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harkovskaia [24]

2 years ago

13

For the meter stick shown in figure 10-4, the force F1 10.0 N acts at 10.0 cm. What is the magnitude of torque due to F1 about a

n axis through point A perpendicular to the page? Is it clockwise, or is it counterclockwise? Show your work and give correct units.

1 answer:

Phantasy [73]2 years ago

8 0

Torque is equal position vector times (r) times force vector (F). Since F= 10 N and r = 0.1 m, so the torque is equal to (10 N) x ( 0.1 m) = 1Nm. The direction of the torque would be into the screen, clockwise rotation.

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Two window washers, Bob and Joe, are on a 3.00 m long, 395 N scaffold supported by two cables attached to its ends. Bob weighs 8

WINSTONCH [101]

Answer:

- the forces on the left hand side is 1.038 kN

- the forces on the right hand side is 1.483 kN

Explanation:

Given the data in the question, as illustrated in the image below;

Length of the scaffold = 3 m

weight of the scaffold = 395 N

Weight of Bob = 805 N and stands 1 m from the left end

weight of washing equipment = 500N and on sits 2 m from the left end

Weight of Joe = 820 N and stand 0.500 m from the right end

so the force on the left cable will be;

$T_{left$ = $\frac{1}{3m}$ [ (805 N)( (3-1) m) + ( 395 N )( $\frac{3}{2} m$ ) + ( 500 N )(1m ) + ( 820 N)( 0.500m ) ]

$T_{left$ = $\frac{1}{3m}$ [ 1610 + 592.5 + 500 + 410 ]

$T_{left$ = $\frac{1}{3m}$ [ 3112.5 ]

$T_{left$ = 1037.5 N

$T_{left$ = 1.038 kN

Therefore, the forces on the left hand side is 1.038 kN

On the right hand side;

$T_{Right$ = ( 805 N + 395 N + 500 N + 820 N ) - 1037.5 N

$T_{Right$ = 2520 N - 1037.5 N

$T_{Right$ = 1482.5 N

$T_{Right$ = 1.483 kN

Therefore, the forces on the right hand side is 1.483 kN

5 0

1 year ago

An electron is moving in the vicinity of a long, straight wire that lies along the z-axis. The wire has a constant current of 8.

viktelen [127]

Answer:

The force that the wire exerts on the electron is $-4.128\times10^{-20}i-6.88\times10^{-20}j+0k$

Explanation:

Given that,

Current = 8.60 A

Velocity of electron $v= (5.00\times10^{4})i-(3.00\times10^{4})j\ m/s$

Position of electron = (0,0.200,0)

We need to calculate the magnetic field

Using formula of magnetic field

$B=\dfrac{\mu I}{2\pi d}(-k)$

Put the value into the formula

$B=\dfrac{4\pi\times10^{-7}\times8.60}{2\pi\times0.200}$

$B=0.0000086\ T$

$B=-8.6\times10^{-6}k\ T$

We need to calculate the force that the wire exerts on the electron

Using formula of force

$F=q(\vec{v}\times\vec{B}$

$F=1.6\times10^{-6}((5.00\times10^{4})i-(3.00\times10^{4})j\times(-8.6\times10^{-6}) )$

$F=(1.6\times10^{-19}\times3.00\times10^{4}\times(-8.6\times10^{-6}))i+(1.6\times10^{-19}\times5.00\times10^{4}\times(-8.6\times10^{-6}))j+0k$

$F=-4.128\times10^{-20}i-6.88\times10^{-20}j+0k$

Hence, The force that the wire exerts on the electron is $-4.128\times10^{-20}i-6.88\times10^{-20}j+0k$

5 0

2 years ago

A block spring system oscillates on a frictionless surface with an amplitude of 10\text{ cm}10 cm and has an energy of 2.5 \text

antoniya [11.8K]

Answer:

The energy of the system is 15 J.

Explanation:

Given that,

Energy E = 2.5 J

Amplitude = 10 cm

We need to calculate the spring constant

Using formula of mechanical energy of the system

$E=\dfrac{1}{2}kA^2$

Put the value into the formula

$2.5=\dfrac{1}{2}k\times(10\times10^{-2})^2$

$k=\dfrac{2.5\times2}{(10\times10^{-2})^2}$

$k=500\ N/m$

If the block is replaced by a block with twice the mass of the original block

Amplitude = 6 cm

We need to calculate the energy

Using formula of mechanical energy

$E=\dfrac{1}{2}kA^2$

Put the value into the formula

$E=\dfrac{1}{2}\times500\times(6\times10^{-2})$

$E=15\ J$

Hence, The energy of the system is 15 J.

8 0

2 years ago

A stone with mass 0.80 kg is attached to one end of a string 0.90 m long. The string will break if its tension exceeds 60.0 N. T

AleksandrR [38]

Answer:

$v=8.2158m/s$

Explanation:

(a) Free-body diagram attached.

(b) The stone attached with the string experiences both centripetal (towards the center) and centrifugal (away from the center) forces. The tension of the string counters the centrifugal force until it breaks.

We know that,

Centrifugal force = $\frac{mv^2}{r}$

where,

$m$ = mass of the stone

$v$ = velocity of the stone

$r$ = length of the string

To find the maximum speed attained by the stone without the string breaking, we must equate:

$\frac{mv^2}{r} =60$

or, $v=\sqrt \frac{{r\times 60}}{m} } =\sqrt{\frac{0.90\times60}{0.80} } =8.2158m/s$

5 0

1 year ago

Snow has been lying on a mountainside. Suddenly, it starts to move down the mountain. Which types of friction are observed in th

Alexus [3.1K]

<h2>Snow starts to move down the mountain </h2>

The energy which is due to position is potential energy. So when the snow is lying on the mountain. It possess potential energy but when suddenly, it starts to move down the mountain, the potential energy is converted into the kinetic energy. Yet some force is exerting on the snow to stop the smooth flow of snow through mountains.

This example of frictional force may be due to presence of rough surface or stones. Generally, there are four types of friction as static, rolling, sliding and fluid friction. Though in this case when snow is lying it possess static friction, when flows then it possesses sliding and fluid friction both.

8 0

2 years ago