Answer:
The value is 
Explanation:
From the question we are told that
The initial speed is 
Generally the total energy possessed by the space probe when on earth is mathematically represented as

Here
is the kinetic energy of the space probe due to its initial speed which is mathematically represented as
=>
=> 
And
is the kinetic energy that the space probe requires to escape the Earth's gravitational pull , this is mathematically represented as

Here
is the escape velocity from earth which has a value 
=> 
=> 
Generally given that at a position that is very far from the earth that the is Zero, the kinetic energy at that position is mathematically represented as

Generally from the law energy conservation we have that
So

=> 
=> 
=> 
Answer:
t = 1.58 s
Explanation:
given,
Speed of ranger, v = 56 km/h
v = 56 x 0.278 = 15.57 m/s
distance, d = 65 m
deceleration,a = 3 m/s²
reaction time = ?
using stopping distance formula


t is the reaction time

t = 1.58 s
hence, the reaction time of the ranger is equal to 1.58 s.
Answer:

Explanation:
<u>Displacement Vector</u>
Suppose an object is located at a position

and then moves at another position at

The displacement vector is directed from the first to the second position and can be found as

If the position is given as magnitude-angle data ( z , α), we can compute its rectangular components as


The question describes the situation where the initial point is the base of the mountain, where both components are zero

The final point is given as a 520 m distance and a 32-degree angle, so


The displacement is

Answer:
"C" I think....
Explanation:
I am really sorry if I am wrong, but if right, I hope this helps!
Answer:
a)F=3 x 10⁻⁷ N
b)x=2.405 m
Explanation:
Given that
m₁=295 kg
m₂=595 kg
d= 4.1 m
a)
m₃=63 kg
r=d/2 = 2.05 m
The force between the mass m₁ and m₃

by putting the values


F₁₃=2.94 x 10⁻⁷ N
The force between the mass m₂ and m₃
by putting the values


F₂₃=5.94 x 10⁻⁷ N
The net force F
F= F₂₃- F₁₃
F=5.94 x 10⁻⁷ N-2.94 x 10⁻⁷ N
F=3 x 10⁻⁷ N
b)
Lest take at distance x from mass m₂ net force is zero.


Form above two equation



x²=2.01(4.1-x)²
x=1.42 (4.1-x)
x=5.82 - 1.42x
x=2.405 m