The short answer is that the displacement is equal tothe area under the curve in the velocity-time graph. The region under the curve in the first 4.0 s is a triangle with height 10.0 m/s and length 4.0 s, so its area - and hence the displacement - is
1/2 • (10.0 m/s) • (4.0 s) = 20.00 m
Another way to derive this: since velocity is linear over the first 4.0 s, that means acceleration is constant. Recall that average velocity is defined as
<em>v</em> (ave) = ∆<em>x</em> / ∆<em>t</em>
and under constant acceleration,
<em>v</em> (ave) = (<em>v</em> (final) + <em>v</em> (initial)) / 2
According to the plot, with ∆<em>t</em> = 4.0 s, we have <em>v</em> (initial) = 0 and <em>v</em> (final) = 10.0 m/s, so
∆<em>x</em> / (4.0 s) = (10.0 m/s) / 2
∆<em>x</em> = ((4.0 s) • (10.0 m/s)) / 2
∆<em>x</em> = 20.00 m
The circulatory system is important in meeting the needs of all cells in the body because it offers means of transport by which cells obtain the materials needed to live and function.
The functions of the circulatory system by which the cells are supported include:
1. Respiration- oxygen is delivered to the cells and carbon dioxide is removed from them.
2. Nutrition- nutrients for energy are supplied to every cell in the body.
3. Waste removal- metabolic waste products are taken away before they accumulate and become harmful to the cells.
4. Cellular communication- hormones are transported to the cells and organs that need them for proper functioning.
5.Thermoregulation- As blood circulates, it keeps body temperature balanced and thus cells are able to carry out processes well.
Gravitational I think would be the answer, Hope this helps!
Answer:
Explanation:
Part A) Using
light intensity I= P/A
A= Area= π (Radius)^2= π((0.67*10^-6m)/(2))^2= 1.12*10^-13 m^2
Radius= Diameter/2
P= power= 10*10^-3=0.01 W
light intensity I= 0.01/(1.12*10^-13)= 9*10^10 W/m^2
Part B) Using
I=c*ε*E^2/2
rearrange to solve for E= 
((I*2)/(c*ε))
c is the speed of light which is 3*10^8 m/s^2
ε=permittivity of free space or dielectric constant= 8.85* 10^-12 F⋅m−1
I= the already solved light intensity= 8.85*10^10 W/m^2
amplitude of the electric field E= (9*10^10 W/m^2)*(2) / (3*10^8 m/s^2)*(8.85* 10^-12 F⋅m−1)
---> E= (1.8*10^11) / (2.66*10^-3) = (6.8*10^13) = 8.25*10^6 V/m
Answer:
= 4.3 × 10 ⁻¹⁴ m
Explanation:
The alpha particle will be deflected when its kinetic energy is equal to the potential energy
Charge of the alpha particle q₁= 2 × 1.6 × 10⁻¹⁹ C = 3.2 × 10⁻¹⁹ C
Charge of the gold nucleus q₂= 79 × 1.6 × 10⁻¹⁹ = 1.264 × 10⁻¹⁷C
Kinetic energy of the alpha particle = 5.28 × 10⁶ × 1.602 × 10⁻¹⁹ J ( 1 eV)
= 8.459 × 10⁻¹³
k electrostatic force constant = 9 × 10⁹ N.m²/c²
Kinetic energy = potential energy = k q₁q₂ / r where r is the closest distance the alpha particle got to the gold nucleus
r = ( 9 × 10⁹ N.m²/c² × 3.2 × 10⁻¹⁹ C × 1.264 × 10⁻¹⁷C) / 8.459 × 10⁻¹³
= 4.3 × 10 ⁻¹⁴ m
