
The average angular acceleration
α = (ωₓ - ωₐ) / T₁
= (7.272205217 x 10⁻⁵ - 7.272205023 x 10⁻⁵) / 31536000.84
α = 6.152 X 10⁻²⁰ rad/s²
Hence the average angular acceleration of the Earth, α = 6.152 X 10⁻²⁰ rad/s²
Final answer:
The average angular acceleration of the Earth from the year 1906 to 2006 would be -5.73 x 10^-20 rad/s^2. This value was obtained by finding the change in angular velocity and then dividing it by the elapsed time.
Explanation:
The question is asking for the average angular acceleration of the Earth from the year 1906 to 2006, during which the Earth's rotation rate decreased, causing the day to increase in duration by about 0.840 seconds.
To find the average angular acceleration, you first need to calculate the change in angular velocity, which can be found from the change in rotation time. One revolution (one day) is 2π radians, so the change in angular velocity is Δω = 2π/86400 s - 2π/(86400+0.840) s = -1.81 x 10^-10 rad/s.
The time interval from 1906 to 2006 is 100 years or about 3.16 x 10^9 seconds. Therefore, the average angular acceleration, α, which is the change in angular velocity divided by time, would be α = Δω/Δt = -1.81 x 10^-10 rad/s / 3.16 x 10^9 s = -5.73 x 10^-20 rad/s^2.
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