Explain why it takes much more effort to

Physics

1 answer:

natali 33 [55]2 years ago

6 0

Answer:

Mass Kinetic Energy and Jules

Explanation:

The train in question is big and heavy and a car is decently heavy but say a train moving at 55 mph can plow through a car and a car driving at 55mph driving at a train will be stopped dead in its tracks. This is because newtons laws of motion specifically an object in motion will stay in motion unless its opposed. The train also has a payload behind it meaning it hurts with force while a car doesn't have to much mass behind it. The train takes loner to stop for as it's acceleration as well as it's deceleration are very slow because its huge and takes a lot of force to stop it while a car is very centralized and compact when it comes to weight and its brakes are usually effective at stopping at 55 mph in about 2 to 6 seconds while a train might stay moving for a good 35 seconds. The force behind the train is immense for as even if the wheels don't spin at all the train will still move since the force behind it is great and a cars tires have a lot of grip and not a lot mass which plays into the force the car has so it can stop simply.

You might be interested in

In order for a person to "see" an object, light waves pass through the ___________.

spayn [35]

I think the correct answer from the choices listed above is the first option. In order for a person to "see" an object, light waves pass through the cornea. The cornea is the transparent layer forming at the front of the eye. Hope this answers the question. Have a nice day.

8 0

3 years ago

Read 2 more answers

Una placa de cobre a 20°C tiene unas dimensiones de 65cm x 78 cm. Encuentra el área de la placa a 400°C; Coeficiente de dilataci

ValentinkaMS [17]

Answer:

El área de la placa es aproximadamente 5102.752 centímetros cuadrados.

Explanation:

Asumamos que el cambio dimensional como consecuencia de la temperatura es pequeña, entonces podemos estimar el área de la placa de cobre en función de la temperatura mediante la siguiente aproximación:

$A_{f} = w\cdot l \cdot [1 + 2\cdot \alpha\cdot (T_{f}-T_{o})]$ (1)

Donde:

$w$ - Ancho de la placa, en centímetros.

$l$ - Longitud de la placa, en centímetros.

$\alpha$ - Coeficiente de dilatación, en $\frac{1}{^{\circ}C}$ .

$T_{o}$ - Temperatura inicial, en grados Celsius.

$T_{f}$ - Temperatura final, en grados Celsius.

Si sabemos que $w = 65\,cm$ , $l = 78\,cm$ , $\alpha = 17\times 10^{-6}\,\frac{1}{^{\circ}C}$ , $T_{o} = 20\,^{\circ}C$ and $T_{f} = 400\,^{\circ}C$ , entonces el área de la placa a la temperatura final: