Answer:
The sled needed a distance of 92.22 m and a time of 1.40 s to stop.
Explanation:
The relationship between velocities and time is described by this equation:
, where
is the final velocity,
is the initial velocity,
the acceleration, and
is the time during such acceleration is applied.
Solving the equation for the time, and applying to the case:
, where
because the sled is totally stopped,
is the velocity of the sled before braking and,
is negative because the deceleration applied by the brakes.
In the other hand, the equation that describes the distance in term of velocities and acceleration:
, where
is the distance traveled,
is the initial velocity,
the time of the process and,
is the acceleration of the process.
Then for this case the relationship becomes:
.
<u>Note that the acceleration is negative because is a braking process.</u>
QUESTION: A pure jet engine propels and aircraft at 340 m/s through air at 45 kPa and -13C. The inlet diameter of this engine is 1.6 m, the compressor pressure ratio is 13, and the temperature at the turbine inlet is 557C. Determine the velocity at the exit of this engines nozzle and the thrust produced.
ANSWER: Due to the propulsion from the inlet diameter of this engine bring 1.6 m allows the compressor rations to radiate allowing thrust propultion above all velocitic rebisomes.
Answerana alyom kint gahda amshi
Explanation:
I can't make sense of this question. Julie's throwing the ball, so it's leaving her rather than arriving at her ???
Answer:
The rate of heat conduction through the layer of still air is 517.4 W
Explanation:
Given:
Thickness of the still air layer (L) = 1 mm
Area of the still air = 1 m
Temperature of the still air ( T) = 20°C
Thermal conductivity of still air (K) at 20°C = 25.87mW/mK
Rate of heat conduction (Q) = ?
To determine the rate of heat conduction through the still air, we apply the formula below.


Q = 517.4 W
Therefore, the rate of heat conduction through the layer of still air is 517.4 W