Well they could go down a hill to gain more kinetic energy.
Answer:
See Explanation
Explanation:
a) We know that;
v = λf
Where;
λ = wavelength of the wave
f = frequency of the wave
v = velocity of the wave
So;
T = 2 * 2.10 s = 4.2 s
Hence f = 1/4.2 s
f = 0.24 Hz
The wavelength = 6.5 m
Hence;
v = 6.5 m * 0.24 Hz
v = 1.56 m/s
b)The amplitude of the wave is;
A = 0.600 m/2 = 0.300 m
c) Since the wave speed does not depend on the amplitude of the wave then the answer in (a) above remains the same
Where d = 0.30 m
A = 0.30 m/2 = 0.15 m
Answer:
Solution given:
height [H]=25m
initial velocity [u]=8.25m/s
g=9.8m/s
now;
a. How long is the ball in flight before striking the ground?
Time of flight =?
Now
Time of flight=
substituting value
- =

- =2.26seconds
<h3>
<u>the ball is in flight before striking the ground for 2.26seconds</u>.</h3>
b. How far from the building does the ball strike the ground?
<u>H</u><u>o</u><u>r</u><u>i</u><u>z</u><u>o</u><u>n</u><u>t</u><u>a</u><u>l</u><u> </u>range=?
we have
Horizontal range=u*
<h3>
<u>The ball strikes 18.63m far from building</u>. </h3>
The car is accelerating at 3 m/s² in the positive direction (to the right). By Newton's second law, the net force on the car in this direction is
∑ F = F[a] - F[f] - F[air] = ma
3100 N - 200 N - F[air] = (650 kg) (3 m/s²)
Solve for F[air] :
F[air] = 3100 N - 200 N - (650 kg) (3 m/s²)
F[air] = 3100 N - 200 N - 1950 N
F[air] = 950 N
Answer:
B) x^2+6x+8
Explanation:
x-4 | x^3+2x^2-16x-32
- x^3-4x^2 <-- (x-4)(x^2)
_________________
6x^2-16x-32
- 6x^2-24x <-- (x-4)(6x)
_________________
8x-32
- 8x-32 <- (x-4)(8)
___________________________
0 | x^2+6x+8
This means the answer is B) x^2+6x+8