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3 years ago

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A 0.050 kg bullet strikes a 5.0 kg wooden block with a velocity of 909 m/s and embeds itself in the block which fies off its sta

nd. what was the final velocity of the bullet?

1 answer:

serg [7]3 years ago

5 0

Answer:

The final velocity of the bullet is 9 m/s.

Explanation:

We have,

Mass of a bullet is, m = 0.05 kg

Mass of wooden block is, M = 5 kg

Initial speed of bullet, v = 909 m/s

The bullet embeds itself in the block which flies off its stand. Let V is the final velocity of the bullet. The this case, momentum of the system remains conserved. So,

$mv=(m+M)V\\\\V=\dfrac{mv}{m+M}\\\\V=\dfrac{0.05\times 909}{0.050+5}\\\\V=9\ m/s$

So, the final velocity of the bullet is 9 m/s.

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Two velcro-covered pucks slide across the ice, collide and stick to one another. Their interaction with the ice is frictionless.

balu736 [363]

Answer:

<em>1. False</em>

<em>2. True</em>

<em>3. False</em>

<em>4. True</em>

Explanation:

<u>Conservation of Momentum</u>

According to the law of conservation of linear momentum, the total momentum of the system formed by both pucks won't change regardless of their interaction if no external forces are acting on the system.

The momentum of an object of mass ma moving at speed va is

$p_a=m_a.v_a$

The total momentum of both pucks at the initial condition is

$p_1=m_a.v_a+m_b.v_b$

Both pucks are moving to the right and puck B has twice the mass of puck A (let's call it m), thus

$m_a=m$

$m_b=2m$

We are given

$v_a=6\ m/s\\v_b=2\ m/s$

The total initial momentum is

$p_1=6m+2(2m)=10m$

At the final condition, both pucks stick together, thus the total mass is 3m and the final speed is common, thus

$p_2=3m.v'$

Equating the initial and final momentum

$10m=3m.v'$

Solving for v'

$v'=10/3\ m/s=3.33\ m/s$

1. Compute the initial kinetic energy:

$\displaystyle K_1=\frac{1}{2}mv_a^2+\frac{1}{2}2mv_b^2$

$\displaystyle K_1=\frac{1}{2}m\cdot 6^2+\frac{1}{2}2m\cdot 2^2$

$K_1=18m+4m=22m$

The final kinetic energy is

$\displaystyle K_2=\frac{1}{2}mv'^2+\frac{1}{2}2mv'^2$

$\displaystyle K_2=\frac{1}{2}m\cdot 3.33^2+\frac{1}{2}2m\cdot 3.33^2$

$K_2=16.63m$

As seen, part of the kinetic energy is lost in the collision, thus the statement is False

2. The initial speed of puck B was 2 m/s and the final speed was 3.33 m/s, thus it increased the speed: True

3. The initial speed of puck A was 6 m/s and the final speed was 3.33 m/s, thus it decreased the speed: False

4. The momentum is conserved since that was the initial assumption to make all the calculations. True

$p_1=10m$

$p_2=3m.v'=3m(10/3)=10m$

Proven

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3 years ago

What is it called when objects become positively charged when they lose electrons and negatively charged when they gain electron

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It is c: electric charge

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3 years ago

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A person pushes on a box with a force of 500N. If the box does not move, what is the force of static friction on the box?

Alexxx [7]

Answer:

c

Explanation:

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2 years ago

An LC circuit consists of a 3.4-µF capacitor and a coil with a self-inductance 0.080 H and no appreciable resistance. At t = 0 t

alexira [117]

Answer

given,

capacitance = C = 3.4-µF

inductance = L = 0.08 H

frequency is expressed as

$f = \dfrac{1}{2\pi\sqrt{LC}}$

time period

$T = \dfrac{1}{f}=2\pi\sqrt{LC}$

after time T/4 current reach maximum

$t = \dfrac{T}{4}$

$t = \dfrac{2\pi\sqrt{LC}}{4}$

$t = \dfrac{2\pi\sqrt{0.08 \times 3.4 \times 10^{-6}}}{4}$

t = 8.2 x 10⁻⁴ s

t = 0.82 ms

b) using law of conservation

$\dfrac{1}{2}CV^2=\dfrac{1}{2}LI^2$

$I^2 = \dfrac{CV^2}{L}$

$I^2 = \dfrac{C}{L}\dfrac{Q^2}{C^2}$

$I =\sqrt{\dfrac{Q^2}{CL}}$

$I =\sqrt{\dfrac{(5.4 \times 10^{-6})^2}{0.08 \times 3.4 \times 10^{-6}}}$

I = 0.010 A

I = 10 mA

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3 years ago

You are loading a refrigerator weighing 2267 N onto a truck, using a wheeled cart. The refrigerator is raised 1.09 m to the truc

meriva

Answer:

a).

Wmin= 2471.03 J

W=1603.01 J

Explanation:

Weight, w= 2267 N

$w= m*g\\W=m*g*h \\W=w*h$

Minimum work 'h' is the distance the refrigerator is raised h=1.09m

$W_{min}=2267 N* 1.09m\\W_{min}=2471.03 J$

The motion is no frictional force so, the magnitude of the force with a angle of 45.0° is find using:

$W=m*g*h'\\h'= sin(45)\\W=2267N*sin(45)\\W=1603.01 J$

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2 years ago