By dropping a ball and seeing how long it takes to hit the ground or throw a ball up and time it as well
Answer:
b) 472HZ, 408HZ
Explanation:
To find the frequencies perceived when the bus approaches and the train departs, you use the Doppler's effect formula for both cases:
fo: frequency of the source = 440Hz
vs: speed of sound = 343m/s
vo: speed of the observer = 0m/s (at rest)
v: sped of the train
f: frequency perceived when the train leaves us.
f': frequency when the train is getTing closer.
Thus, by doing f and f' the subjects of the formulas and replacing the values of v, vo, vs and fo you obtain:
hence, the frequencies for before and after tha train has past are
b) 472HZ, 408HZ
Average speed = (distance covered) / (time to cover the distance)
Tissa covered 60 meters in 10 seconds. Her average speed was
(60 m) / (10 sec) = 6 m/s.
That's the slope of the dotted line.
Lilly covered 60 meters in 8 seconds. Her average speed was
(60 m) / (8 sec) = 7.5 m/s .
That's the slope of the solid line.
Lilly covered the same distance in less time, and both girls
arrived at the finish line together. Technically, in science talk,
we would say that Lilly ran "faster", and her average speed
was "greater".
We can detect that by looking at the graph, because Lilly's line
has the characteristic of being "steeper", and we know that the
slope of the line on a distance/time graph is "speed".
Answer:
Explanation:
When the car is under an accelerating force and hits a tree, the instant force received by the tree is the same force that is accelerating the car.
The accelerating force can be calculated using Newton's second law:
Where m is the mass of the car and a is the acceleration.
Answer:
1.) 11 km/s
2.) 9.03 × 10^-5 metres
Explanation:
Given that an electron enters a region of uniform electric field with an initial velocity of 64 km/s in the same direction as the electric field, which has magnitude E = 48 N/C.
Electron q = 1.6×10^-19 C
Electron mass = 9.11×10^-31 Kg
(a) What is the speed of the electron 1.3 ns after entering this region?
E = F/q
F = Eq
Ma = Eq
M × V/t = Eq
Substitute all the parameters into the formula
9.11×10^-31 × V/1.3×10^-9 = 48 × 1.6×10^-19
V = 7.68×10^-18 /7.0×10^-22
V = 10971.43 m/s
V = 11 Km/s approximately
(b) How far does the electron travel during the 1.3 ns interval?
The initial velocity U = 64 km/s
S = ut + 1/2at^2
S = 64000×1.3×10^-6 + 1/2 × 8.4×10^12 × ( 1.3×10^-9)^2
S =8.32×10^-5 + 7.13×10^-6
S = 9.03 × 10^-5 metres
