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2 years ago

If the applied force is 60 N and the lower mass is 1.0 kg, what is the tension in the string?(frictionless surface)

Physics

1 answer:

Alik [6]2 years ago

3 0

The net force on the larger mass in the horizontal direction (where we take the direction of F to be positive) is

F - T = 2Ma

where T is the magnitude of tension and a is the mass's acceleration.

The net force on the smaller mass in the vertical direction (upward is positive) is

T - Mg = Ma

If M = 1.0 kg and F = 60 N, then we end up with two equations with two unknowns,

60 N - T = (2.0 kg) a

T - (1.0 kg) (9.8 m/s²) = (1.0 kg) a

Eliminate a and solve for T :

(60 N - T) - 2 (T - (1.0 kg) (9.8 m/s²)) = (2.0 kg) a - 2 (1.0 kg) a

60 N - T - 2T + 19.6 N = (2.0 kg) a - (2.0 kg) a

40.4 N - 3T = 0

3T = 40.4 N

T = (40.4 N) / 3

T ≈ 13 N

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Answer:

Part a)

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Part b)

$dF = \frac{2mvdv}{r}$

Part c)

$dT = - \frac{2\pi r}{v^2} dv$

Explanation:

Part a)

As we know that force on the passenger while moving in circle is given as

$F = \frac{mv^2}{r}$

now variation in force is given as

$dF = -\frac{mv^2}{r^2} dr$

here speed is constant

Part b)

Now if the variation in force is required such that r is constant then we will have

$F = \frac{mv^2}{r}$

so we have

$dF = \frac{2mvdv}{r}$

Part c)

As we know that time period of the circular motion is given as

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$dT = - \frac{2\pi r}{v^2} dv$

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3 years ago

What does it mean when it says a scientific question must be supportable

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A good scientific question has certain characteristics. It should have some answers (real answers), should be testable (can be tested by someone through an experiment or measurements), leads to a hypothesis that is falsifiable (means it should generate a hypothesis that can be shown to fail), etc.

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3 years ago

When an airplane flies from Dallas - Ft. Worth International airport to Los Angeles International airport, the time in the air i

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Explanation:

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3 years ago

At the intersection of Texas Avenue and University Drive,

Zielflug [23.3K]

Answer:

The initial speed of the truck is 21.93 m/s, and the initial speed of the car is 19.524 m/s

Explanation:

We can use conservation of momentum to find the initial velocities.

Taking the unit vector $\hat{i}$ pointing north and $\hat{j}$ pointing east, the final velocity will be

$\vec{V}_f = 16.0 \frac{m}{s} \ ( \ cos(24.0 \°) \ , \ sin (24.0 \°) \ )$

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The final linear momentum will be:

$\vec{P}_f = (m_{car}+ m_{truck}) * V_f$

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$\vec{P}_f = (2.850 \ kg \ ) * ( \ 14.617 \frac{m}{s} \ , \ 6.508 \frac{m}{s} \ )$

$\vec{P}_f = ( \ 41,658.45 \frac{ kg \ m}{s} \ , \ 18,547.8 \frac{kg \ m}{s} \ )$

As there are not external forces, the total linear momentum must be constant.

So:

$\vec{P}_0= \vec{P}_f$

As initially the car is travelling east, and the truck is travelling north, the initial linear momentum must be

$\vec{P}_0= ( m_{truck} * v_{truck}, m_{car}* v_{car} )$

so:

$\vec{P}_0= \vec{P}_f$

$( m_{truck} * v_{truck}, m_{car}* v_{car} ) = ( \ 41,658.45 \frac{ kg \ m}{s} \ , \ 18,547.8 \frac{kg \ m}{s} \ )$

$\left \{ {{m_{truck} \ v_{truck} = 41,658.45 \frac{ kg \ m}{s} } \atop {m_{car} \ v_{car}=18,547.8 \frac{kg \ m}{s} }} \right.$

So, for the truck

$m_{truck} \ v_{truck} = 41,658.45 \frac{ kg \ m}{s}$

$1900 \ kg \ v_{truck} = 41,658.45 \frac{ kg \ m}{s}$

$v_{truck} = \frac{41,658.45 \frac{ kg \ m}{s}}{1900 \ kg}$

$v_{truck} = 21.93 \frac{m}{s}$

And, for the car

$950 \ kg \ v_{car}=18,547.8 \frac{kg \ m}{s}$

$v_{car}=\frac{18,547.8 \frac{kg \ m}{s}}{950 \ kg}$

$v_{car}=19.524 \frac{m}{s}$

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Explanation:

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