Refer to the diagram shown below.
Let T = the tension in each wire.
For equilibrium,
2T cos(50°) = 150 N
1.2856T = 150
T = 116.677 N ≈ 117 N
Answer: 117 N
Answer:
a) Height of the antenna (in m) for a radio station broadcasting at 604 kHz = 124.17 m
b)Height of the antenna (in m) for radio stations broadcasting at 1,710 kHz =43.86 m
Explanation:
(a) Radiowave wavelength= λ = c/f
As we know, Radiowave speed in the air = c = 3 x 10^8 m/s
f = frequency = 604 kHz = 604 x 10^3 Hz
Hence, wavelength = (3x10^8/604x10^3) m
λ
= 496.69 m
So the height of the antenna BROADCASTING AT 604 kHz = λ /4 = (496.69/4) m
= 124.17 m
(b) As we know , f = 1710 kHz = 1710 x 10^3 Hz (1kHZ = 1000 Hz)
Hence, wavelength = λ = (3 x 10^8/1710 x 10^3) m
λ= 175.44 m
So, height of the antenna = λ /4 = (175.44/4) m
= 43.86 m
Answer:
a) 2.7s
b) 29 m/s
Explanation:
The equation for the velocity and position of a free fall are the following
-(1)
- (2)
Since the hot-air ballon is <em>descending </em>at 2.1m/s and the camera is dropped at 42 m above the ground:
To calculate the time which it takes to reach the ground we use eq(2) with x=0, and look for the positive solution of t:
t = 2.71996
Rounding to two significant figures:
t = 2.7 s
Now we calculate the velocity the camera had just before it lands using eq(1) with t=2.7s
v = -28.782 m/s
Rounding to two significant figures:
v = -29 m/s
where the minus sign indicates the downwards direction
By the law of momentum conservation:-
=>m¹u¹ + m²u² = m1v1 + m²v² {let East is +ve}
=>u¹ + u² = v¹ + v² {as m1=m2}
=>3.5 - 2.75 = v1-1.5
<span>
=>v¹ = 2.25 m/s (East) </span>
