A wave looses its power as it comes to shore because it gets less deeper every second it gets closer to shore
        
             
        
        
        
Answer:
a) -41.1 Joule
b) 108.38 Kelvin
Explanation:
Pressure = P = 290 Pa
Initial volume of gas = V₁ = 0.62 m³
Final volume of gas = V₂ = 0.21 m³
Initial temperature of gas = T₁ = 320 K
Heat loss = Q = -160 J
Work done = PΔV
⇒Work done = 290×(0.21-0.62)
⇒Work done = -118.9 J
a) Change in internal energy = Heat - Work
ΔU = -160 -(-118.9)
⇒ΔU = -41.1 J
∴ Change in internal energy is -41.1 J
b) V₁/V₂ = T₁/T₂
⇒T₂ = T₁V₂/V₁
⇒T₂ = 320×0.21/0.62
⇒T₂ = 108.38 K
∴ Final temperature of the gas is 108.38 Kelvin
 
        
             
        
        
        
Answer:
<em>The magnitude of the magnetic field will act in a direction towards me.</em>
<em></em>
Explanation:
When a charged particle enters a magnetic field, it is deflected. The direction of travel of the particle is deflected, but the kinetic energy of the particle is not affected. <em>The force experienced by a charged particle as it enters a magnetic field that acts perpendicular to the path of the velocity of the particle, will produce a force that is perpendicular to both the direction of travel of the particle and the direction of the magnetic field.</em> In this case, the proton moves in the y-direction, the magnetic field is in the x-direction, therefore the force experienced by the particle will be towards me.
 
        
             
        
        
        
Answer:
The location of the survey is biased. 
Explanation:
If someone left a movie theater, it is most likely because he likes watching movies there. A better place to interview would be on the streets, as it’s less biased there
 
        
             
        
        
        
Length of the pendulum (l) = 2 m
Acceleration due to gravity = g = 9.8 m/s^2
For small amplitude, the pendulum will undergo simple harmonic motion.
Hence, the time period of the pendulum for small amplitude = 
Now, plug the values of l and g
T = 
T = 2 × 3.14 × 0.451
T = 2.83 seconds
Hence, the time period of the pendulum for small amplitude = 2.83 s