Physics, 21.12.2019 01:31 zyrashanson

A2150 kg car, moving east at 10 m/s, collided and joins with a 3250 kg car. the cars move east together at 5.22 m/s. what is the 3250 kg car's initial velocity calculated to the nearest tenth?

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Answer from: ssssss190

The second car was moving at 2.1 m/s to the east

Explanation:

Conservation Of Linear Momentum

The total momentum of a system of bodies is constant if no external force is applied to it. The momentum of a body with mass m and velocity v is p=mv. When two objects collide and join together afterward, the total mass is the sum of the individual masses, and the final speed is common to both.

Let's say

$\displaystyle m_1,\ m_2\ ,\ v_1,\ v_2\ ,\ p_1,\ p_2$

are the masses of the objects 1 and 2, their speeds, and their linear momentums respectively before the collision, and

$\displaystyle v_1',\ v_2'\ ,\ p_1',\ p_2'$

are the speeds of each object and their linear momentums after the collision.

The principle of conservation of linear momentum states that

$\displaystyle p_1+p_2=p_1'+p_2'$

This means that

$\displaystyle m_1\ v_1+m_2\ v_2=m_1\ v_1'+m_2\ v_2'$

Since both cars remain together after the collision,

$\displaystyle v_1'=v_2'=v'$

The relation becomes

$\displaystyle m_1\ v_1+m_2\ v_2=(m_1+m_2)\ v'$

Solving for v_2

$\displaystyle v_2=\frac{(m_1+m_2)v'-m_1\ v_1}{m_2}$

The given data is

$\displaystyle m_1=2150\ kg\ ,\ m_2=3250\ kg\ ,\ v_y=10\ m/s,\ v'=5.22\ m/s$

Let's assume all the speeds are positive towards East

Replacing those values

$\displaystyle v_2=\frac{(2150+3250)5.22-2150(10)}{3250}$

$\displaystyle v_2=\frac{6688}{3250}=2.06\ m/s$

The second car was moving at 2.1 m/s (to the nearest tenth) to the east

Answer from: Quest

electromagnetic shielding is the practice of reducing the electromagnetic field in a space by blocking the field with barriers made of conductive or magnetic materials. shielding is typically applied to enclosures to isolate electrical devices from their surroundings, and to cables to isolate wires from the environment through which the cable runs. electromagnetic shielding that blocks radio frequency electromagnetic radiation is also known as rf shielding.

the shielding can reduce the coupling of radio waves, electromagnetic fields and electrostatic fields. a conductive enclosure used to block electrostatic fields is also known as a faraday cage. the amount of reduction depends very much upon the material used, its thickness, the size of the shielded volume and the frequency of the fields of interest and the size, shape and orientation of apertures in a shield to an incident electromagnetic

from wikipedia

Answer from: Quest

For the bullet, $m = 19 \text{ g} \cdot \dfrac{1\text{ kg}}{1000 \text{ g}} = 0.019\text{ kg}$ the net force on the bullet: $\sigma \vec{f} = \vec{f}_f = 50 000 \text{ n}$ an fbd would look like a dot with a horizontal "friction" force protruding from one end.by newton's 2nd law: $\sigma \vec{f} = m\vec{a}$ we can solve for acceleration since we know $\sigma \vec{f} = 50000\text{ n}$ and $m = 0.019\text{ kg}$ . $\begin{aligned} \sigma \vec{f} & = m\vec{a}\\ \vec{a} & = \dfrac{ \sigma \vec{f} }{m} \\ \vec{a} & = \dfrac{50000\text{ n}}{0.019\text{ kg}} \\ \vec{a} & = 2631578.95 \text{ m/s}^2 \end{aligned}$ we know that $v_f = 0\text{ m/s}$ (final velocity is bullet not moving), $v_i = 334\text{ m/s}$ . we can use the following kinematic equation to get $d$ , or the displacement. $\begin{aligned} v_f^2 & = v_i^2 + 2ad \\ v_f^2 - v_i^2 & = 2ad \\ d & = \dfrac{v_f^2 - v_i^2}{2a} \\ d & = \dfrac{0 - (334\text{ m/s})^2}{2\cdot 2631578.95 \text{ m/s}^2} \\ d & = -0.02119563997 \text { m} \\ d & = -0.02 \text { m} & & \text{1 significant figure} \end{aligned}$ so it penetrates $0.02$ m.

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A2150 kg car, moving east at 10 m/s, collided and joins with a 3250 kg car. the cars move east toget...

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