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vredina [299]
3 years ago
5

Four point masses of 3.0 kg each are arranged in a square on masslessrods. The length of a side of the square is 0.50m. What is

the rotational inertia for rotation about an axis (a) passing through masses Band C? (b) passing through masses Aand C ? (c) passing through the center of the square and perpendicular to the plane of the square?
Physics
1 answer:
Zigmanuir [339]3 years ago
8 0

Answer:

Part a)

I = 1.5 kg m^2

Part b)

I = 0.75 kg m^2

Part c)

I = 1.5 kg m^2

Explanation:

Part a)

Moment of inertia of the system about an axis passing through B and C is given as

I = mL^2 + mL^2 + m(0) + m(0)

I = 2mL^2

I = 2(3 kg)(0.50^2)

I = 1.5 kg m^2

Part b)

Moment of inertia of the system about an axis passing through A and C is given as

I = m(0^2) + m(\frac{L}{\sqrt2})^2 + m(0) + m(\frac{L}{\sqrt2})^2

I = 2m\frac{L^2}{2}

I = (3 kg)(0.50^2)

I = 0.75 kg m^2

Part c)

Moment of inertia of the system about an axis passing through the center of the square and perpendicular to the plane of the square

I = m(\frac{L}{\sqrt2})^2 + m(\frac{L}{\sqrt2})^2 + m(\frac{L}{\sqrt2})^2 + m(\frac{L}{\sqrt2})^2

I = 4m\frac{L^2}{2}

I = 2(3 kg)(0.50^2)

I = 1.5 kg m^2

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a 1210 kg roller coaster car is moving 6.33 m/s. as it approaches the station, brakes slow it down to 2.38 m/s over a distance o
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Answer:

4960 N

Explanation:

First, find the acceleration.

Given:

v₀ = 6.33 m/s

v = 2.38 m/s

Δx = 4.20 m

Find: a

v² = v₀² + 2aΔx

(2.38 m/s)² = (6.33 m/s)² + 2a (4.20 m)

a = -4.10 m/s²

Next, find the force.

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4 0
3 years ago
Read 2 more answers
At a certain instant a particle is moving in the +x direction with momentum +8 kg m/s. During the next 0.13 seconds a constant f
jeka94

Answer:

The momentum of the particle at the end of the 0.13 s time interval is 7.12 kg m/s

Explanation:

The momentum of the particle is related to force by the following equation:

Δp = F · Δt

Where:

Δp =  change in momentum = final momentum - initial momentum

F = constant force.

Δt = time interval.

Let´s calculate the x-component of the momentum after the 0.13 s:

final momentum - 8 kg m/s = -7 N · 0.13 s

final momentum = -7 kg m/s² · 0.13 s + 8 kg m/s

final momentum = 7.09 kg m/s

Now let´s calculate the y-component of the momentum vector after the 0.13 s. Since the particle wasn´t moving in the y-direction, the initial momentum in this direction is zero:

final momentum = 5 kg m/s² · 0.13 s

final momentum = 0.65 kg m/s

Then, the mometum vector will be as follows:

p = (7.09 kg m/s,  0.65 kg m/s)

The magnitude of this vector is calculated as follows:

|p| = \sqrt{(7.09 kg m/s)^{2} + (0.65 kg m/s)^{2}} = 7.12 kg m/s

The momentum of the particle at the end of the 0.13 s time interval is 7.12 kg m/s

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3 years ago
How can a magnetic field be produced, using a wire, a battery, and and a nail?
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Step 1 - Gather the Materials
One iron nail fifteen centimeters (6 in) long
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What is the frequency of a wave having a period equal to 18 seconds <br>​
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Explanation:

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The relation between wave period and frequency is as follows.

T = \frac{1}{f}T=

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where, T = time period

f = frequency

It is given that wave period is 18 seconds. Therefore, calculate the wave period as follows.

T = \frac{1}{f}T=

f

1

or, f = \frac{1}{T}f=

T

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= \frac{1}{18 sec}

18sec

1

= 0.055 per second (1cycle per second = 1 Hertz)

or, f = 5.5 \times 10^{-2} hertz5.5×10 −2 hertz

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