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Mariana [72]
3 years ago
13

String 1 is pulled on with a force of 50 N to the right. String 1 is attached to a mass of 20 kg on a level surface. String 2 co

nnects the 20 kg mass to the second mass of 40 kg. Friction can be ignored. Find the tension force in string 2.
​
Physics
1 answer:
vladimir1956 [14]3 years ago
7 0

The masses are pulled in only one dimension, so we can ignore the forces acting in the vertical direction. We take the positive direction to be to the right.

For the 20 kg mass, the net force is

∑ <em>F</em>₁ = 50 N - <em>T</em>

where <em>T</em> is the tension in string 2.

For the 40 kg mass, the net force is

∑ <em>F</em>₂ = <em>T</em>

For both masses combined, the net force is

∑ <em>F</em> = 50 N

Use Newton's second law to compute the acceleration of the system:

50 N = (20 kg + 40 kg) <em>a</em>

<em>a</em> = (50 N) / (60 kg) ≈ 0.83 m/s²

Then use the net force equation for the 40 kg mass to compute the tension:

<em>T</em> = (40 kg) <em>a</em> ≈ 33 N

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Answer:

Product, P=9.1321972\times 10^7\times 7.004\times 10^{-3}

P = 639619.091888

Explanation:

In this case, we need to find the product of two numbers. These are as follows :

N_1=91321972=9.13\times 10^7

N_2=0.007004=7.004\times 10^{-3}

On doing calculation, we get the product of two numbers as :

P=N_1\times N_2

P=9.1321972\times 10^7\times 7.004\times 10^{-3}

P = 639619.091888

In scientific notation the product of two number is given by :

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3 years ago
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3 years ago
What is the (magnitude of the) centripetal acceleration (as a multiple of g=9.8~\mathrm{m/s^2}g=9.8 m/s ​2 ​​ ) towards the Eart
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Answer:

The centripetal acceleration as a multiple of g=9.8 m/s^{2} is 1.020x10^{-3}m/s^{2}

Explanation:

The centripetal acceleration is defined as:

a = \frac{v^{2}}{r}  (1)

Where v is the velocity and r is the radius

Since the person is standing in the Earth surfaces, their velocity will be the same of the Earth. That one can be determined by means of the orbital velocity:

v = \frac{2 \pi r}{T}  (2)

Where r is the radius and T is the period.

For this case the person is standing at a latitude 71.9^{\circ}. Remember that the latitude is given from the equator. The configuration of this system is shown in the image below.

It is necessary to use the radius at the latitude given. That radius can be found by means of trigonometric.

\cos \theta = \frac{adjacent}{hypotenuse}

\cos \theta = \frac{r_{71.9^{\circ}}}{r_{e}} (3)

Where r_{71.9^{\circ}} is the radius at the latitude of 71.9^{\circ} and r_{e} is the radius at the equator (6.37x10^{6}m).

r_{71.9^{\circ}} can be isolated from equation 3:

r_{71.9^{\circ}} = r_{e} \cos \theta  (4)

r_{71.9^{\circ}} = (6.37x10^{6}m) \cos (71.9^{\circ})

r_{71.9^{\circ}} = 1.97x10^{6} m

Then, equation 2 can be used

v = \frac{2 \pi (1.97x10^{6} m)}{24h}

Notice that the period is the time that the Earth takes to give a complete revolution (24 hours), this period will be expressed in seconds for a better representation of the velocity.

T = 24h . \frac{3600s}{1h} ⇒ 84600s

v = \frac{2 \pi (1.97x10^{6} m)}{84600s}

v = 146.31m/s

Finally, equation 1 can be used:

a = \frac{(146.31m/s)^{2}}{(1.97x10^{6}m)}

a = 0.010m/s^{2}

Hence, the centripetal acceleration is 0.010m/s^{2}

To given the centripetal acceleration as a multiple of g=9.8 m/s^{2}​ it is gotten:

\frac{0.010m/s^{2}}{9.8 m/s^{2}} = 1.020x10^{-3}m/s^{2}

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