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Afina-wow [57]
3 years ago
9

A spring is hung from the ceiling. When a coffee mug is attached to its end, it stretches 2.5 cm before reaching its new equilib

rium length. The block is then pulled down slightly and released. What is the frequency of oscillation
Physics
1 answer:
densk [106]3 years ago
5 0

Answer:

Explanation:

In equilibrium , weight of mug is equal to restoring force .

mg = kx where m is mass of mug , k is spring constant and x is extension .

k / m = g / x = 9.8 ms⁻² / .025 m

= 392

frequency of oscillation n = \frac{1}{2\pi}\sqrt{\frac{k}{m} }

n=\frac{1}{2\pi}\sqrt{392 }

= 4.46 per second.

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Match each term with the appropriate definition.
Ratling [72]

Answer:

mass- the amount of matter in an object

balance- tool used to measure mass

scale- a tool used to measure weight

weight- the downward pull on an object due to gravity

8 0
3 years ago
To practice Problem-Solving Strategy 21.1 Coulomb's Law. Three charged particles are placed at each of three corners of an equil
Salsk061 [2.6K]

Answer:

The force felt by charge 3 is F=(-5.6*10⁻⁶,3.36⁻⁵)N

Explanation:

As the superposition principle applies to static charges, we can find the net electric force as the sum of the two forces felt by q3.

Looking at the drawing and knowing that they form an equilateral triangle of lenght 4 we can conclude that each internal angle is 60°.

So, the positions in our coordinate system are:

r_1=(0,0)\\r_2=(4\ cos(60\°),4\ sin(60\°))\\r_3=(4,0)\\

Now  using Coulomb's force:

F_{ij}=\frac{-kq_iq_j}{d^2}(\vv{r}_j-\vv{r}_i)

Where d=4, q1 = -7.8*10⁻⁹C, q2 = -15.6 *10⁻⁹C, q3 = 8.0 *10⁻⁹C, k=8.98*10⁹, e0=8.8*10¹⁰:

Replacing we get 2 equations:

F_{13}=\frac{-kq_1q_3}{d^2}(\vv{r}_1-\vv{r}_3)=\frac{-kq_1q_2}{d^2}(-0.04\ cos(60\°),-0.04\ sin(60\°))\\\\F_{23}=\frac{-kq_2q_3}{d^2}(\vv{r}_1-\vv{r}_3)=\frac{-kq_1q_2}{d^2}(0.04-0.04\ cos(60\°),-0.04\ sin(60\°))\\

To work with the sam

F=∑F_i=3.5*10⁻⁴(0.023,0.032)+7*10⁻⁴(-0.016,0.032)=

=((3.5*10⁻⁴-7*10⁻⁴)*0.016,(3.5*10⁻⁴+7*10⁻⁴)*0.032)=

F=(-5.6*10⁻⁶,3.36⁻⁵)N

7 0
3 years ago
Two wave pulses move toward each other along a rope.
mojhsa [17]

Answer:

The answer is B. It is the one that has the triangle pointing up on the topside of the rope and seems to be a bit bigger than the other triangles.

Explanation:

7 0
3 years ago
Read 2 more answers
Consider a particle moving along the x-axis where x(t) is the position of the particle at time t, x'(t) is its velocity, and x''
Lady_Fox [76]

Answer:

We know that the acceleration of the particle is defined as

a(t)=\frac{dv}{dt}

Since it is given that

v(t)=\frac{5}{\sqrt{t}}\\\\\therefore a(t)=\frac{d(\frac{5}{\sqrt{t}})}{dt}\\\\a(t)=5\times \frac{dt^{-\frac{1}{2}}}{dt}\\\\=\frac{-5}{2}t^{\frac{-1}{2}-1}\\\\\therefore a(t)=x''(t)=\frac{-2.5}{t^{\frac{3}{2}}}

Now by definition of velocity we have

v(t)=\frac{dx(t)}{dt}\\\\\Rightarrow dx(t)=v(t)dt

Integrating on both sides we get

v(t)=\frac{dx(t)}{dt}\\\\\int dx(t)=\int v(t)dt\\\\x(t)=\int v(t)dt

Applying values we get

v(t)=\frac{dx(t)}{dt}\\\\\int dx(t)=\int v(t)dt\\\\x(t)=\int \frac{5}{t^{\frac{1}{2}}}dt\\\\\therefore x(t)=\frac{5}{0.5}\int t^{-0.5}dt\\\\x(t)=10\sqrt{t}+c

To find the constant we note that at t=1 the particle is at x=12 Thus applying values in the above equation we get

c=12-10\\\therefore c=2\\\\x(t)=10\sqrt{t}+2

6 0
3 years ago
When a bouncy ball is dropped from a second floor window, at what point is the kinetic energy highest?
Dmitry [639]

Answer:

Explanation:

Because the total energy available to the ball doesn't change whatsoever during its entire trip from the window to the ground,

TE = KE + PE which says that the total energy available to a system is equal to the kinetic energy of the system plus the potential energy, and that this value will not ever change (because energy cannot be created or destroyed. Sound familiar?) If the ball is being held still before it is dropped from some height off the ground, it is here that the total energy can be determined, and that total energy at this point is all potential, since the ball is not moving while someone is holding it and getting ready to drop it. The SECOND it starts to fall, the potential energy begins to be converted to kinetic. As the potential energy is losing value, the kinetic is gaining it at the same rate (again, energy doesn't just disappear; it has to go somewhere. Here, it goes from potential to kinetic a little at a time). When the ball finally hits the ground, or an INSTANT BEFORE it hits the ground, the potential energy is 0 because the height of something on the ground has a height of 0. At this instant, right before the ball hits the ground, is where the KE is the greatest. All the energy at that point has been converted from potential to kinetic.

Long story short, choice B is the one you want.

6 0
3 years ago
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