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Scilla [17]
3 years ago
14

An element

Physics
1 answer:
adell [148]3 years ago
8 0
<u />The correct answer is:  [D]:
_______________________________________________________
    "<span>cannot be divided into smaller substances by chemical means<span>" .
</span>_______________________________________________________
     An <u><em>element</em></u><u><em /> </u><em> </em>:  "</span>cannot be divided into smaller substances by chemical means."
_______________________________________________________
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What is amperage?
Shtirlitz [24]

question one b

question 2 i think a

3  d

4 c

5 not sure but wanting to say d

6  letter b

7 not sure

8 idk

9 i have no idea

3 0
3 years ago
A rock is thrown from a bridge at an angle 30∘ below horizontal.immediately after the rock is released, is the magnitude of its
Wittaler [7]
<span>The magnitude of the rock is equal to g. After the rock is released, there are no more forces acting on it, yet gravity remains. The initial inputs, on a bridge, at an angle of 30 deg below horizontal do not matter after the release.</span>
4 0
3 years ago
I need help please it’s for a lab
krok68 [10]

Answer:

idk

Explanation:

5 0
3 years ago
A motorcycle, which has an initial linear speed of 9.7 m/s, decelerates to a speed of 4.0 m/s in 4.4 s. Each wheel has a radius
Morgarella [4.7K]

Hi there!

We can begin by solving for the linear acceleration as we are given sufficient values to do so.

We can use the following equation:

vf = vi + at

Plug in given values:

4 = 9.7 + 4.4a

Solve for a:

a = -1.295 m/s²

We can use the following equation to convert from linear to angular acceleration:

a = αr

a/r = α

Thus:

-1.295/0.61 = -2.124 rad/sec² ⇒ 2.124 rad/sec² since counterclockwise is positive.

Now, we can find the angular displacement using the following:

θ = ωit + 1/2αt²

We must convert the initial velocity of the tire (9.7 m/s) to angular velocity:

v = ωr

v/r = ω

9.7/0.61 = 15.9 rad/sec

Plug into the equation:

θ = 15.9(4.4) + 1/2(2.124)(4.4²) = 20.56 rad

6 0
2 years ago
Use the dot product to find the magnitude of u if u = 6i - 3j
LuckyWell [14K]
Vector u :
u = 6 i - 3 j
The magnitude of vector u :
| u | = \sqrt{6 ^{2}+(-3) ^{2}  } = \sqrt{36+9}= \\  \sqrt{45}= \sqrt{9*5}=3  \sqrt{5}
Answer:
The magnitude of vector u is 3√5. 
3 0
2 years ago
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