Answer:
M₀ = 5i - 4j - k
Explanation:
Using the cross product method, the moment vector(M₀) of a force (F) is about a given point is equal to cross product of the vector A from the point (r) to anywhere on the line of action of the force itself. i.e
M₀ = r x F
From the question,
r = i + j + k
F = 1i + 0j + 5k
Therefore,
M₀ = (i + j + k) x (1i + 0j + 5k)
M₀ = ![\left[\begin{array}{ccc}i&j&k\\1&1&1\\1&0&5\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Di%26j%26k%5C%5C1%261%261%5C%5C1%260%265%5Cend%7Barray%7D%5Cright%5D)
M₀ = i(5 - 0) -j(5 - 1) + k(0 - 1)
M₀ = i(5) - j(4) + k(-1)
M₀ = 5i - 4j - k
Therefore, the moment about the origin O of the force F is
M₀ = 5i - 4j - k
Explanation:
The given data is as follows.
Velocity of bullet,
= 814.8 m/s
Observer distance from marksman, d = 24.7 m
Let us assume that time necessary for report of rifle to reach the observer is t and will be calculated as follows.
t =
(velocity in air = 343 m/s)
= 0.072 sec
Now, before the observer hears the report the distance traveled by the bullet is as follows.

= 
= 58.66
= 59 (approx)
Thus, we can conclude that each bullet will travel a distance of 59 m.
Answer:
230kg would be the best answer
Explanation:
Answer:
The height reached by the material on Earth is 91 km.
Explanation:
Given that,
Mass 
Radius = 1821 km
Height 
Suppose we need to find that how high would this material go on earth if it were ejected with the same speed as on Io?
We need to calculate the acceleration due to gravity on Io
Using formula of gravity

Put the value into the formula


Let v be the speed at which the material is ejected.
We need to calculate the height
Using the formula of height

Using ratio of height of earth and height of Io


Put the value into the formula





Hence, The height reached by the material on Earth is 91 km.
mass of the ball m = 0.63 kg
initial height h = 1.8 m
final height h ' = 3.03 m
initial speed v = 7.09 m / s
final speed v ' = 4.21 m / s
Let the work done on the ball by air resistance W = ?
we know from law of conservation of energy ,
total energy at height h + work done by air = total energy at height h '
mgh + ( 1/ 2) mv^ 2 + W = mgh ' + ( 1/ 2) mv'^ 2
0.630*9.8*1.8 + 0.63*7.09^2 + W = mgh ' + ( 1/ 2) mv'^ 2
From there you can find W
if there is negative sign indicates it work opposite direction to motion