Answer: Earth's magnetic field has flipped its polarity many times over the ... Earth has settled in the last 20 million years into a pattern of a pole reversal about ... And while reversals have happened more frequently in "recent" years, when ... per year, as opposed to about 10 miles per year in the early 20th century.
Explanation:
Answer:
4.D.all of them are correct
5. B.Making observation
Answer:
a)ω₂=127.64 rad/s ( min)
ω₁= 333.32 rad/s ( max)
b) At d₂= 11.75 cm ,ω₂=127.64 rad/s
d₁=4.5 cm,ω₁= 333.32 rad/s
c)α = 7.1 x 10⁻³ rad/s²
Explanation:
Given that
r₂= 11.75 cm
r₁=4.5 cm
v= 7.5 m/s
a)
We know that
v=ω r
ω =Angular speed
r= radius
v= velocity
When d₂= 11.75 cm :
v=ω r
7.5 x 2 =ω₂ x 0.1175
ω₂=127.64 rad/s ( min)
N₂=1219.4 rpm
When d₁=4.5 cm :
v=ω r
7.5 x 2=ω₁ x 0.045
ω₁= 333.32 rad/s ( max)
N₁=3184.58 rpm
The average angular acceleration α given as
ω₁ = ω₂ + α t
333.32 = 127.64 + α x 8 x 60 x 60
α = 7.1 x 10⁻³ rad/s²
We know the equation
weight = mass × gravity
To work out the weight on the moon, we will need its mass, and the gravitational field strength of the moon.
Remember that your weight can change, but mass stays constant.
So using the information given about the earth weight, we can find the mass by substituting 100N for weight, and we know the gravity on earth is 10Nm*2 (Use the gravitational field strength provided by your school, I am assuming yours in 10Nm*2)
Therefore,
100N = mass × 10
mass= 100N/10
mass= 10 kg
Now, all we need are the moon's gravitational field strength and to apply this to the equation
weight = 10kg × (gravity on moon)
Here is the answer of the given problem above.
Use this formula: <span>P = FV = ma*at = ma^2 t
</span><span>Substitute the values, and therefore, we got m(a0)^2t = m(x)^2 (2t)
then, solve for x which is the acceleration at 2t.
</span>The <span>answer would be a0/sqrt(2).
Hope this answers your question. Thanks for posting.
</span>