The internal energy increases 1.5 times
Explanation:
the internal energy of a mono atomic gas is given by E= 3/2kT
T1= 27⁰C= 27+273=300 K
T2=177⁰C= 177+273=450 K
E1= 3/2 k (300)
E2=3/2 k (450)
so E2/E1=[3/2 k (450)][3/2 k (300)]
E2/E1=450/300
E2= 1.5 E1
so the internal energy increases by a factor of 1.5
Answer:
ΔX = 0.0483 m
Explanation:
Let's analyze the problem, the car oscillates in the direction y and advances with constant speed in the direction x
The car can be described with a spring mass system that is represented by the expression
y = A cos (wt + φ)
The speed can be found by derivatives
= dy / dt
= - A w sin (wt + φ
So that the amplitude is maximum without (wt + fi) = + -1
= A w
X axis
Let's reduce to the SI system
vₓ = 15 km / h (1000 m / 1 km) (1h / 3600s) = 4.17 m / s
As the car speed is constant
vₓ = d / t
t = d / v
ₓ
t = 4 / 4.17
t = 0.96 s
This is the time between running two maximums, which is equivalent to a full period
w = 2π f = 2π / T
w = 2π / 0.96
w = 6.545 rad / s
We have the angular velocity we can find the spring constant
w² = k / m
m = 1200 + 4 80
m = 1520 m
k = w² m
k = 6.545² 1520
k = 65112 N / m
Let's use Newton's second law
F - W = 0
F = W
k x = W
x = mg / k
Case 1 when loaded with people
x₁ = 1520 9.8 / 65112
x₁ = 0.22878 m
Case 2 when empty
x₂ = 1200 9.8 / 65112
x₂ = 0.18061 m
The height variation is
ΔX = x₁ -x₂
ΔX = 0.22878 - 0.18061
ΔX = 0.0483 m
Answer:
0.53 N, 25.6°
Explanation:
side of triangle, a = 1.2 m
q = 7 μC
q1 = - 8 μC
q2 = - 6 μC
Let F1 be the force between q and q1
By using the coulomb's law
F1 = 0.35 N
Let F2 be the force between q and q2
By using the coulomb's law
F2 = 0.26 N
Write the forces in the vector form
Net force
Magnitude of the force
F = 0.53 N
Direction of force with x axis
θ = 25.6°
The correct answer is:
<span>B.) At terminal velocity there is no net force
In fact, when the parachutist reaches the terminal velocity, his velocity does not change any more. It means that the acceleration acting on the parachutist is zero, and for Newton's second law, this means the net force acting on him is zero:
</span>
<span>because the acceleration is zero: a=0.
This also means that the two relevant forces acting on the parachutist (gravity, downward, and air resistance, upward) are balanced to produce a net force equal to zero.</span>
Answer
Correct Answer are
Electron 1, neutron -6, photon-7, ground state-5, protons-3, element-2,orbital-4
Explanation:
Defination of terms is given in front of terms
Electron 1. negatively charged
Neutron 6.neutral subatomic particles found in the nucleus of the atom
photon 7.packet of energy of specific size
ground state 5.lowest energy position of an electron in an atom
protons 3. positively charged, subatomic particles found in the nucleus of the atom
orbital 2.the path of an electron around the nucleus of an atom
element 4.substance with only one type of atom