As per kinematics equation we are given that
now we are given that
a = 2.55 m/s^2
now we need to find x
from above equation we have
so it will cover a distance of 93.2 m
Distinguish between the terms strength, power, and endurance as they are used in weightlifting
2 answers:
You might be interested in
Complete Question
An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The electric field in the wire changes with time as E(t)=0.0004t2−0.0001t+0.0004 newtons per coulomb, where time is measured in seconds.
I = 1.2 A at time 5 secs.
Find the charge Q passing through a cross-section of the conductor between time 0 seconds and time 5 seconds.
Answer:
The charge is
Explanation:
From the question we are told that
The diameter of the wire is
The radius of the wire is
The resistivity of aluminum is
The electric field change is mathematically defied as
Generally the charge is mathematically represented as
Where A is the area which is mathematically represented as
So
Therefore
substituting values
From the question we are told that t = 5 sec
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>