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3 years ago

8

When a mortar shell is ﬁred with an initial

1 answer:

Shtirlitz [24]3 years ago

5 0

The equation y = Vo cos(alfa) t - 16t^2 implies that the mortar landed at the same level that it was fire and that fire angle is also 75°.

With that said, let us work on the parametric equations

y = 0 = Vo sin(alfa) t - 16t^2, which by factoring =>

t (Vo sin(alfa) - 16t) = 0 => t = Vo sin(alfa) / 16 .....(1)

x = 2500 = Vo cos(alfa) t => t = 2500 / [Vo cos(alfa)] ..... (2)

Now make (1) equal to (2)

Vo sin(alfa) / 16 = 2500 / [Vo cos(alfa)] =>

Vo^2 sin(alfa)cos(alfa) = 2500ft*16ft/s^2 =>

Vo^2 = 2500*16 / [sin(75°)cos(75)] = 2500*16/0.25 = 160,000 ft^2/s^2

Vo = √(160,000) ft/s = 400 ft/s

Answer: option 5. 400 ft/s

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A very long uniform line of charge has charge per unit length λ1 = 4.68 μC/m and lies along the x-axis. A second long uniform li

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Answer:

$E_{net} = 6.44 \times 10^5 N/C$

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As we know that electric field due to infinite line charge distribution at some distance from it is given as

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now we need to find the electric field at mid point of two wires

So here we need to add the field due to two wires as they are oppositely charged

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$E_{net} = \frac{2k\lambda_1}{r} + \frac{2k\lambda_2}{r}$

now plug in all data

$\lambda_1 = 4.68 \muC/m$

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now we have

$E_{net} = \frac{2k}{r}(4.68 + 2.48)$

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A 150 kg boy and his bike are traveling 12 m/s when he slams on his breaks and stop at his friend’s house. How much impulse is r

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Answer:

J = 1800 kg-m/s

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Given that,

Mass of a boy, m = 150 kg

Initial velocity of a boy, u = 12 m/s

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We need to find the impulse is required to produce this change in momentum. We know that impulse is equal to the change in momentum. So,

$J=m(v-u)\\\\=150\times (0-12)\\\\=-1800\ kg-m/s\\\\|J|=1800\ kg-m/s$

So, the impulse is equal to 1800 kg-m/s

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It is found that the most probable speed of molecules in a gas at equilibrium temperature

kaheart [24]

Answer:

$\frac{T_2}{T_1} = 1$

Explanation:

The root mean square velocity of the gas at an equilibrium temperature is given by the following formula:

$v = \sqrt{\frac{3RT}{M} }$

where,

v = root mean square velocity of molecules:

R = Universal Gas Constant

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M = Molecular Mass of the Gas

Therefore,

For T = T₁ :

$v = \sqrt{\frac{3RT_1}{M} }$

For T = T₂ :

$v = \sqrt{\frac{3RT_2}{M} }$

Since both speeds are given to be equal. Therefore, comparing both equations, we get:

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