A 10.0-kg mass is placed on a 25.0o incline and friction keeps it from sliding. The coefficient of static friction in this case
1 answer:
The frictional force while the mass is sliding will be 46.2 N.
<h3>What is friction force?</h3>
Opposition forces on the surface cause heat loss during the motion of an object known as the friction force.
Given data:
m(mass)= 10.0-kg
Θ (Inclination angle)=25.0o
Coefficient of sliding friction,=0.520
Coefficient of static friction,
The friction force, F=?
Resolve the force in the inclined plane;
Hence, the frictional force while the mass is sliding will be 46.2 N.
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The wavenumber and (b) the wavelength of the radiation used by an fm radio transmitter broadcasting at 92. 0 mhz will be 31.25 *
and 0.032 *
m respectively
Forms of electromagnetic radiation like radio waves, light waves or infrared (heat) waves make characteristic patterns as they travel through space. Each wave has a certain shape and length. The distance between peaks (high points) is called wavelength.
Wavenumber, also called wave number, a unit of frequency, often used in atomic, molecular, and nuclear spectroscopy, equal to the true frequency divided by the speed of the wave and thus equal to the number of waves in a unit distance.
wavelength = ?
frequency = 92 m Hz = 92 * Hz
speed of light = 3 * m/s
speed of light = frequency * wavelength
wavelength = speed of light / frequency
= 3 * / 92 *
= 0.032 * m
wavenumber = 1 / wavelength
= 1 / 0.032 * m
= 31.25 *
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Answer:
solution:
dT/dx =T2-T1/L
&
q_x = -k*(dT/dx)
<u>Case (1) </u>
dT/dx= (-20-50)/0.35==> -280 K/m
q_x =-50*(-280)*10^3==>14 kW
Case (2)
dT/dx= (-10+30)/0.35==> 80 K/m
q_x =-50*(80)*10^3==>-4 kW
Case (2)
dT/dx= (-10+30)/0.35==> 80 K/m
q_x =-50*(80)*10^3==>-4 kW
Case (3)
q_x =-50*(160)*10^3==>-8 kW
T2=T1+dT/dx*L=70+160*0.25==> 110° C
Case (4)
q_x =-50*(-80)*10^3==>4 kW
T1=T2-dT/dx*L=40+80*0.25==> 60° C
Case (5)
q_x =-50*(200)*10^3==>-10 kW
T1=T2-dT/dx*L=30-200*0.25==> -20° C
note:
all graph are attached
