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earnstyle [38]
3 years ago
9

Which would produce mechanical waves?

Physics
2 answers:
lisabon 2012 [21]3 years ago
6 0

Answer:

C

Explanation:

think about energy transfers when you turn on flashlight

Please give good review

IRISSAK [1]3 years ago
4 0
The answer is C. Hope this helped!
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The velocity of an object is given by the expression v(t) = 3.00 m/s + ( 4.00 m/s^3)t^2, where t is in seconds. Determine the po
bulgar [2K]

Answer:

Position of object is;

s(t) = 4t³/3 + 3t + 1

Explanation:

We are told that the velocity has an expression;

v(t) = 3.00 m/s + ( 4.00 m/s³)t²

Now, to get the expression for the position(s(t)) of the object, we have to integrate the velocity expression. Thus;

s(t) = ∫3 + 4t²

s(t) = 3t + 4t³/3 + c

Now, we were told that at x = 1.00 m, time t = 0.000 s

Thus, plugging the values in;

1 = 3(0) + 4(0³/3) + c

c = 1

Thus,the expression for the position of the object is;

s(t) = 4t³/3 + 3t + 1

6 0
3 years ago
B) A skilled jet fighter flies a stunt plane in a vertical circle of 1200 ft
adell [148]

The jet fighter's acceleration at the highest point is 17.56 ft/s²

The jet fighter's acceleration at the lowest point is 35.12 ft/s²

The given parameters;

radius of the circular path, r = 1200 ft

constant velocity, v = 140 mi/h = 205.3 ft/s

The formula for centripetal acceleration in a circular path is given as;

a_c = \frac{v^2}{r}

The jet fighter's acceleration at the highest point is calculated as;

h = 2r = 2 x 1200 ft = 2,400 ft

a_c = \frac{v^2}{r} \\\\a_c = \frac{(205.3)^2}{2400} \\\\a_c = 17.56 \ ft/s^2

The jet fighter's acceleration at the lowest point is calculated as;

h = r = 1,200 ft

a_c = \frac{v^2}{r} \\\\a_c = \frac{(205.3)^2}{1200} \\\\a_c = 35.12\ ft/s^2

Learn more here:brainly.com/question/11700262

6 0
3 years ago
A pot of water is heated on a gas-flame stove and begins to boil. Which two
Tatiana [17]

Answer:a

Explanation: itz A hoOOO

8 0
3 years ago
Read 2 more answers
Which statement most accurately describes the second law of thermodynamics?
balandron [24]

Answer:

Energy cannot be changed from one form to another without a loss of usable energy

Explanation:

Second law of thermodynamics states that the total entropy or the randomness of the system remains constant over time.  It also states that the net entropy will remain the same or it will increase.

Entropy of a system is given by heat absorbed divided by temperature. It is given by :

\Delta S=\dfrac{\Delta Q}{T}

So, the correct option is (A) "Energy cannot be changed from one form to another without a loss of usable energy".

4 0
3 years ago
What is the net force exerted by these two charges on a third charge q3 = 49.0nC placed between q1 and q2 at x3 = -1.085m ?Your
densk [106]

Answer:

The net force exerted by these two charges on a third charge is 5.468\times10^{-6}\ N

Explanation:

Given that,

Third charge q_{3}=49.0\ nC

Distancex_{3}=-1.085\ m

Suppose The magnitude of the force F between two particles with charges Q and Q' separated by a distance d. Consider two point charges located on the x axis one charge, q₁ = -12.5 nC , is located at x₁ = -1.650 m, the second charge, q₂ = 31.5 nC , is at the origin.

We need to calculate the total force will be the vector sum of two forces

Using Coulomb's law,

F_{13}=\dfrac{kq_{1}q_{3}}{(x_{1}-x_{3})^2}

Put the value into the formula

F_{13}=\dfrac{9\times10^{9}\times(-12.5\times10^{-9})\times49\times10^{-9}}{(-1.650-(-1.085))^2}

F_{13}=-17268.3\times10^{-9}\ N

We need to calculate the force will be to the negative charge with opposite charges

Using Coulomb's law,

F_{23}=\dfrac{kq_{2}q_{3}}{(x_{2}-x_{3})^2}

Put the value into the formula

F_{23}=\dfrac{9\times10^{9}\times(31.5\times10^{-9})\times49\times10^{-9}}{(-1.085)^2}

F_{23}=11800.2\times10^{-9}\ N

The force also will be to the negative side, charges with same charge sign

We need to calculate the net force exerted by these two charges on a third charge

Using formula of net force

F_{net}=F_{13}+F_{23}

F_{net}=-17268.3\times10^{-9}+11800.2\times10^{-9}

F_{net}=-0.0000054681\ N

F_{net}=-5.468\times10^{-6}\ N

Negative sign shows the negative direction.

Hence, The net force exerted by these two charges on a third charge is 5.468\times10^{-6}\ N

5 0
3 years ago
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