Answer:
T=280.41 °C
Explanation:
Given that
At T= 24°C Resistance =Ro
Lets take at temperature T resistance is 2Ro
We know that resistance R given as
R= Ro(1+αΔT)
R-Ro=Ro αΔT
For copper wire
α(coefficient of Resistance) = 3.9 x 10⁻³ /°C
Given that at temperature T
R= 2Ro
Now by putting the values
R-Ro=Ro αΔT
2Ro-Ro=Ro αΔT
1 = αΔT
1 = 3.9 x 10⁻³ x ΔT
ΔT = 256.41 °C
T- 24 = 256.41 °C
T=280.41 °C
So the final temperature is 280.41 °C.
<h2>Answer: It becomes an Ion
</h2>
When an atom has gained or lost electrons (negative charge), it becomes an ion.
In this sense:
<h2>I
ons are atoms that have <u>
gained or lost</u>
electrons in their electronic cortex.
</h2><h2>
</h2>
If a neutral atom <u>loses electrons</u>, it remains with an excess of positive charge and transforms into a positive ion or <u>cation</u>, whereas if a neutral atom <u>gains electrons</u>, it acquires an excess of negative charge and transforms into a negative ion or <u>anion</u>.
It is then how ions form bonds with other atoms differently depending on the number of electrons they have.
Answer:Half-life is the amount of time it takes for the initial mass of the isotope to decompose, by half, into other lighter atoms.
Explanation:Different radioactive isotopes have different half-lives. For example, the element technetium-99m has a half life of 6 hours. This means that is 100 kg of the element is left to decay, in 6 hours, 50kg of the mass will have changed into other elements/atoms. The half-life of uranium-238 is 4.5 billion years while that of polonium-216 is only 0.145 seconds.
Answer:
The resultant force would (still) be zero.
Explanation:
Before the 600-N force is removed, the crate is not moving (relative to the surface.) Its velocity would be zero. Since its velocity isn't changing, its acceleration would also be zero.
In effect, the 600-N force to the left and 200-N force to the right combines and acts like a 400-N force to the left.
By Newton's Second Law, the resultant force on the crate would be zero. As a result, friction (the only other horizontal force on the crate) should balance that 400-N force. In this case, the friction should act in the opposite direction with a size of 400 N.
When the 600-N force is removed, there would only be two horizontal forces on the crate: the 200-N force to the right, and friction. The maximum friction possible must be at least 200 N such that the resultant force would still be zero. In this case, the static friction coefficient isn't known. As a result, it won't be possible to find the exact value of the maximum friction on the crate.
However, recall that before the 600-N force is removed, the friction on the crate is 400 N. The normal force on the crate (which is in the vertical direction) did not change. As a result, one can hence be assured that the maximum friction would be at least 400 N. That's sufficient for balancing the 200-N force to the right. Hence, the resultant force on the crate would still be zero, and the crate won't move.
Answer:
Q1: 3.2km
Q2: 4.8K
Explanation:
Q1:
So db is the distance of bird, and dr is the distance of runner
db = 2vr and the distance of bird is going to be 2 times greater than the runner.
formulas: db = 2vr & db = 2dr
- db = 2dr
- L + (L - x) = 2x
- 2L - x = 2x
- 2L = 3x
- x =
L
Insert it in x = L
(2.4km) = 1.6km
Now we use formula db = 2dr
- db = 2L - x
- db = 2(2.4km) - 1.6km
- <u>db = 3.2km</u>
Q2:
Formulas: Vr = L /Δt & Vb = db/Δt
- Vr = L/ Δt ⇒ Δt =
(Km cancel each other)
- Vb = db/Δt ⇒ db = VbΔt
- 13.6km/hr
- <u>4.8km</u>
(hr cancel each other)
Hope it helps you :)
