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aliya0001 [1]
4 years ago
11

A horizontal spring with stiffness 0.4 N/m has a relaxed length of 11 cm (0.11 m). A mass of 21 grams (0.021 kg) is attached and

you stretch the spring to a total length of 27 cm (0.27 m). The mass is then released from rest. What is the speed of the mass at the moment when the spring returns to its relaxed length of 11 cm (0.11 m)?
Physics
1 answer:
riadik2000 [5.3K]4 years ago
5 0

Answer:

0.6983 m/s

Explanation:

k = spring constant of the spring = 0.4 N/m

L₀ = Initial length = 11 cm = 0.11 m

L = Final length = 27 cm = 0.27 m

x = stretch in the spring = L - L₀ = 0.27 - 0.11 = 0.16 m

m = mass of the mass attached = 0.021 kg

v = speed of the mass

Using conservation of energy

Kinetic energy of mass = Spring potential energy

(0.5) m v² = (0.5) k x²

m v² = k x²

(0.021) v² = (0.4) (0.16)²

v = 0.6983 m/s

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3 years ago
A wire carrying a 29.0 A current passes between the poles of a strong magnet such that the wire is perpendicular to the magnet's
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Answer:

2.59 T

Explanation:

Parameters given:

Current flowing through the wire, I = 29 A

Angle between the magnetic field and wire, θ = 90°

Magnetic force, F = 2.25 N

Length of wire, L = 3 cm = 0.03 m

The magnetic force, F, is related to the magnetic field, B, by the equation below:

F = I * L * B * sinθ

Inputting the given parameters:

2.25 = 29 * 0.03 * B * sin90

2.25 = 0.87 * B

=> B = 2.25/0.87

B = 2.59 T

The magnetic field strength between the poles is 2.59 T

4 0
3 years ago
Help meh in this question plzzz <br>​
iragen [17]

The Moment of Inertia of the Disc is represented by I = \frac{15}{32}\cdot M\cdot R^{2}. (Correct answer: A)

Let suppose that the Disk is a Rigid Body whose mass is uniformly distributed. The Moment of Inertia of the element is equal to the Moment of Inertia of the entire Disk minus the Moment of Inertia of the Hole, that is to say:

I = I_{D} - I_{H} (1)

Where:

  • I_{D} - Moment of inertia of the Disk.
  • I_{H} - Moment of inertia of the Hole.

Then, this formula is expanded as follows:

I = \frac{1}{2}\cdot M\cdot R^{2} - \frac{1}{2}\cdot m\cdot \left(\frac{1}{2}\cdot R^{2} \right) (1b)

Dimensionally speaking, Mass is directly proportional to the square of the Radius, then we derive the following expression for the Mass removed by the Hole (m):

\frac{m}{M} = \frac{R^{2}}{4\cdot R^{2}}

m = \frac{1}{2}\cdot M

And the resulting equation is:

I = \frac{1}{2}\cdot M\cdot R^{2} -\frac{1}{2}\cdot \left(\frac{1}{4}\cdot M \right) \cdot \left(\frac{1}{4}\cdot R^{2} \right)

I = \frac{1}{2} \cdot M\cdot R^{2} - \frac{1}{32}\cdot M\cdot R^{2}

I = \frac{15}{32}\cdot M\cdot R^{2}

The moment of inertia of the Disc is represented by I = \frac{15}{32}\cdot M\cdot R^{2}. (Correct answer: A)

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3 years ago
The energy emitted from the sun is a product of ________.
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Q. The energy emitted from the sun is a product of ________.

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3 years ago
Two airplanes leave an airport at the same time. The velocity of the first airplane is 730 m/h at a heading of 44.3 ◦ . The velo
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Answer:

So airplane will be 1324.9453 m apart after 2.9 hour

Explanation:

So if we draw the vectors of a 2d graph we see that the difference in angles is  = 83 - 44.3 = 83-44.3=38.7^{\circ}

Distance traveled by first plane = 730×2.9 = 2117 m

And distance traveled by second plane = 590×2.9 = 1711 m

We represent these distances as two sides of the triangle, and the distance between the planes as the side opposing the angle 38.7.

Using the law of cosine, d^2 representing the distance between the planes, we see that:

d^2=2117^2 + 1711^2 -2\times (2117)\times (1711)cos(38.7)=1755480.2482

d = 1324.9453 m

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