Answer:
the answer might be the first one A. safe
If a car crashes into another car like this, the wreck should go nowhere. Besides this being an unrealistic question, the physics of it would look like this:
Momentum before and after the collision is conserved.
Momentum before the collision:
p = m * v = 50000kg * 24m/s + 55000kg * 0m/s = 50000kg * 24m/s
Momentum after the collision:
p = m * v = (50000kg + 55000kg) * v
Setting both momenta equal:
50000kg * 24m/s = (50000kg + 55000kg) * v
Solving for the velocity v:
v = 50000kg * 24m/s/(50000kg + 55000kg) = 11,43m/s
The heat flows into my body and I start with a warming feeling going down my throat as it starts there I get this warm feeling and I break out in a coldsweat from consuming a warm drink a cold drink will help cool me down :
Explanation:
If drink warm will make u warm drinking cool will make u cool down faster
Answer:
The gauge pressure is 
Explanation:
From the question we are told that
The height of the water contained is 
The height of liquid in the cylinder is 
At the bottom of the cylinder the gauge pressure is mathematically represented as

Where
is the pressure of water which is mathematically represented as

Now
is the density of water with a constant values of 
substituting values


While
is the pressure of oil which is mathematically represented as

Where
is the density of oil with a constant value

substituting values


Therefore


Answer:
1.25 m
Explanation:
From the question given above, the following data were obtained:
Force ratio = 2.5
Distance of load from the fulcrum = 0.5 m
Distance of effort =.?
The distance of the effort from the fulcrum can be obtained as illustrated below:
Force ratio = Distance of effort / Distance of load
2.5 = Distance of effort / 0.5
Cross multiply
Distance of effort = 2.5 × 0.5
Distance of effort = 1.25 m
Therefore, the distance of the effort from the fulcrum is 1.25 m