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.
Explanation:
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Answer:
The horizontal component of displacement is d' = 1422.7 m
Explanation:
Given data,
The distance covered by the truck, d = 1430 m
The angle formed with the horizontal, Ф = 5.76°
The displacement is a vector quantity.
The horizontal component of displacement is given by,
d' = d cos Ф
= 1430 cos 5.76°
= 1422.7 m
Hence, the horizontal component of displacement is d' = 1422.7 m
The age of the galaxy when we look back is 13.97 billion years.
The given parameters:
- <em>distance of the galaxy, x = 2,000 Mpc</em>
According Hubble's law the age of the universe is calculated as follows;
v = H₀x
where;
H₀ = 70 km/s/Mpc

Thus, the age of the galaxy when we look back is 13.97 billion years.
Learn more about Hubble's law here: brainly.com/question/19819028
Answer:
R = 98304.75 m = 98.3 km
Explanation:
The density of an object is given as the ratio between the mass of that object and the volume occupied by that object.
Density = Mass/Volume
Now, it is given that the density of Earth has become:
Density = 1 x 10⁹ kg/m³
Mass = Mass of Earth (Constant) = 5.97 x 10²⁴ kg
Volume = 4/3πR³ (Volume of Sphere)
R = Radius of Earth = ?
Therefore,
1 x 10⁹ kg/m³ = (5.97 x 10²⁴ kg)/[4/3πR³]
4/3πR³ = (5.97 x 10²⁴ kg)/(1 x 10⁹ kg/m³)
R³ = (3/4)(5.97 x 10¹⁵ m³)/π
R = ∛[0.95 x 10¹⁵ m³]
<u>R = 98304.75 m = 98.3 km</u>