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3 years ago

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A bullet with momentum of 2.8 kg·m/s [E] is traveling at a speed of 187 m/s. The mass of the bullet is: a) 67 g b) 15 g c) 0.067

g d) 0.015 g

1 answer:

Vesna [10]3 years ago

4 0

Answer:

The mass of the bullet is 15 g.

(b) is correct option.

Explanation:

Given that,

Momentum = 2.8 kg m/s

Speed = 187 m/s

We need to calculate the mass of the bullet

Using formula of momentum

$P = mv$

$m = \dfrac{P}{v}$

Where, P = momentum

v = speed

Put the value into the formula

$m = \dfrac{2.8}{187}$

$m = 0.015\ kg$

$m = 15\ g$

Hence, The mass of the bullet is 15 g.

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3 years ago

A 1.00 kg particle has the xy coordinates (-1.20 m, 0.500 m) and a 4.50 kg particle has the xy coordinates (0.600 m, -0.750 m).

just olya [345]

Answer:

a) The x coordinate of the third mass is -1.562 meters.

b) The y coordinate of the third mass is -0.944 meters.

Explanation:

The center of mass of a system of particles ( $\vec r_{cm}$ ), measured in meters, is defined by this weighted average:

$\vec r_{cm} = \frac{\Sigma_{i=1}^{n}\,m_{i}\cdot \vec r_{i}}{\Sigma_{i=1}^{n}\,m_{i}}$ (1)

Where:

$m_{i}$ - Mass of the i-th particle, measured in kilograms.

$\vec r_{i}$ - Location of the i-th particle with respect to origin, measured in meters.

If we know that $\vec r_{cm} = (-0.500\,m,-0.700\,m)$ , $m_{1} = 1\,kg$ , $\vec r_{1} = (-1.20\,m, 0.500\,m)$ , $m_{2} = 4.50\,kg$ , $\vec r_{2} = (0.600\,m, -0.750\,m)$ and $m_{3} = 4\,kg$ , then the coordinates of the third particle are:

$(-0.500\,m, -0.700\,m) = \frac{(1\,kg)\cdot (-1.20\,m,0.500\,m)+(4.50\,kg)\cdot (0.600\,m,-0.750\,m)+(4\,kg)\cdot \vec r_{3}}{1\,kg+4.50\,kg+4\,kg}$

$(-4.75\,kg\cdot m, -6.65\,kg\cdot m) = (-1.20\,kg\cdot m, 0.500\,kg\cdot m) + (2.7\,kg\cdot m, -3.375\,kg\cdot m) +(4\cdot x_{3},4\cdot y_{3})$

$(4\cdot x_{3}, 4\cdot y_{3}) = (-6.25\,kg\cdot m,-3.775\,kg\cdot m)$

$(x_{3},y_{3}) = (-1.562\,m,-0.944\,m)$

a) The x coordinate of the third mass is -1.562 meters.

b) The y coordinate of the third mass is -0.944 meters.

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3 years ago

The volume occupied by a sample of gas is 480 mL when the pressure is 115 kPa.What pressure must be applied to the gas to make i

balandron [24]

Answer:

The answer is

<h2>84.9 kPa</h2>

Explanation:

Using Boyle's law to find the final pressure

That's

$P_1V_1 = P_2V_2$

where

P1 is the initial pressure

P2 is the final pressure

V1 is the initial volume

V2 is the final volume

Since we are finding the final pressure

$P_2 = \frac{P_1V_1}{V_2}$

From the question

P1 = 115 kPa

V1 = 480 mL

V2 = 650 ml

So we have

$P_2 = \frac{115000 \times 480}{650} = \frac{55200000}{650} \\ = 84923.076923...$

We have the final answer as

<h3>84.9 kPa</h3>

Hope this helps you

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