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Verdich [7]
2 years ago
11

Help me <3 please Thank you :)

Physics
1 answer:
NARA [144]2 years ago
4 0

Answer:

11,890

Explanation:

First we need to know what is considered a significant figure.

A significant figure is a value that is not a zero at the start OR end of a value.

Which means, the 0 in the value of 90 or 0.363 are not considered a significant figure.

The 0 in the value of 3056 is considered a significant figure.

So from the table, we can deduce:

0.275 has 3 significant figures

750 has 2 significant figures

10.4 \times  {10}^{5}  = 1040000

has 3 significant figures.

11,890 has 4 significant figures.

320,050 has 5 significant figures.

So from the above, we can already see the answer.

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You place a book on top of a spring and push down, compressing the spring by 10 cm. When you let go of the book, it is pushed up
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What is the potential energy of a 45 kg object resting on the ground?
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Answer:The potential energy is zero

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3 years ago
A small object with momentum 7.0 kg∙m/s approaches head-on a large object at rest. The small object bounces straight back with a
EastWind [94]

Answer:

The magnitude of the large object's momentum change is 3 kilogram-meters per second.

Explanation:

Under the assumption that no external forces are exerted on both the small object and the big object, whose situation is described by the Principle of Momentum Conservation:

p_{S,1}+p_{B,1} = p_{S,2}+p_{B,2} (1)

Where:

p_{S,1}, p_{S,2} - Initial and final momemtums of the small object, measured in kilogram-meters per second.

p_{B,1}, p_{B,2} - Initial and final momentums of the big object, measured in kilogram-meters per second.

If we know that p_{S,1} = 7\,\frac{kg\cdot m}{s}, p_{B,1} = 0\,\frac{kg\cdot m}{s} and p_{S, 2} = 4\,\frac{kg\cdot m}{s}, then the final momentum of the big object is:

7\,\frac{kg\cdot m}{s} + 0\,\frac{kg\cdot m}{s} = 4\,\frac{kg\cdot m}{s}+p_{B,2}

p_{B,2} = 3\,\frac{kg\cdot m}{s}

The magnitude of the large object's momentum change is:

p_{B,2}-p_{B,1} = 3\,\frac{kg\cdot m}{s}-0\,\frac{kg\cdot m}{s}

p_{B,2}-p_{B,1} = 3\,\frac{kg\cdot m}{s}

The magnitude of the large object's momentum change is 3 kilogram-meters per second.

4 0
2 years ago
You are standing at the top of a cliff that has a stairstep configuration. There is a vertical drop of 6 m at your feet, then a
Zigmanuir [339]

Answer:

4.5 m/s

Explanation:

The rock must barely clear the shelf below, this means that the horizontal distance covered must be

d_x = 5 m

while the vertical distance covered must be

d_y = 6 m

The rock is thrown horizontally with velocity v_x, so we can rewrite the horizontal distance as

d_x = v_x t

where t is the time of flight. Re-arranging the equation,

t=\frac{d_x}{v_x} (1)

The vertical distance covered instead is

d_y = \frac{1}{2}gt^2

where we omit the term ut since the initial vertical velocity is zero. From this equation,

t=\sqrt{\frac{2d_y}{g}} (2)

Equating (1) and (2), we can solve the equation to find v_x:

\frac{d_x}{v_x}=\sqrt{\frac{2d_y}{g}}\\\frac{d_x^2}{v_x^2}=\frac{2d_y}{g}\\v_x = d_x \sqrt{\frac{g}{2d_y}}=5\sqrt{\frac{9.8}{2(6)}}=4.5 m/s

6 0
3 years ago
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