Answer:
An ultra intense laser is one with which intensities greater than 1015 W cm-2 can be achieved.
Explanation:
This intensity, which was the upper limit of lasers until the invention of the Chirped Pulse Amplification, CPA technique, is the value around which nonlinear effects on the transport of radiation in materials begin to appear.
Currently, the most powerful lasers reach intensities of the order of 1021W cm-2 and powers of Petawatts, PW, in each pulse. This range of intensities has opened the door for lasers to a multitude of disciplines and scientific areas traditionally reserved for accelerators and nuclear reactors, applying as generators of high-energy electron, ion, neutron and photon beams, without the need for expensive infrastructure.
Answer:
[ 2.67 , 1 ] m
Explanation:
Given:-
- The side lengths of the rods are as follows:
a = 4 m , b = 4 m , c = 5 m
a = Base , b = Perpendicular , c = Hypotenuse
- All rods are made of same material with uniform density. With
Find:-
Find the coordinates of the center of mass of the triangle.
Solution:-
- The center of mass of any triangle is at the intersection of its medians.
- So let’s say we have a triangle with vertices at points (0,0) , (a,0) , and (0,b).
- Median from (0,0) to midpoint (a/2,b/2) of opposite side has equation:
bx−ay=0
- Median from (a,0) to midpoint (0,b/2) of opposite side has equation:
bx+2ay=ab
- Median from (0,b) to midpoint (a/2,0) of opposite side has equation:
2bx+ay=ab
- Solve all three equations simultaneously:
bx−ay=0 , bx = ay
ay + 2ay = ab , 3ay = ab , y = b/3
bx = b/3
x = a / 3
- So the distance from the median to each leg of the triangle is 1/3 length of other leg.
- So the coordinates of the centroid for right angle triangle would be:
[ 2a/3 , b/3 ]
[ 2.67 , 1 ] m
Answer:
90-100meters
Explanation:
The overall length limitation of a UTP cable is 90-100meters. Once this limitation is reached, a repeater is employed to transfer data.
Mary and her younger brother Alex decide to ride the carousel at the State Fair, Mary's and Alex's angular speed M and tangential speed vM is mathematically given as
Mary's and Alex's angular speed=1.43
Tangential speed mary=3.22 m/s
Tangential speed alex =2.260m/s
<h3>What is Mary's and Alex's angular
speed M and tangential speed vM?</h3>
Generally, the equation for angular speed is mathematically given as
w = 1.61 rev/see 3.9
Centripetal acc mary = v^2/r
Centripetal acc mary = w^2r
Centripetal acc mary = w^2x 2m
Centripetal acc. of Alex = w²x L.u
Therefore
Hence
tang. speed V=Wr
tang. speed of mary = 1.61x2 = 3.22 m/s
tang. speed of Alex: 1.61X1·4 =2.260m/s
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Answer:
a) 29.4 J
b) - 29.4 J
Explanation:
Given:
Mass of the book, m = 2 kg
Height above the floor, h = 1.5 m
Now,
the work done by the person will be = Force applied on the book × displacement of the book
thus,
Work done by the person = mg × h
where, g is the acceleration due to gravity
thus, on substituting the values, we get
Work done by the person = 2 × 9.8 × 1.5 = 29.4 J
now,
for the force applied by the gravitational pull (downwards) the displacement is in opposite direction (upwards) to the force of the gravity.
Thus,
work done by the gravity will be negative
therefore, the work done by the gravity = - mg × h
or
work done by the gravity = - 29.4 J
