Answer:
Answer:
Speed of the wave in the string will be 3.2 m/sec
Explanation:
We have given frequency in the string fixed at both ends is 80 Hz
Distance between adjacent antipodes is 20 cm
We know that distance between two adjacent anti nodes is equal to half of the wavelength
So \frac{\lambda }{2}=20cm
2
λ
=20cm
\lambda =40cmλ=40cm
We have to find the speed of the wave in the string
Speed is equal to v=\lambda f=0.04\times 80=3.2m/secv=λf=0.04×80=3.2m/sec
So speed of the wave in the string will be 3.2 m/sec
In other words, it would take Deep Space 1 more than 81,000 years to travel the 4.24 light-years between Earth and Proxima Centauri at its top speed of 56,000 km/h. In relation to human history, that would be more than 2,700 generations.
Nearly 40 trillion kilometers, or 4.4 light-years, separate us from Alpha Centauri. The NASA-Germany Helios probes, the fastest spacecraft to date to be launched into orbit, flew at a speed of 250,000 kilometers per hour. The probes would need 18,000 years to travel at such pace to arrive at the sun's nearest neighbor. The calculations reveal that it is almost impossible to reach the nearest star in a human lifetime, even with the most futuristic technologies.
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Answer:
b. 600,000 J
Explanation:
Applying the law of conservation of energy,
The thermal energy created = Kinetic energy of the suv.
Q' = 1/2(mv²)............... Equation 1
Where Q' = Thermal energy, m = mass of the suv, v = velocity of the suv.
From the question,
Given: m = 3000 kg, v = 20 m/s
Substitute these values into equation 1
Q' = 1/2(3000×20²)
Q' = 600000 J
Hence the right option is b. 600,000 J
Answer:
The magnification is -6.05.
Explanation:
Given that,
Focal length = 34 cm
Distance of the image =2.4 m = 240 cm
We need to calculate the distance of the object
Where, u = distance of the object
v = distance of the image
f = focal length
Put the value into the formula
The magnification is
Hence, The magnification is -6.05.
