Answer:
density is Mg/µL
Explanation:
given data
density of nuclear = kg/m³
1 ml = 1 cm³
to find out
density of nuclear matter in Mg/µL
solution
we know here
1 Mg = 1000 kg
so
1 m³ is equal to cm³
and here 1 cm³ is equal to 1 mL
so we can say 1 mL is equal to 10³ µL
so by these we can convert density
density = kg/m³
density = kg/m³ ×
Mg/µL
density = Mg/µL
Answer:
The freight train would take 542.265 second to increase the speed of the train from rest to 80.0 kilometers per hour.
Explanation:
Statement is incomplete. Complete description is presented below:
<em>A freight train has a mass of </em><em>. The wheels of the locomotive push back on the tracks with a constant net force of </em>
<em>, so the tracks push forward on the locomotive with a force of the same magnitude. Ignore aerodynamics and friction on the other wheels of the train. How long, in seconds, would it take to increase the speed of the train from rest to 80.0 kilometers per hour?</em>
If locomotive have a constant net force (), measured in newtons, then acceleration (
), measured in meters per square second, must be constant and can be found by the following expression:
(1)
Where is the mass of the freight train, measured in kilograms.
If we know that and
, then the acceleration experimented by the train is:
Now, the time taken to accelerate the freight train from rest (), measured in seconds, is determined by the following formula:
(2)
Where:
- Final speed of the train, measured in meters per second.
- Initial speed of the train, measured in meters per second.
If we know that ,
and
, the time taken by the freight train is:
The freight train would take 542.265 second to increase the speed of the train from rest to 80.0 kilometers per hour.
Answer:
<em>a) 318.2 W/m^2</em>
<em>b) 2.5 x 10^-4 J</em>
<em>c) 1.55 x 10^-8 v/m</em>
<em></em>
Explanation:
Power of laser P = 1 mW = 1 x 10^-3 W
exposure time t = 250 ms = 250 x 10^-3 s
If beam diameter = 2 mm = 2 x 10^-3 m
then
cross-sectional area of beam A = = (3.142 x
)/4
A = 3.142 x 10^-6 m^2
a) Intensity I = P/A
where P is the power of the laser
A is the cros-sectional area of the beam
I = ( 1 x 10^-3)/(3.142 x 10^-6) = <em>318.2 W/m^2</em>
<em></em>
b) Total energy delivered E = Pt
where P is the power of the beam
t is the exposure time
E = 1 x 10^-3 x 250 x 10^-3 = <em>2.5 x 10^-4 J</em>
<em></em>
c) The peak electric field is given as
E =
where I is the intensity of the beam
E is the electric field
c is the speed of light = 3 x 10^8 m/s
= 8.85 x 10^9 m kg s^-2 A^-2
E = = <em>1.55 x 10^-8 v/m</em>