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3 years ago

How many neutrons does element X have if its atomic number is 48 and its mass number is 167?

1 answer:

6 0

If its atomic number is 48, then it has 48 protons in the nucleus
of each atom. Any more mass than that is supplied by the neutrons
that are mixed in there with the protons.

If the mass is 167, and 48 of those are protons, then there are

(167 - 48) = 119 neutrons

in each nucleus.

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3 years ago

Part A

7nadin3 [17]

Answer:

2.5 m/s²

Explanation:

Using the formula, v = u + at ( v = Final velocity; u = Initial velocity; t = Time; a = Acceleration)

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2 years ago

A heat engine uses thermal energy that flows from a heat source with a temperature of 325 K to a cold sink, which has a temperat

Ilya [14]

The answer is "58.5"

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2 years ago

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What is everyone's take on time travel?

Andrew [12]

Answer:

Its not really possible I don't think. UNLESS! You fall into a manhole then find a wirling vortex in the sewers! : )

Explanation:

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2 years ago

A T-shirt cannon can shoot a 0.085 kg T-shirt at nearly 30 m/s. The T-shirt cannon has a mass of 33 kg. If the initial net momen

IgorLugansk [536]

Answer:

Approximately $0.077\; {\rm m\cdot s^{-1}}$ (assuming that external forces on the cannon are negligible.)

Explanation:

If an object of mass $m$ is moving at a velocity of $v$ , the momentum $p$ of that object would be $p = m\, v$ .

Momentum of the t-shirt:

$\begin{aligned} p(\text{t-shirt}) &= m(\text{t-shirt}) \, v(\text{t-shirt}) \\ &= 0.085\; {\rm kg} \times 30\; {\rm m \cdot s^{-1}} \\ &= 2.55 \; {\rm kg \cdot m \cdot s^{-1}} \end{aligned}$ .

If there is no external force (gravity, friction, etc.) on this cannon, the total momentum of this system should be conserved. In other words, if $p(\text{cannon})$ denote the momentum of this cannon:

$p(\text{t-shirt}) + p(\text{cannon}) = 0$ .

$p(\text{cannon}) = -p(\text{t-shirt}) = -2.55\; {\rm kg \cdot m \cdot s^{-1}}$ .

Rewrite $p = m\, v$ to obtain $v = (p / m)$ . Since the mass of this cannon is $m(\text{cannon}) = 33\; {\rm kg}$ , the velocity of this cannon would be:

$\begin{aligned} v(\text{cannon}) &= \frac{p(\text{cannon})}{m(\text{cannon})} \\ &= \frac{-2.55\; {\rm kg \cdot m \cdot s^{-1}}}{33\; {\rm kg}} \\ &\approx 0.077\; {\rm m \cdot s^{-1}}\end{aligned}$ .

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1 year ago