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3 years ago

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A loaded gun is dropped on a frozen lake. The gun fires, with the bullet going horizontally in one direction and the gun sliding

on the ice in the other direction. The bullet's mass is 0.04 kg, and its speed is 285 m/s. If the gun's mass is 2.3 kg, what is its speed (in m/s)?

1 answer:

omeli [17]3 years ago

5 0

Answer:

4.96m/s

Explanation:

Using mv = m'v'

m = 0.04kg

v = 285 m/s

m' = 2.3kg

v'= ?

0.04×285 = 2.3v'

V' = 11.4/2.3

= 4.96m/s

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A 4.87-kg ball of clay is thrown downward from a height of 3.21 m with a speed of 5.21 m/s onto a spring with k = 1570 N/m. The

Yuki888 [10]

Answer:

Approximately $0.560\; {\rm m}$ , assuming that:

the height of $3.21\; {\rm m}$ refers to the distance between the clay and the top of the uncompressed spring.
air resistance on the clay sphere is negligible,
the gravitational field strength is $g = 9.81\; {\rm m\cdot s^{-2}}$ , and
the clay sphere did not deform.

Explanation:

Notations:

Let $k$ denote the spring constant of the spring.
Let $m$ denote the mass of the clay sphere.
Let $v$ denote the initial speed of the spring.
Let $g$ denote the gravitational field strength.
Let $h$ denote the initial vertical distance between the clay and the top of the uncompressed spring.

Let $x$ denote the maximum compression of the spring- the only unknown quantity in this question.

After being compressed by a displacement of $x$ , the elastic potential energy $\text{PE}_{\text{spring}}$ in this spring would be:

$\displaystyle \text{PE}_{\text{spring}} = \frac{1}{2}\, k\, x^{2}$ .

The initial kinetic energy $\text{KE}$ of the clay sphere was:

$\displaystyle \text{KE} = \frac{1}{2}\, m \, v^{2}$ .

When the spring is at the maximum compression:

The clay sphere would be right on top of the spring.
The top of the spring would be below the original position (when the spring was uncompressed) by $x$ .
The initial position of the clay sphere, however, is above the original position of the top of the spring by $h = 3.21\; {\rm m}$ .

Thus, the initial position of the clay sphere ( $h = 3.21\; {\rm m}$ above the top of the uncompressed spring) would be above the max-compression position of the clay sphere by $(h + x)$ .

The gravitational potential energy involved would be:

$\text{GPE} = m\, g\, (h + x)$ .

No mechanical energy would be lost under the assumptions listed above. Thus:

$\text{PE}_\text{spring} = \text{KE} + \text{GPE}$ .

$\displaystyle \frac{1}{2}\, k\, x^{2} = \frac{1}{2}\, m\, v^{2} + m\, g\, (h + x)$ .

Rearrange this equation to obtain a quadratic equation about the only unknown, $x$ :

$\displaystyle \frac{1}{2}\, k\, x^{2} - m\, g\, x - \left[\left(\frac{1}{2}\, m\, v^{2}\right)+ (m\, g\, h)\right] = 0$ .

Substitute in $k = 1570\; {\rm N \cdot m^{-1}}$ , $m = 4.87\; {\rm kg}$ , $v = 5.21\; {\rm m\cdot s^{-1}}$ , $g = 9.81\; {\rm m \cdot s^{-2}}$ , and $h = 3.21\; {\rm m}$ . Let the unit of $x$ be meters.

$785\, x^{2} - 47.775\, x - 219.453 \approx 0$ (Rounded. The unit of both sides of this equation is joules.)

Solve using the quadratic formula given that $x \ge 0$ :

$\begin{aligned}x &\approx \frac{-(-47.775) + \sqrt{(-47.775)^{2} - 4 \times 785 \times (-219.453)}}{2 \times 785} \\ &\approx 0.560\; {\rm m}\end{aligned}$ .

(The other root is negative and is thus invalid.)

Hence, the maximum compression of this spring would be approximately $0.560\; {\rm m}$ .

5 0

3 years ago

Could someone answer this?

Nadusha1986 [10]

Answer:

Here the circuit in which a 4Ω resistor resistor is connected in series and two 8Ω resistor resistors are connected in parallel. Also, ammeter and voltmeter connected in series and parallel circuit respectively.

Now,

The maximum power of each resistance is 16 W

The 4Ω resistor is linked in series with the circuit.

so, P o w e r = I

two

R, here i is the current through the resistor resistor R

1 6 = I

two

∗ 4 Ω

i = 2A

Now 2A passes through parallel resistors of 8Ω resistance.

we know that, in parallel, the potential difference must be constant,

the current is divided into two parts, because the same resistance current in each resistance will be half. then the current through each resistor in parallel is

2 A

two

.

= 1 A

So finally the current through the 4Ω resistor = 2 A

current through each 8Ω resistor = 1 A

Explanation:

I hope this answer has helped you

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What is the oxidation number of the element Se

maria [59]

Answer:

The oxidation number of a monatomic (composed of one atom) ion is the same as the charge of the ion. For example, the oxidation numbers of K+, Se2−, and Au3+ are +1, -2, and +3, respectively. The oxidation number of oxygen in most compounds is −2.

Explanation:

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Answer:

stars are made of matter

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Nat2105 [25]

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4 years ago

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