Heat required to raise the temperature of water is given as
here we have
m = 100 g = 0.100 kg
s = 4183 J/kg C
now we can use the above equation
so here it requires 20920 J heat to raise the temperature of 100 g water by 50 degree C
Which of the following is made from a Crystaline solid that conducts a current only under certain circumstances?
1 answer:
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1) 9.4 m/s
First of all, we can calculate the work done by the horizontal force, given by
W = Fd
where
F = 34.6 N is the magnitude of the force
d = 12.9 m is the displacement of the cart
Solving ,
W = (34.6 N)(12.9 m) = 446.3 J
According to the work-energy theorem, this is also equal to the kinetic energy gained by the cart:
Since the cart was initially at rest, , so
(1)
where
m is the of the cart
v is the final speed
The mass of the cart can be found starting from its weight, :
So solving eq.(1) for v, we find the final speed of the cart:
2)
The work done on the train is given by
W = Fd
where
F is the magnitude of the force
d is the displacement of the train
In this problem,
So the work done is
3)
According to the work-energy theorem, the change in kinetic energy of the train is equal to the work done on it:
where
W is the work done
is the change in kinetic energy
Therefore, the change in kinetic energy is
4) 37.2 m/s
According to the work-energy theorem,
where
is the final kinetic energy of the train
is the initial kinetic energy of the train, which is zero since the train started from rest
Re-writing the equation,
where
m = 36300 kg is the mass of the train
v is the final speed of the train
Solving for v, we find
Answer:
a) I = 1,944 10⁻⁴⁶ Kg m²
, b) I = 7,915 10⁻⁵¹ Kg m²
Explanation:
a) The moment of inertia of point masses is
I = m r²
The nuclei have a very small size (10-15 m) so we can consider them punctual.
The distance to the center of mass passing through the middle of the two nuclei of equal mass
r = 0.6 10⁻¹⁰ m
The moment of total inertia
I = I₁ + I₂ = 2 I₀
I = 2 2.7 10⁻²⁶ (0.6 10⁻¹⁰)²
I = 1,944 10⁻⁴⁶ Kg m²
The kinetic energy of the rotation is
w = h / 2π
K = ½ I w²
K = ½ 1,944 10⁻⁴⁶ (h / 2π)² = ½ 1.944 10⁻⁴⁶ (6.63 10⁻³⁴ / 2π)²
K = 2.16 10⁻¹¹⁴ J (1eV / 1.6 10⁻¹⁹ J)
K = 2.16 10⁻⁹⁵ eV
B) the moment of inertia of the electron in the orbit, we can calculate it with the parallel axis theorem
I = + m R²
I = r² +
R²
Where R we calculate it by Pythagoras
R² = (0.6 10⁻¹⁰)² + (0.5 10⁻¹⁰) 2
R = √ (0.61 10⁻²⁰)
R = 0.78 10⁻¹⁰ m
I = 9.1 10⁻³¹ (0.5 10⁻¹⁰)² + 9.1 10⁻³¹ (0.78 10⁻¹⁰)²
I = (2,275 +5.54) 10⁻⁵¹
I = 7,915 10⁻⁵¹ Kg m²
The rotation energy of the electron
K = ½ I w²
If the angular velocity is the electrons outside the core, its kinetic energy is much lower by an order 10⁵, but the angular velocity of the electrons is much higher.
Answer:
a) the answer is t=4 seconds
b) acceleration is zero
c) displacement= 142 m
Explanation:
Given the position of the particle
a) the time required when velocity
now we solve for time
b) acceleration when
acceleration is the time derivative of velocity i.e
c) the net displacement of the particle during the interval t = 1 s to t = 4 s is