Answer:
The carrier lengthen is 0.08436 m.
Explanation:
Given that,
Length = 370 m
Initial temperature = 2.0°C
Final temperature = 21°C
We need to calculate the change temperature
Using formula of change of temperature
We need to calculate the carrier lengthen
Using formula of length
Put the value into the formula
Hence, The carrier lengthen is 0.08436 m.
Answer:
a) 95950 pascals
b) 137642.5 pascals
Explanation:
The absolute pressure (Pabs) on a fluid is:
(1)
With Pgauge the pressure due depth on the fluid and Patm the atmospheric pressure. Pgauge is equal to:
(2)
with ρ the fluid density, g the gravitational acceleration and h the depth on the fluid. Using (2) on (1) and solving for Patm:
b) Here we're going to use again (1) but now we have another value of density because it's other liquid, to know that value we should use the fact that specific gravity (S.G) for liquids is the ratio between fluid density and water density:
so:
Answer:
Check the explanation
Explanation:
given
R = 1.5 cm
object distance, u = 1.1 cm
focal length of the ball, f = -R/2
= -1.5/2
= -0.75 cm
let v is the image distance
use, 1/u + 1/v = 1/f
1/v = 1/f - 1/u
1/v = 1/(-0.75) - 1/(1.1)
v = -0.446 cm <<<<<---------------Answer
magnification, m = -v/u
= -(-0.446)/1.1
= 0.405 <<<<<<<<<---------------Answer
The image is virtual
The image is upright
given
R = 1.5 cm
object distance, u = 1.1 cm
focal length of the ball, f = -R/2
= -1.5/2
= -0.75 cm
let v is the image distance
use, 1/u + 1/v = 1/f
1/v = 1/f - 1/u
1/v = 1/(-0.75) - 1/(1.1)
v = -0.446 cm <<<<<---------------Answer
magnification, m = -v/u
= -(-0.446)/1.1
= 0.405 <<<<<<<<<---------------Answer
Kindly check the diagram in the attached image below.
Objects having both kinetic and potential energy:
water flowing downstream
a child swinging on a swing
a bouncing ball
a plane in flight at 30,000 feet
Explanation:
The kinetic energy of an object is the energy possessed by the object due to its motion. It is given by
where
m is the mass of the object
v is its speed
Therefore, an object has kinetic energy when its speed is non-zero (so, whenever it is moving).
The potential energy of an object is the energy possessed by the object due to its position in the gravitational field. It is given by
where
m is the mass
g is the acceleration of gravity
h is the heigth of the object relative to the ground
Therefore, an object has potential energy whenever it is located at a certain height above the ground.
So in this problem, the objects that have both kinetic and potential energy are:
a rock at the edge of a cliff --> NO, because the rock is at rest (so KE = 0)
water flowing downstream --> YES, because the water is moving AND it is at a certain height above the ground
a child swinging on a swing --> YES, because the child is moving AND it is at a certain height above the ground
a bouncing ball --> YES, because the ball is moving AND it is at a certain height above the ground
water behind a dam --> NO, because the water is at rest (so KE=0)
a car moving on a level road --> NO, because the car is at ground level (so PE=0)
a plane in flight at 30,000 feet --> YES, because the plane is moving AND it is at a certain height above the ground
a compressed spring --> NO, because the spring is at rest (so KE=0)
Learn more about kinetic and potential energy:
#LearnwithBrainly
Answer:
See explanation
Explanation:
Solution:-
- Here we will assume that the grating has the line density ( N ) defined by the number of lines per mm.
- The angle that each fringe forms on the screen is defined by ( θ ).
- The order of bright/dark spot is defined by an integer ( n )
- The wavelength of the incident light is ( λ )
- Here we will use the relation given for diffraction grating by the Young's Experiment as follows:
- The above given formulation is for constructive interference.
- We will inspect the effect of increasing the distance between the screen and the grating. Consider the length ( L ) from the center of the grating to the center of the screen. The distance ( yn ) will denote the distance between each fringe in vertical direction on the screen.
- For small angles ( θ ) we can make an approximation of sin ( θ ) ≈ tan ( θ ). Where,
sin ( θ ) ≈ tan ( θ ) = [ yn / L ]
- Substitute the above approximation in the given relation of diffraction gratings as follows:
- To double the distance between the screen and grating we will use the above relation with ( 2L ):
yn ∝ L
Result: The distance between each order of bright and dark fringe is doubled. The interference pattern would have twice the spread! This also means that less number of bright spots would be seen on the screen as the coverage area would require a larger screen to accommodate the entire interference pattern. The spread also reduces the intensity/contrast between the bright and dark fringes because the distance travelled by each ray of light has increased. The intensity is inversely proportional to the square of distance travelled.
- Similarly, the line density of the grating ( N ) was doubled. Then,
yn ∝ N
Result: The distance between each order of bright and dark fringe is doubled. The interference pattern would have twice the spread!This also means that less number of bright spots would be seen on the screen as the coverage area would require a larger screen to accommodate the entire interference pattern.