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Rus_ich [418]
2 years ago
11

If a students clothing catches fire the student should

Physics
1 answer:
Roman55 [17]2 years ago
4 0

Answer:

stop, drop and roll.

Explanation:

This is because rolling on the ground can help put out the fire by depriving it of oxygen.

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A small object carrying a charge of -4.00 nC is acted upon by a downward force of 24.0 nN when placed at a certain point in an e
PIT_PIT [208]

Answer:

E  = -6 \  N/C

Generally given that the electric field is negative it mean that its direction is opposite to that of the force    

Explanation:

From the question we are told that

   The charge on the small object is Q = -4.00 \ nC =  -4.00 *10^{-9} \  C

   The force is  F  =  24  \ nN  =  24 *10^{-9} \  N

    Generally the magnitude of the electric  field is mathematically represented as

       E  =  \frac{F}{Q}

=>    E  =  \frac{ 24 *10^{-9}} {-4 *10^{-9 }}

=>     E  = -6 \  N/C

Generally given that the electric field is negative it mean that its direction is opposite to that of the force    

 

6 0
3 years ago
What is definition of population?
Oksanka [162]
All the inhabitants of a particular place or town, area or country
5 0
3 years ago
A 0.50-kg red cart is moving rightward with a speed of 40 cm/s when it collides
lukranit [14]

The momentum of the red cart before the collision is 0.2 kgm/s and the blue cart is 0.

The momentum of the red cart after the collision is 0.05 kgm/s and the blue cart is 0.15 kgm/s.

The change in momentum of the system of the carts is 0.

<h3>Initial momentum of the carts before collision</h3>

The momentum of the carts before the collision is calculated as follows;

P(red) = 0.5 kg x 0.4 m/s = 0.2 kgm/s

P(blue) = 1.5 x 0 = 0

<h3>Momentum of the carts after collision</h3>

The momentum of the carts after the collision is calculated as follows;

P(red) = 0.5 x 0.1 = 0.05 kgm/s

P(blue) = 1.5 0.1 = 0.15 kgm/s

<h3>Change in momentum of the carts</h3>

\Delta P = P_f - P_i

ΔP = (0.05 + 0.15) - (0.2)

ΔP = 0

Learn more about momentum here: brainly.com/question/7538238

8 0
2 years ago
Se deja caer una moneda desde cierta altura. Si se desprecian los efectos del aire, ¿cómo varía la fuerza neta sobre la moneda a
forsale [732]

Answer:

Ok, primero pensemos en una situación normal.

La moneda comienza a caer, pero la moneda esta inmersa en una sustancia, el aire. El aire comienza a aplicar una resistencia al movimiento de la moneda, y esta resistencia incremente a medida que la velocidad de la moneda incremente. Llega un punto en el que esta nueva fuerza es igual a la fuerza gravitatoria, y en sentido opuesto, lo que causa que la fuerza neta sea 0, y que la moneda caiga a velocidad constante hasta que esta impacta con el suelo.

Ahora, en este caso tenemos que ignorar los efectos del aire, entonces no hay ninguna fuerza que se oponga a la fuerza gravitatoria, entonces la fuerza neta no cambia a medida que cae (La fuerza neta cambia cuando la moneda impacta el suelo).

También se puede analizar el caso en el que, como la fuerza gravitatoria decrece con el radio al cuadrado, a medida que la moneda cae, la fuerza gravitatoria incrementa. El tema es que en para estas dimensiones, ese cambio en la fuerza gravitacional es imperceptible,

3 0
3 years ago
Please help with the working of this question
Vilka [71]

Answer:

Principle of conservation of linear momentum is state as provided no external force acts on a system of colliding bodies the total linear momentum of the bodies (in a given direction) remains constant

8 0
3 years ago
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