Answer:
17.2 seconds
Explanation:
Given:
v₀ = 0 m/s
a₁ = 10.0 m/s²
t₁ = 3.0 s
a₂ = 16 m/s²
t₂ = 5.0 s
a₃ = -12 m/s²
v₃ = 0 m/s
Find: t
First, find v₁:
v₁ = a₁t₁ + v₀
v₁ = (10.0 m/s²) (3.0 s) + (0 m/s)
v₁ = 30 m/s
Next, find v₂:
v₂ = a₂t₂ + v₁
v₂ = (16 m/s²) (5.0 s) + (30 m/s)
v₂ = 110 m/s
Finally, find t₃:
v₃ = a₃t₃ + v₂
(0 m/s) = (-12 m/s²) t₃ + (110 m/s)
t₃ = 9.2 s
The total time is:
t = t₁ + t₂ + t₃
t = 3.0 s + 5.0 s + 9.2 s
t = 17.2 s
Round as needed.
Answer:
Mercury and Venus lie closer to Sun than position of Earth
Explanation:
As we know that all planets around the sun in the range of their distance can be arranged as following:
1). Mercury.
2). Venus.
3). Earth.
4). Mars.
5). Jupiter.
6). Saturn.
7). Uranus.
8). Neptune.
Since Earth lie at 3rd position from sun so two closer planets are Mercury and Venus
Answer:
Transverse wave- Back and forth at right angles to the direction of the wave arrow.
longitudinal wave- bask and forth in the direction of the motion of the motion of the wave.
electromagnetic wave- two alternating waves moving at right angles to each other.
Explanation:
In a longitudinal wave, the particles vibrate at right angles in reference to the wave motion.
In a transverse wave, the particles vibrate parallel to the wave motion
Electromagnetic waves occur as a result of the interaction between two waves and are normally transverse in nature.
Answer:
High boiling and melting points: Hydrogen bonds increase the amount of energy required for phase changes to occur, thereby raising the boiling and melting points.
High specific heat: Hydrogen bonds increase the amount of energy required for molecules to increase in speed, thereby raising the specific heat.
Lower density as a solid than as a liquid: Hydrogen bonds increase the volume of the solid by holding molecules apart, thereby decreasing the density
High surface tension: Hydrogen bonds produce strong intermolecular attractions, which increase surface tension
Explanation:
Answer:
I = 1.875 A
Explanation:
For this exercise we use Ampere's law
∫ B . ds = μ₀ I
We use a circular path around the wire whereby B and ds are parallel, whereby the dot product is reduced to the algebraic product
ds = 2π dr
B (2πr) = μ₀ I
I = B 2π R /μ₀
r= 7.5 cm = 0.075 m
calculate
I = (50 μ₀ /π) 2π 0.075 /μ₀
I = 1.875 A
