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4 years ago

8

Please help me :) ty

1 answer:

slega [8]4 years ago

8 0

1. Physical Property
2. Chemical Change
3. Chemical Change
4. Chemical Property
5. Physical Property
6. Physical Change
7. Chemical Change
8. Physical Change
9. Chemical Change
10. Physical Property
11. Physical Property
12. Chemical Change
13. Chemical Change
14. Chemical Property
15. Chemical Change

Explanation:
Physical Property: A property of matter that can be seen, felt, tasted, smelled, or heard.
Physical Change: The change of an object of matter's physical property/properties.
Chemical Property: An element/object of matter's internal property/properties
Chemical Change: An element/object of matter's internal property/properties being altered or changed

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A worker at the top of a 588-m-tall television transmitting tower accidentally drops a heavy tool. If air resistance is negligib

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The final velocity of the tool is 107.4 m/s

Explanation:

We can solve this problem by using the principle of conservation of energy.

In fact, if air resistance is negligible, the total mechanical energy of the tool is conserved during the fall, so we can write:

$K_i + U_i = K_f + U_f$

where

$K_i = 0$ is the kinetic energy of the tool at the top (zero since it is at rest)

$U_i = mgh$ is the gravitational potential energy of the tool at the top, where

m is the mass of the tool

g is the acceleration of gravity

h is the heigth of the tool

$K_f = \frac{1}{2}mv^2$ is the kinetic energy of the tool just before hitting the ground, where

v is the final speed of the tool

$U_f = 0$ is the gravitational potential energy of the tool at the bottom (zero since the height is zero)

Re-arranging the equation,

$mgh=\frac{1}{2}mv^2$

where we have

$g=9.8 m/s^2\\h=588 m$

And solving for v,

$v=\sqrt{2gh}=\sqrt{2(9.8)(588)}=107.4 m/s$

Learn more about kinetic energy and potential energy:

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3 years ago

Imagine you use Nitrogen as your gas. If you have the cold side as cold as you can without liquefying it (78 K), and run the hot

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Answer:

The efficiency of a Stirling engine is 74%

Explanation:

Given:

Temperature of gas when it is cold $T_{1} = 78$ K

Temperature of gas when it is hot $T_{2} = 300$ K

The efficiency of a stirling engine,

$\eta =1 - \frac{T_{1} }{T_{2} }$

$\eta = 1- \frac{78}{300}$

$\eta = 1-0.26$

$\eta = 0.74$

∴ $\eta = 74 \%$

Therefore, the efficiency of a Stirling engine is 74%

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3 years ago

To withstand ‘g-forces’ of up to 10 g’s, caused by suddenly pulling out of a steep dive, fighter jet pilots train on a ‘human ce

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Answer:

34.292 m/s

Explanation:

When plane dives the gravitational acceleration becomes 10 times extra i-e

acceleration a = g= 10g's=10 ×9.8= 98 m/sec2

radius of dive circle = 12 m

v= speed= ?

Using

$a= \frac{v^{2} }{r}$

==> $v^{2}$ = a × r = 98 × 12 = 1176 $(\frac{m}{s}) ^{2}$

==> v= $\sqrt{1176(\frac{m}{s} )^{2} }$

==> v= 34.292 m/s

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Answer:

A

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Answer:

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Explanation:

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