Answer:
0.650
Explanation:
Move the decimal point three times to the left.
Answer:
a) correct answer is C
, b) 14º from the west to the north, c) v_{1g} = 300.79 km / h
Explanation:
This is a relative speed exercise using the addition of speeds.
1) when it is not specified regarding what is being measured, the medicine is carried out with respect to the Z Earth, therefore the correct answer is C
2 and 3) In this case we must compose the speed using the Pythagorean Theorem.
² =
² +
²
where v_{1a} is the speed of the airplane with respect to the air, v_{1g} airplane speed with respect to the Earth, v_{ag} air speed with respect to the Earth
in this case let's clear the speed of the airplane with respect to the Earth
v_{1g} = √(v_{1a}² - v_{ag}²)
v_{1g} = √ (310² - 75²)
v_{1g} = 300.79 km / h
we find the direction of the airplane using trigonometry
sin θ = v_{ag} / v_{1a}
θ = sin⁻¹ (v_{ag} /v_{1a})
θ = sin⁻¹ (75/310)
θ= 14º
the pilot must direct the aircraft at an angle of 14º from the west to the north
Answer:
The value is 
Explanation:
From the question we are told that
The period of the asteroid is 
Generally the average distance of the asteroid from the sun is mathematically represented as
![R = \sqrt[3]{ \frac{G M * T^2 }{4 \pi} }](https://tex.z-dn.net/?f=R%20%3D%20%5Csqrt%5B3%5D%7B%20%5Cfrac%7BG%20M%20%2A%20T%5E2%20%7D%7B4%20%5Cpi%7D%20%7D)
Here M is the mass of the sun with a value

G is the gravitational constant with value 
![R = \sqrt[3]{ \frac{6.67 *10^{-11} * 1.99*10^{30} * [5.55 *10^{9}]^2 }{4 * 3.142 } }](https://tex.z-dn.net/?f=R%20%3D%20%5Csqrt%5B3%5D%7B%20%5Cfrac%7B6.67%20%2A10%5E%7B-11%7D%20%20%2A%201.99%2A10%5E%7B30%7D%20%2A%20%5B5.55%20%2A10%5E%7B9%7D%5D%5E2%20%7D%7B4%20%2A%203.142%20%7D%20%7D)
=> 
Generally

So

=> 
=> 
Answer:
0.20kg-m^2
Explanation:
Let the linear velocity of the rope(=of pulley) is v m/s
Using kinematic equation
=> v = u + at
=>v = 0 + 4.9a
=>v = 4.9a ------------ eq1
By v^2 = u^2 + 2as
=>v^2 = 0 + 2 x v/4.9 x 1.2
=>4.9v^2 - 2.4v = 0
=>v(4.9v - 2.4) = 0
=>v = 2.4/4.9 = 0.49 m/s
Thus by v = r x omega
=>omega = v/r = 0.49/0.02 = 24.49 rad/sec
BY W = F x s = 50 x 1.2 = 60 J
=>KE(rotational) = W = 1/2 x I x omega^2
=>60 = 1/2 x I x (24.49)^2
=>I = 0.20 kg-m^2