Answer:
The toy car
Explanation:
the real car is parked so yeah but maybe in some way technically the real car has more "momentum"
Answer:
High speed optical communication technology
To be able to communicate from the space to the earth and from earth to space is one of the most essential features required during space exploration.
Explanation:
Space exploration involves going into the space, beyond the earth's atmosphere. Landing on other planets and studying their details, going into deeper space beyond the planets to discover new cosmic events or structures is all a part of space exploration.
The key to analyse the studies and observations is being able to communicate the data collected, photos taken etc to the launch centers or space centers on earth. The space centers on earth should also be able to communicate with the persons or the satellites in space.
This is made possible using the optical communication technology which involves the use of optical fibers, lasers etc, since high speeds are more efficient during communication
Answer:
41.8m/s^2
Explanation:
Since the dragster starts from rest, initial velocity (u) = 0m/s, final velocity (v) = 25.9m/s, time (t) = 0.62s
From the equations of motion, v = u + at
a = (v - u)/t = (25.9 - 0)/0.62 = 25.9/0.62 = 41.8m/s^2
Answer:
Q= -6900 J
Explanation:
use the formula Q=mC(T_2 - T_1) and sub in givens
Q=mC(T_2 - T_1)
Q= (200 g)(0.444 J/g°C)(22-100)
Q= -6900 J
The negative sign means heat is lost, which agrees with the decrease in temperature
Clever problem.
We know that the beat frequency is the DIFFERENCE between the frequencies of the two tuning forks. So if Fork-A is 256 Hz and the beat is 6 Hz, then Fork-B has to be EITHER 250 Hz OR 262 Hz. But which one is it ?
Well, loading Fork-B with wax increases its mass and makes it vibrate SLOWER, and when that happens, the beat drops to 5 Hz. That means that when Fork-B slowed down, its frequency got CLOSER to the frequency of Fork-A ... their DIFFERENCE dropped from 6 Hz to 5 Hz.
If slowing down Fork-B pushed it CLOSER to the frequency of Fork-A, then its natural frequency must be ABOVE Fork-A.
The natural frequency of Fork-B, after it gets cleaned up and returns to its normal condition, is 262 Hz. While it was loaded with wax, it was 261 Hz.
