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Leno4ka [110]
3 years ago
13

In an intense battle, gunfire can so concentrated that bullets from opposite sides collide in midair. Suppose that a bullet (wit

h mass ????=5.12 g moving to the right at a speed ????=208 m/s directed 21.3∘ above the horizontal) collides and fuses with another with mass m=3.05 g moving to the left at a speed ????=282 m/s directed 15.4∘ above the horizontal.
Physics
1 answer:
Vesnalui [34]3 years ago
6 0

Answer:

velocity

a). v_{f}=235.2 (m/s)

direction

b).\beta =71.3

kinetic energy

c).E_{K}=225.97N

Explanation:

Those kind of question have two typical questions inside to response the final velocity and the direction finding the angle and finally the kinetic energy so each the information you need can find there so

a).

m_{b}*v_{b}+m_{t}*v_{t}=(m_{b}+m_{t})*v_{f}

solve to vf

v_{fx}=\frac{m_{b}*v_{b}*cos(21.3)+m_{t}*v_{t}*cos(15.4)}{m_{b}+m_{t}}=\frac{5.12x10^{-3}kg*208(m/s)0.931+3.05x10^{-3}kg*228(m/s)*0.964}{(5.12x10^{-3}+3.05x10^{-3}kg)}

v_{fx}=222.94(m/s)

v_{fy}=\frac{m_{b}*v_{b}*sin(21.3)+m_{t}*v_{t}*sin(15.4)}{m_{b}+m_{t}}=\frac{5.12x10^{-3}kg*208(m/s)0.363+3.05x10^{-3}kg*228(m/s)*0.265}{(5.12x10^{-3}+3.05x10^{-3}kg)}

v_{fy}=75.21(m/s)

V_{f}=\sqrt{V_{fx}^2+V_{fy}}=\sqrt{222.94^2+75.21^2}

V_{f}=235.28\frac{m}{s}

b).

Tan(\beta)=\frac{v_{fx}}{v_{fy}}

solve to β

\beta =tan^-1*(\frac{222.94}{75.4})

\beta=tan^{-1}*2.956

\beta =71.3

c).

Final kinetic energy

E_{k}=\frac{1}{2}m_{t}*V_{f}^2

E_{k}=\frac{1}{2}(5.12x10^{-3}+3.05x10^{-3})kg*(235.2m/s)^2

E_{K}=225.97N

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Explanation:

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3 years ago
Analogue signals transmit information for such things as _____________.
ivann1987 [24]

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AM and FM radio are an interesting subject.  They're all still analog.
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7 0
3 years ago
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Water has a specific heat of 4.184
LuckyWell [14K]

Answer:

Water.

Explanation:

This means:

1) For the temperature of water to raise at any point to the next degree by 1°C, will require a specific heat capacity of  4.184  J/Kg°C

2) For the temperature of wood to raise at any point to the next degree by 1°C, will require a specific heat capacity of  1.760  J/Kg°C

Note that: specific heat is directly proportional to energy, therefore the higher the heat capacity, the higher the energy.

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8 0
3 years ago
A V = 108-V source is connected in series with an R = 1.1-kΩ resistor and an L = 34-H inductor and the current is allowed to rea
soldi70 [24.7K]

Answer:

Explanation:

Given an RL circuit

A voltage source of.

V = 108V

A resistor of resistance

R = 1.1-kΩ = 1100 Ω

And inductor of inductance

L = 34 H

After he inductance has been fully charged, the switch is open and it connected to the resistor in their own circuit, so as to discharge the inductor

A. Time the inductor current will reduce to 12% of it's initial current

Let the initial charge current be Io

Then, final current is

I = 12% of Io

I = 0.12Io

I / Io = 0.12

The current in an inductor RL circuit is given as

I = Io ( 1—exp(-t/τ)

Where τ is time constant and it is given as

τ = L/R = 34/1100 = 0.03091A

So,

I = Io ( 1—exp(-t/τ))

I / Io = ( 1—exp(-t/τ))

Where I/Io = 0.12

0.12 = 1—exp(-t/τ)

0.12 — 1 = —exp(-t/τ)

-0.88 = -exp(-t/0.03091)

0.88 = exp(-t/0.03091)

Take In of both sides

In(0.88) = In(exp(-t/0.03091)

-0.12783 = -t/0.030901

t = -0.12783 × 0.030901

t = 3.95 × 10^-3 seconds

t = 3.95 ms

B. Energy stored in inductor is given as

U = ½Li²

So, the current at this time t = 3.95ms

I = Io ( 1—exp(-t/τ))

Where Io = V/R

Io = 108/1100 = 0.0982 A

Now,

I = Io ( 1—exp(-t/τ))

I = 0.0982(1 — exp(-3.95 × 10^-3 / 0.030901))

I = 0.0982(1—exp(-0.12783)

I = 0.0982 × 0.12

I = 0.01178

I = 11.78mA

Therefore,

U = ½Li²

U = ½ × 34 × 0.01178²

U = 2.36 × 10^-3 J

U = 2.36 mJ

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Answer:

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