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Nostrana [21]
3 years ago
8

38 Points + Brainlyest!!!

Physics
2 answers:
Damm [24]3 years ago
8 0
The answer is C. Flammability because if you were to test flammability of an object, you would e making a chemical change to the object.
Zina [86]3 years ago
6 0

A chemical change is when you mix one or more substances to form a new one. A physical change is when you can apply something to a substance but it doesn't change the substance.

Changing the density, color, and mass of an object it doesn't change the type of substance that object is.

Changing the flammability of something changes the type of substance it is. Well, because its irreversable.

The correct answer is [C] Flammability

Best of Luck!

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Based on Kepler's work, which best describes the orbit If a planet around the Sun?
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an ellipse with the Sun at one focus  or D

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A bike travels at a constant speed for 4.00 m/s for 5.00 seconds. How far does it go
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Explanation:

4.00 m/s x 5 seconds

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If you set the cruise control of your car to a certain speed and take a turn, the speed of the car will remain the same. Is the
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Yes, the car has acceleration.

Explanation:

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How is Einstein's equation E = mc^2 used in medicine?
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8 0
3 years ago
A 2.45-kg frictionless block is attached to an ideal spring with force constant 355 N/m. Initially the spring is neither stretch
ANTONII [103]

Answer:

A.    A = 0.913 m

B.    amax = 132.24m/s^2

C.    Fmax = 324.01N

Explanation:

When the block is moving at the equilibrium point , its velocity is maximum.

A. To find the amplitude of the motion you use the following formula for the maximum velocity:

v_{max}=A\omega          (1)

vmax = maximum velocity = 11.0 m/s

A: amplitude of the motion = ?

w: angular frequency = ?

Then, you have to calculate the angular frequency of the motion, by using the following formula:

\omega=\sqrt{\frac{k}{m}}           (2)

k: spring constant = 355 N/m

m: mass of the object = 2.54 kg

\omega = \sqrt{\frac{355N/m}{2.45kg}}=12.03\frac{rad}{s}

Next, you solve the equation (1) for A and replace the values of vmax and w:

A=\frac{v}{\omega}=\frac{11.0m/s}{12.03rad/s}=0.913m

The amplitude of the motion is 0.913m

B. The maximum acceleration of the block is given by:

a_{max}=A\omega^2 = (0.913m)(12.03rad/s)^2=132.24\frac{m}{s^2}

The maximum acceleration is 132.24 m/s^2

C. The maximum force is calculated by using the second Newton law and the maximum acceleration:

F_{max}=ma_{max}=(2.45kg)(132.24m/s^2)=324.01N

It is also possible to calculate the maximum force by using:

Fmax = k*A = (355N/m)(0.913m) = 324.01N

The maximum force exertedbu the spring on the object is 324.01 N

4 0
3 years ago
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