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fgiga [73]
4 years ago
7

A truck covers 40.0 m in 7.10 s while smoothly slowing down to a final velocity of 2.75 m/s.

Physics
1 answer:
makkiz [27]4 years ago
6 0
V = u + at
75 = u + 7.1 a
s = ut + 1/2 at^2

You can find it using this two equation method.  The original speed will be the u

hope this helps
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A space station shaped like a giant wheel has a radius of a radius of 153 m and a moment of inertia of 4.16 × 10⁸ kg·m² (when it
Naya [18.7K]

Answer:

a = 1.709g

Explanation:

Given the absence of external forces being applied in the space station, it is possibly to use the Principle of Angular Momentum Conservation, which states that:

I_{o} \cdot \omega_{o} = I_{f} \cdot \omega_{f}

The required initial angular speed is obtained herein:

g= \omega_{o}^{2}\cdot R_{ss}

\omega_{o}=\sqrt{\frac{g}{R_{ss}} }

\omega_{o}= \sqrt{\frac{9.807\,\frac{m}{s^{2}} }{153\,m} }

\omega_{o} \approx 0.253\,\frac{rad}{s}

The initial moment of inertia is:

I_{o} =I_{ss}+n\cdot m_{person}\cdot R_{ss}^{2}

I_{o} = 4.16\times 10^{8}\,kg\cdot m^{2}+(150)\cdot (65\,kg)\cdot (153\,m)^{2}

I_{o} = 6.442\times 10^{8}\,kg\cdot m^{2}

The final moment of inertia is:

I_{f} =I_{ss}+n\cdot m_{person}\cdot R_{ss}^{2}

I_{f} = 4.16\times 10^{8}\,kg\cdot m^{2}+(50)\cdot (65\,kg)\cdot (153\,m)^{2}

I_{f} = 4.921\times 10^{8}\,kg\cdot m^{2}

Now, the final angular speed is obtained:

\omega_{f} = \frac{I_{o}}{I_{f}}\cdot \omega_{o}

\omega_{f} = \frac{6.442\times 10^{8}\,{kg\cdot m^{2}}}{4.921\times 10^{8}\,kg\cdot m^{2}} \cdot (0.253\,\frac{rad}{s} )

\omega_{f} = 0.331\,\frac{rad}{s^}

The apparent acceleration is:

a_{f} = \omega_{f}^{2}\cdot R_{ss}

a_{f} = (0.331\,\frac{rad}{s} )^{2}\cdot (153\,m)

a_{f} = 16.763\,\frac{m}{s^{2}}

This is approximately 1.709g.

4 0
4 years ago
on a part-time job, you are asked to bring a cylindrical iron rod of density 7800 kg/m3 , length 92.8 cm and diameter 2.05 cm fr
emmainna [20.7K]

On a part-time job, you are asked to bring a cylindrical iron rod of density 7800 kg/m3 , length 92.8 cm and diameter 2.05 cm from a storage room to a machinist, the  weight of rod w is 23.41 N .

<h3>How is the weight calculated?</h3>

Given ,

density (d) = 7800 kg/m³

D = 2.05 cm = 0.0205 m

r = D/2 = 0.0205 / 2 =0.01025 m

h = 92.8 cm = 0.928 m

The weight is calculated by ,

W = m × g

But, to find mass (m)

m = d × V

The volume of the rod, because is cylindrical, is,

V = π × r² × h

∴ V = 3.14 × (0.01025)² × 0.928

∴ V = 3.14 × 1.051 × 10⁻⁴ × 0.928

∴ V = 3.0625 × 10⁻⁴

Now to calculate mass,

m = d × V

∴ m = 7800 × 3.0625 × 10⁻⁴

∴ m = 2.3887 Kg

Now, for calculating weight

W = m × g

∴ W =  2.3887 × 9.8

∴ W = 23.41 N

∴ The weight of the rod W assuming the free-fall acceleration is 23.41 N .

<h3>What is weight?</h3>
  1. The weight of an object is the force exerted on it by gravity.
  2. It is define weight as a vector quantity, the force of gravity acting on an object.
  3. Some define weight as a scalar quantity that is the magnitude of gravity.

Can learn more about weight calculation from brainly.com/question/15685221

#SPJ4

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1 year ago
Saturn's moon Mimas has an orbital period of 82,800 s at a distance of 1.87x10^8m from Saturn. Using m central m= (4n^2d^3/GT^2)
FromTheMoon [43]
5.65×10^26kg here you go
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4 years ago
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Which of the following is a graph of the velocity of an object as it falls fromrest if drag is not ignored? Explain your choice
MariettaO [177]

The air drag is a force that depends on the speed of an object relative to the wind. Under certain conditions, it can be modeled as:

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As a falling object reaches a speed so that its weight is cancelled out by the air drag, the object will reach a maximum velocity.

In a speed vs time gaph, the speed would approach the maximum speed like an asymptote.

On the other hand, since the object falls from rest, the initial speed on the graph must be zero.

Taking these considerations into account, the correct graph for the movement of an object that falls from rest if air drag is not ignored, is option B.

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1 year ago
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Law Incorporation [45]

Answer:

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8 0
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