Recall the formula,
∆<em>θ</em> = <em>ω</em>₀ <em>t</em> + 1/2 <em>α</em> <em>t</em> ²
where ∆<em>θ</em> = angular displacement, <em>ω</em>₀ = initial angular speed (which is zero because the disk starts at rest), <em>α</em> = angular acceleration, and <em>t</em> = time. Solve for the acceleration with the given information:
50 rad = 1/2 <em>α</em> (5 s)²
<em>α</em> = (100 rad) / (25 s²)
<em>α</em> = 4 rad/s²
Now find the angular speed <em>ω </em>after 3 s using the formula,
<em>ω</em> = <em>ω</em>₀ + <em>α</em> <em>t</em>
<em>ω</em> = (4 rad/s²) (3 s)
<em>ω</em> = 12 rad/s
Answer:
a) m = 993 g
b) E = 6.50 × 10¹⁴ J
Explanation:
atomic mass of hydrogen = 1.00794
4 hydrogen atom will make a helium atom = 4 × 1.00794 = 4.03176
we know atomic mass of helium = 4.002602
difference in the atomic mass of helium = 4.03176-4.002602 = 0.029158
fraction of mass lost =
= 0.00723
loss of mass for 1000 g = 1000 × 0.00723 = 7.23
a) mass of helium produced = 1000-7.23 = 993 g (approx.)
b) energy released in the process
E = m c²
E = 0.00723 × (3× 10⁸)²
E = 6.50 × 10¹⁴ J
A is the only logical answer
Answer:
Electrons are negatively charged and are deflected on a curving path towards the positive plate. Neutrons have no charge, and continue on in a straight line.
<h2>
Answer: size</h2>
Gel electrophoresis is called to the technique used by scientists for analytical purposes, in life sciences laboratories to separate macromolecules (DNA, RNA, and proteins from various sources).
The process consists of separating the molecules according to their <u>size</u> and <u>electric charge</u>. This is done with a gel (a gelatinous substance extracted from seaweed, called <em>agarose</em>) of controllable porosity placed in an ionic buffer environment. This is how the gel acts as a molecular sieve that separates larger molecules from the smaller ones, because each molecule has different size and charge and will move through the gel at different speeds.
That is, the smaller molecules move more quickly through the gel while the larger ones are left behind.