This implies that stopping distance and impact force grow as a function of speed. The best ways to improve manoeuvrability and lessen crash severity are to drive at an appropriate pace and to slow down as soon as you spot dangers in front of you.
Keep in mind that stopping distance increases with speed; at 50 mph, it is four times longer than at 25 mph, and at 75 mph, the force of impact is nine times greater.
<h3>What is the impact of speed on kinetic energy ?</h3>
When your car expends or absorbs energy to speed up or slow down, you may feel a pull or a jolt, called impulse. Impulse increases as the energy or force increases, and increases as the duration of the force decreases. You'll feel a harder jolt if you speed up or slow down suddenly.
- Consider: coming to a stop from 60 mph in ten seconds doesn't hurt you or your vehicle because the force of this event is spread out over a long time. But if you hit a wall and come to a stop in just half a second, you'll feel twenty times the impulse, causing severe damage.
Learn more about Kinetic energy here:
brainly.com/question/25959744
#SPJ4
Answer:
Día Internacional del Deporte para el Desarrollo y la Paz, 6 de abril. El deporte ha desempeñado históricamente un papel importante en todas las sociedades, ya sea en forma de competiciones deportivas, de actividades físicas o de juegos. ... De hecho, el deporte es un socio natural para el sistema de la ONU.
Explanation:
Answer:
15.13 m/s
Explanation:
The wave speed of the stretched rope can be calculated using the following formula
where 
is the tension on the rope and 
is the density of the rope per unit length
Answer:
0.267 × 10⁻³
Explanation:
Given:
maximum permissible workday dose = 18 mrem
Weight of the laboratory technician = 54 kg
gamma rays absorbed = 2.6 mJ of 0.30 MeV
relative biological effectiveness (RBE) for gamma rays = 1.00
Now,
Radiation Absorbed Dose 
also,
Roentgen Equivalent Man = rad × Q
here Q is the quality factor = 1 for gamma rays
thus,
Roentgen Equivalent Man = 4.8 × 10⁻³ × 1 = 4.8 × 10⁻³ rem
Therefore,
the ratio of the equivalent dosage received by the technician to the maximum permissible equivalent dosage will be
= 
or
= 0.267 × 10⁻³
