Answer: Electromagnetic waves are generated by moving electrons. An electron generates an electric field which we can visualize as lines radiating from the electron Figure 10a. If the electron moves, say it vibrates back and forth, then this motion will be transferred to the field lines and they will become wavy Figure 10b.
The temperature of the gas is 41.3 °C.
Answer:
The temperature of the gas is 41.3 °C.
Explanation:
So on combining the Boyle's and Charles law, we get the ideal law of gas that is PV=nRT. Here P is the pressure, V is the volume, n is the number of moles, R is gas constant and T is the temperature. The SI unit of pressure is atm. So we need to convert 1 Pa to 1 atm, that is 1 Pa = 9.86923×
atm. Thus, 171000 Pa = 1.6876 atm.
We know that the gas constant R = 0.0821 atmLMol–¹K-¹. Then the volume of the gas is given as 50 L and moles are given as 3.27 moles.
Then substituting all the values in ideal gas equation ,we get
1.6876×50=3.27×0.0821×T
Temperature = 
So the temperature is obtained to be 314.3 K. As 0°C = 273 K,
Then 314.3 K = 314.3-273 °C=41.3 °C.
Thus, the temperature is 41.3 °C.
Answer:
A
B

C

D

Explanation:
Considering the first question
From the question we are told that
The spring constant is 
The potential energy is 
Generally the potential energy stored in spring is mathematically represented as 
=>
=>
=>
Considering the second question
From the question we are told that
The mass of the dart is m = 0.050 kg
Generally from the law of energy conservation

=> 
=> 
Considering the third question
The height at which the dart was fired horizontally is 
Generally from the law of energy conservation

Here KE is kinetic energy of the dart which is mathematical represented as

=> 
=> 
=> 
Considering the fourth question
Generally the total time of flight of the dart is mathematically represented as

=> 
=> 
Generally the horizontal distance from the equilibrium position to the ground is mathematically represented as

=> 
=> 
Answer:
f1 = -3.50 m
Explanation:
For a nearsighted person an object at infinity must be made to appear to be at his far point which is 3.50 m away. The image of an object at infinity must be formed on the same side of the lens as the object.
∴ v = -3.5 m
Using mirror formula,
i/f1 = 1/v + 1/u
Where f1 = focal length of the contact lens, v = image distance = -3.5 m, u = object distance = at infinity(∞) = 1/0
∴ 1/f1 = (1/-3.5) + 1/infinity
Note that, 1/infinity = 1/(1/0) = 0/1 =0.
∴ 1/f1 = 1/(-3.5) + 0
1/f1 = 1/(-3.5)
Solving the equation by finding the inverse of both side of the equation.
∴ f1 = -3.50 m
Therefore a converging lens of focal length f1 = -3.50 m
would be needed by the person to see an object at infinity clearly
Answer:
it will be curved as in deceleration
Explanation: