Answer:
a) nuclear
Explanation:
A wave is sent down a thick rope that is attached to a thinner rope. What happens when the wave reaches the junction? Explain in words, or draw a diagram.
Answer:
A fraction of the wave will be reflected at the junction and the remainder will be transmitted through the junction into the thin rope
Explanation:
When the junction of the thin and the thick rope is reached by the wave, the wave behaves in the following two ways;
1) A fraction of the incident wave from the thick rope to the junction/boundary of the thin rope will be reflected back to in the direction of the thick rope with a lower amplitude and upright
2) A fraction of the incident wave from the thick rope to the junction/boundary of the thin rope will be transmitted through the thin rope such that the transmitted pulse on the thin rope will have an higher amplitude than that of the thick rope due to the available energy in the wave pulse and the lesser weight of the thin rope.
Ahoy starting from rest moves with a distance of 200m in 10 seconds...Find the magnitude of acceleration
Answer:
The magnitude of acceleration is 2m/s² .
Explanation:
First, you have to find the velocity as it starts to travel 200m in 10s. The formula of velocity is V = Distance/Time, then you have to substitute the values into the formula :
[tex]v = d \div t[/tex]
[tex]let \: d = 200 \\ let \: t = 10[/tex]
[tex]v = 200 \div 10[/tex]
[tex]v = 20m/s[/tex]
Next, you have to substitute the values into the acceleration formula, a = (v-u)/t where v represents final velocity, u is initial velocity and t is time taken :
[tex]a = \frac{v - u}{t} [/tex]
[tex]let \: v = 20 \\ let \: u = 0 \\ let \: t = 10[/tex]
[tex]a = \frac{20 - 0}{10} [/tex]
[tex]a = \frac{20}{10} [/tex]
[tex]a = 2m/ {s}^{2} [/tex]