Answer:
1) The force Christian can exert on his bicycle before picking up the the cargo is 529.74 N
2) The force Christian can exert on his bicycle after picking up the the cargo is 647.46 N
Therefore, Christian has to exert more force on his bike after picking up the cargo
Explanation:
The given parameters are;
The mass of Christian and his bicycle = 54 kg
The mass of the cargo = 12 kg
1) The force Christian can exert on his bicycle before picking up the the cargo = Mass of Christian and his bicycle × Acceleration due to gravity
∴ The force Christian can exert on his bicycle before picking up the the cargo = 54 kg × 9.81 m/s² = 529.74 N
2) The force Christian can exert on his bicycle after picking up the the cargo = (54 + 12) kg × 9.81 m/s² = 647.46 N
Therefore, Christian has to exert more force on his bike after picking up the cargo.
Answer:
The answer is A
Explanation:
When a rockets thrusters push on the ground the ground pushes back on the rocket with equal force in the opposite direction. Hence the rocket takes off.
Newtons third law of motion states, for every action there is an equal and opposite reaction.
Answer:

Explanation:
The heaviside function is defined as:

so we see that the Heaviside function "switches on" when
, and remains switched on when 
If we want our heaviside function to switch on when
, we need the argument to the heaviside function to be 0 when 
Thus we define a function f:

The
term inside the heaviside function makes sure to displace the function 5 units to the right.
Now we just need to add a scale up factor of 240 V, because thats the voltage applied after the heaviside function switches on. (
when
, so it becomes just a 1, which we can safely ignore.)
Therefore our final result is:

I have made a sketch for you, and added it as attachment.
300N/25 kg= divide them for the answer
Answer:
a) -2.038 m/s²
b) 40.33 mph
c) 312.5 m
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration

Acceleration of the boat is -2.083 m/s² if the boat will stop at 150 m.

Speed of the boat by when it will hit the dock is 18.03 m/s
Converting to mph



Speed of the boat by when it will hit the dock is 40.33 mph

The distance at which the boat will have to start decelerating is 312.5 m