Displacement

Three guns are aimed at the center of a circle, and each fires a bullet simultaneously. The directions in which they fire are 12

ikadub [295]

Answer:

The unknown mass of the bullet is $m1=3.751x10^{-3} kg$

Explanation:

According to Newton's laws of motion, when a net external force acts on a body of mass <u><em>m</em></u> , it results in change in momentum of the body and is given by:

$F=\frac{P}{dt}$

Where:

P

is the linear momentum of the body

As a consequence, when there are no external forces acting on the body the total momentum remains conserved i.e.

Given:

$m_{2}=5.79x10^{-3}kg \\m_{3}=5.79x10^{-3}kg\\v_{2}=v_{3}=392 \frac{m}{s}\\$

For momentum along the y-direction to be zero, it is achieved when the equal masses are moving at angles of

θ1=180°, θ2=60°, θ3=-60°

Therefore, from conservation of momentum along x - direction:

$m_{1}*v_{1}*cos(180)+m_{2}*v_{2}*cos(60)+m_{3}*v_{3}*cos(-60)=0\\$ $m_{1}*605\frac{m}{s}*cos(180)+5.79x10^{-3}kg *392\frac{m}{s}*cos(60)+5.79x10^{-3}kg*392\frac{m}{s}*cos(-60)=0\\$

$-m_{1}*605+5.79x10^{-3}kg*196\frac{m}{s}+5.79x10^{-3}kg*196\frac{m}{s}=0\\m_{1}*605kg= 2.26968\frac{kg*m}{s}\\m_{1}=3.75 x10^{-3} kg$

3 0

3 years ago

Sean pushes a box with a force of 100 N. Christian pushes an identical box with a force of

kap26 [50]

Answer:

Christian

Explanation:

Given that

Force applied by Sean ,F₁=100 N

Force applied by Christian ,F₂=200 N

Lets take mass of the box = m kg

As we know that from second law of Newton's

F= m a

F=Force,m=mass ,a=acceleration

$F_1=ma_1$

$100=m\times a_1$

$a_1=\dfrac{100}{m}\ m/s^2$

$F_2=ma_2$

$200=m\times a_2$

$a_2=\dfrac{200}{m}\ m/s^2$

From the above we can say that acceleration a₂ is greater than a₁ is for the same mass of the box.

Therefore Christian will acceleration more for the box.

5 0

3 years ago

A que profundidad esta nadando una persona dentro de una alberca si la presión absoluta sobre ésta es de 156kPa?

lara31 [8.8K]

Answer:

La persona está nadando en la alberca a una profundida de 5.575 metros.

Explanation:

La presión absoluta ( $P_{tot}$ ) experimentada por la persona es la suma de la presión atmosférica ( $P_{atm}$ ) y la presión hidrostática de la columna de agua de la alberca ( $P_{h}$ ), medidas en kilopascales. Es decir,

$P_{tot} = P_{atm}+P_{h}$ (1)

$P_{tot} = P_{atm} + \frac{\rho\cdot g \cdot z}{1000}$ (2)

Donde:

$\rho$ - Densidad del fluido de la alberca, medida en kilogramos por metro cúbico.

$g$ - Aceleración gravitacional, medida en metros por segundo al cuadrado.

$z$ - Profundidad de la persona en la alberca, medida en metros.

Si sabemos que $P_{atm} = 101.325\,kPa$ , $\rho = 1000\,\frac{kg}{m^{3}}$ , $g = 9.807\,\frac{m}{s^{2}}$ y $P_{tot} = 156\,kPa$ , entonces la profundidad de la persona en la alberca es: