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3 years ago

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Una caja de 40.0 kg se desliza 4.00 m hacia arriba por un plano inclinado de 26° con respecto a la acción de una fuerza de 340 N

paralela al plano. Si el coeficiente de fricción cinético entre las superficies es 0.250, calcula el trabajo neto realizado sobre el objeto.

1 answer:

otez555 [7]3 years ago

6 0

Answer:

W = 320.30 J

Explanation:

To calculate the net work done over the block you take into account all implied forces:

$\Sigma F=F+F_gsin\theta-F_fcos\theta$ (1)

The gravitational force and friction force are against the applied force F.

θ = 26°

F: applied force = 340N

Fg: gravitational force = Mg = (40.0kg)(9.8m/s^2) = 392N

Ff: friction force = $\mu N=\mu Mg=(0.250)(392N)=98N$

Next, you replace to obtain the net force:

$F_N=(340N)-(392N)sin(26\°)-(98N)cos(26\°)\\\\F_N=80.07N$

Finally, the net work, for 4 m, is:

$W_N=F_Nd=(80.07N)(4m)=320.30J$

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A spherically-spreading EM wave comes from a 104.0 W source. At a distance of 9.6 m, what is the intensity of the wave?

Nookie1986 [14]

Answer:

Approximately 0.0898 W/m².

Explanation:

The intensity of light measures the power that the light delivers per unit area.

The source in this question delivers a constant power of $\rm 104.0\; W$ . If the source here is a point source, that $\rm 104.0\; W$ of power will be spread out evenly over a spherical surface that is centered at the point source. In this case, the radius of the surface will be 9.6 meters.

The surface area of a sphere of radius $r$ is equal to $4\pi r^{2}$ . For the imaginary 9.6-meter sphere here, the surface area will be:

$\rm 4\pi \times (9.6\; m)^{2} \approx 1158.12\; m^{2}$ .

That $\rm 104.0\; W$ power is spread out evenly over this 9.6-meter sphere. The power delivered per unit area will be:

$\displaystyle\rm \frac{104.0\; W}{1158.12\; m^{2}}\approx 0.0898\; W\cdot m^{-2}$ .

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A space station in the shape of a 100 m-diameter (50m radius) wheel is spinning so as to impart a linear tangential speed of 22.

zalisa [80]

Answer:

correct option is b. 31.3 m/s

Explanation:

given data

artificial gravity a1 = 1 g

artificial gravity a2 = 2 g

diameter = 100 m

radius r= 50 m

speed v1 = 22.1 m/s

solution

As acceleration is ∝ v²

so we can say

$\frac{a2}{a1} = \frac{v2}{v1}$ .....................1

put here value

$\frac{2}{1} = \frac{v2}{22.1}$

solve it

v2 = $\sqrt{2 }$ × 22.1

v2 = 31.25 m/s

so correct option is b. 31.3 m/s

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3 years ago

A wire 2.80 m in length carries a current of 5.20 A in a region where a uniform magnetic field has a magnitude of 0.430 T. Calcu

galina1969 [7]

Question:

A wire 2.80 m in length carries a current of 5.20 A in a region where a uniform magnetic field has a magnitude of 0.430 T. Calculate the magnitude of the magnetic force on the wire assuming the following angles between the magnetic field and the current.

(a)60 (b)90 (c)120

Answer:

(a)5.42 N (b)6.26 N (c)5.42 N

Explanation:

From the question

Length of wire (L) = 2.80 m

Current in wire (I) = 5.20 A

Magnetic field (B) = 0.430 T

Angle are different in each part.

The magnetic force is given by

$F=I \times B \times L \times sin(\theta)$

So from data

$F = 5.20 A \times 0.430 T \times 2.80 sin(\theta)\\\\F=6.2608 sin(\theta) N$

Now sub parts

(a)

$\theta=60^{o}\\\\Force = 6.2608 sin(60^{o}) N\\\\Force = 5.42 N$

(b)

$\theta=90^{o}\\\\Force = 6.2608 sin(90^{o}) N\\\\Force = 6.26 N$

(c)

$\theta=120^{o}\\\\Force = 6.2608 sin(120^{o}) N\\\\Force = 5.42 N$

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Two long, straight wires are separated by a distance of 9.15 cm . One wire carries a current of 2.79 A , the other carries a cur

Dafna1 [17]

Answer:

The force is the same

Explanation:

The force per meter exerted between two wires carrying a current is given by the formula

$\frac{F}{L}=\frac{\mu_0 I_1 I_2}{2\pi r}$

where

$\mu_0$ is the vacuum permeability

$I_1$ is the current in the 1st wire

$I_2$ is the current in the 2nd wire

r is the separation between the wires

In this problem

$I_1=2.79 A\\I_2=4.36 A\\r = 9.15 cm = 0.0915 m$

Substituting, we find the force per unit length on the two wires:

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However, the formula is the same for the two wires: this means that the force per meter exerted on the two wires is the same.

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