Answer:
Explanation:
Given
Em wave is in the form of
where 
Wave constant for EM wave k is
Wavelength of wave 
Answer:
T₂ =602 °C
Explanation:
Given that
T₁ = 227°C =227+273 K
T₁ =500 k
Gauge pressure at condition 1 given = 100 KPa
The absolute pressure at condition 1 will be
P₁ = 100 + 100 KPa
P₁ =200 KPa
Gauge pressure at condition 2 given = 250 KPa
The absolute pressure at condition 2 will be
P₂ = 250 + 100 KPa
P₂ =350 KPa
The temperature at condition 2 = T₂
We know that
T₂ = 875 K
T₂ =875- 273 °C
T₂ =602 °C
KE = kinetic energy
PE = potential energy
GPE = gravitational potential energy
energy is always measured in Joules (J)
KE = (0.5) times the mass times the velocity^2
square the velocity first
Mass = (KE x 2) / v^2
square the velocity first, then double the kinetic energy, then divide
mass is measured in kg
velocity = sqrt(KE x 2 / m)
velocity can be called speed, like in the the second problem
remember to find the square root after you double the KE and divide that by the mass.
for example: if after you doubled KE and divided it by the mass you got sqrt(20), the answer would be about 4.47
GPE = mass x gravitational pull (about 9.8 m/s^2 on earth) x height
height = (PE) / (g x m)
do g x m first
So for question 1:
KE = (0.5)0.1 x 1.1^2
always square the velocity first:
KE = (0.5)0.1 x 1.21
KE = 0.0605
so if you rounded it to the nearest hundreths you would get KE = 0.06 J
don't forget the unit of energy is in Joules
Answer:
The moment of inertia is 0.7500 kg-m².
Explanation:
Given that,
Mass = 2.2 kg
Distance = 0.49 m
If the length is 1.1 m
We need to calculate the moment of inertia
Using formula of moment of inertia
Where, m = mass of rod
l = length of rod
x = distance from its center
Put the value into the formula
Hence, The moment of inertia is 0.7500 kg-m².
Answer:
(a) 29 cm
(b) 43.5 cm
Explanation:
(a) when loop A is slack, there are three forces acting on the metre rule.
-0.9 N at 50 cm mark
T at 70 cm mark
-2 N at x
Taking the sum of the torques about B:
∑τ = Iα
(-0.9 N) (50 cm − 70 cm) + (-2 N) (x − 70 cm) = 0
18 Ncm − 2 N (x − 70 cm) = 0
2 N (x − 70 cm) = 18 Ncm
x − 70 cm = 9 cm
x = 79 cm
The distance from the center is |50 cm − 79 cm| = 29 cm.
(b) when loop B is slack, there are three forces acting on the metre rule.
-0.9 N at 50 cm mark
T at 20 cm mark
-2 N at x
Taking the sum of the torques about A:
∑τ = Iα
(-0.9 N) (50 cm − 20 cm) + (-2 N) (x − 20 cm) = 0
-27 Ncm − 2 N (x − 20 cm) = 0
2 N (x − 20 cm) = -27 Ncm
x − 20 cm = -13.5 cm
x = 6.5 cm
The distance from the center is |50 cm − 6.5 cm| = 43.5 cm
