Answer:
To achieve the velocity of 40 m/sec height will become 4 times
Explanation:
We have given initially truck is at rest and attains a speed of 20 m/sec
Let the mass of the truck is m
At the top of the hill potential energy is mgh and kinetic energy is 
So total energy at the top of the hill 
At the bottom of the hill kinetic energy is equal to
and potential energy will be 0
So total energy at the bottom of the hill is equal to 
Form energy conservation 
, for v = 20 m/sec

Squaring both side

h = 20.408 m
Now if velocity is 0 m/sec


h = 81.63 m
So we can see that to achieve the velocity of 40 m/sec height will become 4 times
Her magnitude of deceleration on the ice would be 15.126m/s
Answer:
The velocity is 
Explanation:
From the question we are told that
The first distance is 
The first speed is 
The second distance is 
The second speed is 
Generally the time taken for first distance is



The time taken for second distance is



The total time is mathematically represented as

=> 
=> 
Generally the constant velocity that would let her finish at the same time is mathematically represented as

=> 
=> 
The major shortcoming of Rutherford's model was that it was incomplete. It did not explain how the atom's negatively charged electrons are distrubuted in the space surronding its positively charged nucleus. A form of energy that exhibits wavelike behavior as it travels through space
Answer:
C. -12 ab
Explanation:
The restoring force on a spring is given by Hooke's law:

where
k is the spring constant
x is the stretched (or compressed) displacement of the spring
In this problem we have:
k = 4a
x = 3b
Substituting into the equation, we find:

And the negative sign means that the direction of the force (negative) is opposite to the direction of the displacement (positive).