Answer:
C) It increases to 2.0 cm
Explanation:
In a double-slit diffraction experiment, the distance on the screen between two adjacent maxima is given by
where
is the wavelength of the wave
D is the distance of the screen from the slits
d is the separation between the slits
In this problem, the initial distance between adjacent maxima is 1.0 cm. Later, the slit separation is cut in a half, which means that the new slit separation is
Substituting into the equation, we find that the new separation between the maxima is
So, the distance increases by a factor 2: therefore, the new separation between the maxima will be 2.0 cm.
Answer:
The volume is decreasing at 160 cm³/min
Explanation:
Given;
Boyle's law, PV = C
where;
P is pressure of the gas
V is volume of the gas
C is constant
Differentiate this equation using product rule:
Given;
(increasing pressure rate of the gas) = 40 kPa/min
V (volume of the gas) = 600 cm³
P (pressure of the gas) = 150 kPa
Substitute in these values in the differential equation above and calculate the rate at which the volume is decreasing ( );
(600 x 40) + (150 x ) = 0
Therefore, the volume is decreasing at 160 cm³/min
Earthquakes happen<span> when rock below the Earth's surface moves abruptly. </span>Usually, the rock is moving along large cracks in Earth's crust called faults. Most earthquakes happen<span> at or near the </span>boundaries<span> between Earth's </span>tectonic plates <span>because that's where there is </span>usually<span> a large concentration of faults</span>
Answer:
A) OA, AB, BC
B) 25m/s^2
C) see explanation
D) 25
E) Rest
Explanation:
From the Velocity time graph shown:
The positive slope = OA ; This is positive because, it is the point of uniform acceleration on the graph.
Constant slope = AB, the slope here is constant because, AB on the graph is the point of constant velocity.
-ve slope = BC
B) Acceleration of body in path OA.
Acceleration = change in Velocity / time
Acceleration = (150 - 0) / 6
Acceleration = 150/6 = 25m/s^2
C) Path AB is Parallel to the because it marks the period of constant velocity (that is Velocity does not increase or decrease during the time interval).
D) Length of BC
BC corresponds to the distance moved, that velocity / time
Velocity = 150 ; time = 6
Therefore Distance (BC) = 150/6 = 25
E.) Velocity =0 ; Hence body is at rest