A shadow forms on the side of an object that faces away from the sun. The length of shadows changes as Earth rotates. In the morning, the sun is low in the eastern sky and shadows are long. As time passes in the morning, the sun seems to move higher in the sky.
the answer is b) magnetic current
Answer:
a) i = -9.63 cm
, h ’= .0.24075 cm erect
b) i = 259.74 cm
,
Explanation:
For this exercise let's start by finding the focal length of the lens
1 / f = (n-1) (1 / R₁ - 1 / R₂)
1 / f = (1.70 -1)) 1 / ∞ - 1/13)
1 / f = 0.0538
f = - 18.57 cm
Now we can use the constructor equation
1 / f = 1 / o + 1 / i
1 / i = 1 / f - 1 / o
1 / i = -1 / 18.57 -1/20
1 / i = -0.1038 cm
I = -9.63 cm
For the height of the
image let's use magnification
m = h '/ h = - i / o
h ’= -h i / o
h ’= - 0.5 (-9.63) / 20
h ’= .0.24075 cm
b) we invert the lens
The focal length is
1 / f = (1.70 -1) (1/13 - 1 / int)
1 / f = 0.0538
f = 18.57 cm
1 / i = 1 / f -1 / o
1 / I = 1 / 18.57 - 1/20
1 / I = 3.85 10-3
i = 259.74 cm
h ’= - 0.5 259.74 / 20
h ’= 6.4935 cm
Answer:
The impulse applied by the stick to the hockey park is approximately 7 kilogram-meters per second.
Explanation:
The Impulse Theorem states that the impulse experimented by the hockey park is equal to the vectorial change in its linear momentum, that is:
(1)
Where:
- Impulse, in kilogram-meters per second.
- Mass, in kilograms.
- Initial velocity of the hockey park, in meters per second.
- Final velocity of the hockey park, in meters per second.
If we know that
,
and
, then the impulse applied by the stick to the park is approximately:
![I = (0.2\,kg)\cdot \left(35\,\hat{i}\right)\,\left[\frac{m}{s} \right]](https://tex.z-dn.net/?f=I%20%3D%20%280.2%5C%2Ckg%29%5Ccdot%20%5Cleft%2835%5C%2C%5Chat%7Bi%7D%5Cright%29%5C%2C%5Cleft%5B%5Cfrac%7Bm%7D%7Bs%7D%20%5Cright%5D)
![I = 7\,\hat{i}\,\left[\frac{kg\cdot m}{s} \right]](https://tex.z-dn.net/?f=I%20%3D%207%5C%2C%5Chat%7Bi%7D%5C%2C%5Cleft%5B%5Cfrac%7Bkg%5Ccdot%20m%7D%7Bs%7D%20%5Cright%5D)
The impulse applied by the stick to the hockey park is approximately 7 kilogram-meters per second.
Answer:
True
Explanation:
Standard heat of formation is the heat change that deals with the formation of 1mole at standard rates and states of the given reactants . Standard heat of formation is the difference between the enthalpy change of reactants and products.