Although many characteristics are common<span> throughout the </span>group<span>, the heavier metals such as Ca, Sr, Ba, and Ra are almost as reactive as the </span>Group<span> 1 Alkali Metals. All the </span>elements<span> in </span>Group 2 have two<span> electrons in their valence shells, giving them an oxidation state of +</span><span>2.</span>
Answer: Arrow B
Explanation:
Arrow B best best represents the path the ball follows after the string breaks.
This is because the described situation is related to <u>uniform circular motion</u>, in which the tangential velocity is the linear velocity vector that is <u>always tangent to the trajectory</u> and is the distance traveled by the ball in its circular motion in a period of time.
Hence, if this circular motion suddenly stops, the ball will fly in a direction that is tangent to that circle.
When the box IS on the shelf 2m above the ground,
its potential energy is
(weight) x (height) = (3 N) x (2 m) = 6 joules .
THAT's the work you have to do, to lift the box up to there.
Let us first find out the radioactive constant of Phosphorus 32.
Radioactive constant, λ =
Here, is half life of phosphorus 32 = 14.3 days
λ =
The amount of 4 mg of phosphorus remain after 71.5 days can be found using the formula,
m=m₀e⁻(λt)
=
=
=
=
=
= 0.125 mg
The mass of 4 mg of phosphorus 32 remains after 71.5 days will be 0.125 mg.
Answer:
F = 2,894 N
Explanation:
For this exercise let's use Newton's second law
F = m a
The acceleration is centripetal
a = v² / r
Angular and linear variables are related.
v = w r
Let's replace
F = m w² r
The radius r and the length of the rope is related
cos is = r / L
r = L cos tea
Let's replace
F = m w² L cos θ
Let's reduce the magnitudes to the SI system
m = 101.7 g (1 kg / 1000g) = 0.1017 kg
θ = 5 rev (2π rad / rev) = 31,416 rad
w = θ / t
w = 31.416 / 5.1
w = 6.16 rad / s
F = 0.1017 6.16² 0.75 cos θ
F = 2,894 cos θ
The maximum value of F is for θ equal to zero
F = 2,894 N