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2 years ago

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ball of mass 0.4 kg is attached to the end of a light stringand whirled in a vertical circle of radius R = 2.9 m abouta fixed po

int. Find the magnitude of the tension when themass is at the top if its speed at the top is 8.5 m/s.(

1 answer:

Levart [38]2 years ago

8 0

Answer:

$6.046N$

Explanation:

The net force exerted on the mass is the sum of tension force and the external force of gravity.

$F_n_e_t=F_g+F_t$

$F_t$ is the tension force. $F_g=9.8N/kg$ is the force of gravity.

$F_n_e_t=ma_c=mv^2/r\\$

where $r$ is the rope's radius from the fixed point.

From the net force equation above:

$F_t=F_n_e_t-F_g\\=mv^2/r-mg\\=0.4\times(8.5^2/2.9)-0.4\times9.8\\=6.046N$

Hence the tension force is 6.046N

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2 years ago

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Sound is caused by vibrations. These can pass through a solid, liquid, and gas. But not through vacuum because there are no particles

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A factory worker pushes a 30.0-kg crate a distance of 4.5 m along a level floor at constant velocity by pushing horizontally on

SIZIF [17.4K]

(a) 73.5 N

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$\sum F = ma$

the resultant of the forces must be zero: $\sum F = 0$ (1)

The motion is along the horizontal direction, so we are only interested in the forces acting along this direction. There are two of them:

$F$ , the push applied by the worker

$F_f=-\mu mg$ , the force of friction, with $\mu=0.25$ being the coefficient of friction, $m=30.0 kg$ being the mass of the crate, and $g=9.8 m/s^2$ . The negative sign is due to the fact that the friction acts in the opposite direction to the motion. Eq.(1) then becomes

$F-\mu mg=0\\F=\mu mg=(0.25)(30.0 kg)(9.8 m/s^2)=73.5 N$

So, this is the force that the worker must apply.

(b) 330.8 J

The work done by the pushing force of the worker on the crate is given by:

$W=Fd cos \theta$

where

F = 73.5 N is the force

d = 4.5 m is the displacement

$\theta=0^{\circ}$ is the angle between the direction of the force and the displacement (0 degrees, since they are in same direction)

Substituting, we have

$W=(73.5 N)(4.5 m)(cos 0^{\circ})=330.8 J$

(c) -330.8 J

To calculate the work done by friction, we apply the same formula:

$W=F_f d cos \theta$

where

$F_f = \mu mg=(0.25)(30.0 kg)(9.8 m/s^2)=73.5 N$ is the magnitude of the force of friction

d = 4.5 m is the displacement

$\theta=180^{\circ}$ is the angle between the direction of the force of friction and the displacement (it is 180 degrees since the two are into opposite directions)

Substituting, we find

$W=(73.5 N)(4.5 m)(cos 180^{\circ})=-330.8 J$

So, the work done by friction is negative.

(d) 0 J

As before, the work done by any force on the crate is

$W=F_f d cos \theta$

We notice that both gravity and normal force are perpendicular to the displacement: therefore, $\theta=90^{circ}$ , and so

$cos \theta=0$

which means that the work done by both forces is zero.

(e) 0 J

The total work done on the crate is the sum of the work done by the four forces acting on it, so:

$W=W_{push} + W_{friction}+W_{gravity}+W_{normal}=330.8J-330.8J+0+0=0$

And this is in accordance with the work-energy theorem, which states that the variation of kinetic energy of the crate is equal to the work done on it: since the crate is moving at constant velocity, its variation of kinetic energy is zero, as well as the work done on it.

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2 years ago

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Vanyuwa [196]

A build up of charges on a sock from a dryer

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2 years ago

Consider a disk of radius 2.6 cm with a uni- formly distributed charge of +6.7.1C. Compute the magnitude of the electric field a

MrRissso [65]

Answer:

E=15.3*10¹³ N/C : approximate

Explanation:

We use the following formula to calculate the electric field due to a disk with uniform surface charge at a point P that is along the central perpendicular axis of the disk and at a distance x from the center of the disk:

E= 2*π*k*σ*F Fórmula ( 1 )

$F= 1-\frac{x}{\sqrt{R^{2}+x^{2} } }$

E: Electric field at point P (N/C)

σ: surface charge density (C/m²)

R: disk radio (m)

x :distance from the center of the disk to the point P located on the axis of the disk (m)

K: Coulomb constant ( N*m²/C)

Equivalences

1cm = 10⁻²m

1mm = 10⁻³m

Data

R =2.6 cm= 2.6*10⁻²m = 0.026m

x=3.7 mm = 3.7* 10⁻³m = 0,0037 m

Q=+6.71C.

k= 8.98774 * 10⁹ N* m²/C

Calculation of surface charge density (σ )

σ= Q/A

Q: uniformly distributed charge (C)

A: disk area (m²) = π*R²

σ= +6.71C/π*(2.6*10⁻²)²m² = 3159.56C/m²

Calculation of the electric field at point P

We apply formula (1) and replace data

E= 2*π*3159.56*8.98774*10⁹ *F

$F= 1-\frac{x}{\sqrt{R^{2}+x^{2} } }$

E=15.3*10¹³ N/C : approximate

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2 years ago