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vampirchik [111]

3 years ago

14

A child pushes horizontally on a box of mass m which moves with constant speed v across a horizontal floor. The coefficient of f

riction between the box and the floor is Î¼. At what rate does the child do work on the box?

1 answer:

dexar [7]3 years ago

4 0

Answer:

Explanation:

Given

mass of box is m

speed of box is v

coefficient of friction $\mu =\frac{1}{4}=0.25$

Box is moving at a constant speed i.e. Net Force on box is zero

i.e. Frictional Force and Force Applied are Equal

$F=f_r=\mu mg$

The Rate at which power is generated is

$P=F\cdot v$

$P=\mu mg\cdot v$

$P=0.25mgv$

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A charge q1 = +5.00 nC is placed at the origin of an xy-coordinate system, and a charge q2 = -2.00 nC is placed on the positive

Ivahew [28]

Answer:

a

The x- and y-components of the total force exerted is

$F_{31 +32} = (8.64i - 5.52 j) *10^{-5}$

b

The magnitude of the force is

$|F_{31 +32}| = 10.25 *10^{-5} N$

The direction of the force is

$\theta =327.43 ^o$ Clockwise from x-axis

Explanation:

From the question we are told that

The magnitude of the first charge is $q_1 = +5.00nC = 5.00*10^{-9}C$

The magnitude of the second charge is $q_2 = -2.00nC = -2.00*10^{-9}C$

The position of the second charge from the first one is $d_{12} = 4.00i \ cm = \frac{4.00i}{100} = 4.00i *10^{-2} m$

The magnitude of the third charge is $q_3 = +6.00nC = 6.00*10^{-9}C$

The position of the third charge from the first one is $\= d_{31} = (4i + 3j) cm = \frac{ (4i + 3j)}{100} = (4i + 3j) *10^{-2}m$

$|d_{31}| =(\sqrt{4 ^2 + 3^2}) *10^{-2} m$

$|d_{31}| =5 *10^{-2} m$

The position of the third charge from the second one is

$\= d_{32} = 3j cm = 3j *10^{-2}m$

$|d_{32}| =(\sqrt{ 3^2}) *10^{-2} m$

$|d_{32}| =3 *10^{-2} m$

The force acting on the third charge due to the first and second charge is mathematically represented as

$F_{31 +32} = \frac{kq_3 q_1}{|d_{31}| ^3} *\= d_{31} + \frac{kq_3 q_2}{|d_{32}| ^3} *\= d_{32}$

Substituting values

$F_{31 +32} = \frac{9 *10^9 * 6 *10^{-9} * 5*10^{-9} }{(5*10^{-2}) ^3} * (4i + 3j ) *10^{-2} \\ \ + \ \ \ \ \ \ \ \ \ \frac{9 *10^9 * 6 *10^{-9} * -2*10^{-9} }{(5*10^{-2}) ^3} * (4i + 3j ) *10^{-2}$

$F_{31 +32} = 2.16 *10^{-5} (4i + 3j) - 12*10^{-5} j$

$F_{31 +32} = (8.64i - 5.52 j) *10^{-5}$

The magnitude of $F_{31 +32}$ is mathematically evaluated as

$|F_{31 +32}| = \sqrt{(8.64^2 + 5.52 ^2) } *10^{-5}$

$|F_{31 +32}| = 10.25 *10^{-5} N$

The direction is obtained as

$tan \theta = \frac{-5.52 *10^{-5}}{8.64 *10^{-5}}$

$\theta = tan ^{-1} [-0.63889]$

$\theta = - 32.57 ^o$

$\theta = 360 - 32.57$

$\theta =327.43 ^o$

5 0

3 years ago

When an object reflects all the light waves that strike it looks white or black?

egoroff_w [7]

Light waves can be from any color, depends on what it is bouncing no or reflecting off of.

3 0

3 years ago

Gemiola [76]

Answer:

5.25 J

Explanation:

W = PE = (f)(x)

PE = 35N*0.15m

PE = 5.25 N*m

1 N*m = 1 J

PE = 5.25 J

7 0

2 years ago

Un prisma rectangular de cobre, de base igual a 36 centímetros cuadrados y una altura de 10 cm se sumerge hasta la mitad por med

butalik [34]

La respuesta correcta es 180 centímetros cúbicos o 180 $cm^{3}$

Explicación:

El primer paso para saber cuanto alcohol desaloja el prisma es calcular el volumen total del prisma. El volumen se puede encontrar usando la formula V (Volumen) = B (base) x h (altura). El proceso se muestra a continuación:

V = B x h

V = 36 $cm^{2}$ x 10 cm

V= 360 $cm^{3}$

Finalmente, el volumen total del prisma debe dividirse en 2 considerando que solo la mitad del prisma fue sumergida y esta mitad equivale al volumen del alcohol desplazado.

360 $cm^{3}$ ÷ 2 = 180 $cm^{3}$

5 0

2 years ago

What are the factors that keep a satellite in orbit

Ber [7]

They are the same factors that keep planets, asteroids, and comets in orbit. The factors are gravity.

8 0

3 years ago