Answer:
The speed of the wave on the longer wire is 580m/s
Explanation:
The velocity possessed by a stretched string is directly proportional to the tension in the string and inversely proportional to the mass per unit length of the string. Mathematically,
V = √T/m
Where V is the velocity of wave in the wire
T is the tension in the wire
M is the mass per unit length of the wire
Let m1 and m2 be the mass per unit length of the wires
Let T1 and T2 be their respective tensions
Since the tension and mass of the wire is the same
m1= m2= m.
T1=T2=T
Let m1 =M/l
m2 =M/4l( since the second is tour times as far apart)
V1 = 290m/s(velocity in shorter wire)
V2 is the velocity of the longer wire.
V1 = √T/(m/l)
290 = √Tl/m
290² = Tl/m... 1
V2 = √4Tl/m
V2²= 4Tl/m... 2
Dividing equation 1 by 2 we have;
290²/V2² = {Tl/m}/{4Tl/m}
290²/V2² = Tl/m × m/4Tl
290²/V2² = 1/4
Cross multiplying we have;
V2² = 290²×4
V2 = √290²×4
V2 = 290×2
V2 = 580m/s
Answer:
Wavelength,
Explanation:
It is given that,
Frequency, f = 99.5 MHz = 99.5 × 10⁶ Hz
We need to find the wavelength of the radio waves from an FM station operating at above frequency. The relationship between the frequency and the wavelength is given by :
c = speed of light
So, the wavelength of the radio waves from an FM station is 3.01 m. Hence, this is the required solution.
Answer:
A) Speed
Explanation:
The medium of transmission(air) does not change before and after reflection. The sound still travels in air after the reflection. i.e the medium of transmission of sound is constant.
Since the medium of transmission remains constant, speed also remains constant. Speed only changes when the medium of transmission changes.
Answer:
a) 16.32 m/s
b) 640 N
Explanation:
A) mass of rocket m_r = 1000 g = 1 kg
initial speed of rocket u_r = 15 m/sec
initial speed of ball is u_b = 18 m/sec
final speed of ball is v_b = 40 m/sec
Let m_b be the mass of the ball= 60 g and v_r be the final velocity of the rocket
from law of conservation of momentum
momentum of the system remains zero
m_r×(u_r-v_r)+m_b(16-42) = 0
1×(15-v_r) = -0.060(18-40)
15-v_r = -1.32
v_r = 15+1.32 = 16.32 m/sec.
B) Average force that the rocket exert's on the ball is F_avg can be calculated as
contact time t=7.00 ms
F_avg = m(v-u)/t = 0.06×(40+18)/0.007 = 640 N