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3 years ago

14

If one replaces the conducting cube with one that has positive charge carriers, in what direction does the induced electric fiel

d point

1 answer:

Grace [21]3 years ago

3 0

Answer:

There will be no change in the direction of the electric field .

Explanation:

The direction will remain the same because the sign of the charges has no effect on it.

When one replaces the conducting cube with one that has positive charge carriers there will be no change in the direction of the field as there is no defined relationship between the direction of the electric field and sign of the charge.

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A mass on a horizontal surface is connected to the spring and pulled to the right along the surface stretching the spring by 25

solniwko [45]

Answer:

320 N/m

Explanation:

From Hooke's law, we deduce that

F=kx where F is applied force, k is spring constant and x is extension or compression of spring

Making k the subject of formula then

$k=\frac {F}{x}$

Conversion

1m equals to 100cm

Xm equals 25 cm

25/100=0.25 m

Substituting 80 N for F and 0.25m for x then

$k=\frac {80}{0.25}=320N/m$

Therefore, the spring constant is equal to 320 N/m

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3 years ago

How does the force affect the rocks during collisions? *

Jet001 [13]

In a collision, there is a force on both objects that causes an acceleration of both objects; the forces are equal in magnitude and opposite in direction. For collisions between equal-mass objects, each object experiences the same acceleration.

3 0

2 years ago

Points A (-5,6), B (2,-2), and C (-6,-3) are placed in three different quadrants of a Cartesian coordinate system. Convert each

AURORKA [14]

Answer: A ( $\sqrt{61}$ ,309.8°)

B (2 $\sqrt{2}$ , 315°)

C ( $3\sqrt{5}$ , 26.56°)

Explanation: To transform rectangular coordinates into polar coordinates use:

$r=\sqrt{x^{2}+y^{2}}$ and $\theta=tan^{-1}(\frac{y}{x})$

For point A:

$r=\sqrt{(-5)^{2}+6^{2}}$

$r=\sqrt{61}$

$\theta=tan^{-1}(\frac{6}{-5})$

$\theta=tan^{-1}(-1.2)$

$\theta=-50.2$ °

Point A is in the II quadrant, so we substract the angle for 360° since it is in degrees:

$\theta=360-50.2$

$\theta=$ 309.8°

Polar coordinates for point A is ( $\sqrt{61}$ , 309.8°)

For point B:

$r=\sqrt{2^{2}+(-2)^{2}}$

$r=\sqrt{8}$

$r=2\sqrt{2}$

$\theta=tan^{-1}(\frac{-2}{2} )$

$\theta=tan^{-1}(1)$

$\theta=-45$ °

Point B is in IV quadrant, so:

$\theta=360-45$

$\theta=$ 315°

Polar coordinates for point B is ( $2\sqrt{2}$ , 315°)

For point C:

$r=\sqrt{(-6)^{2}+(-3)^{2}}$

$r=\sqrt{45}$

$r=3\sqrt{5}$

$\theta=tan^{-1}(\frac{-3}{-6} )$

$\theta=tan^{-1}(0.5)$

$\theta=$ 26.56°

Polar coordinates for point C is ( $3\sqrt{5}$ , 26.56°)

3 0

3 years ago

Where is the potential energy equal to zero?

s2008m [1.1K]

Answer:

im sure your already past this but it's E.

Explanation:

This is because in this case potential energy is linear to height, which means that the higher the more potential energy.

4 0

2 years ago

A wave has a speed of 30 m/s, a frequency of 6 Hz, and a wavelength of 5 m. If the wavelength remains constant, and the frequenc

mixer [17]

Answer: 60m/s

Explanation:

The wavespeed is the distance covered by the wave in one second. It is measured in metre per second, and represented by the symbol V

Wavespeed (V) = Frequency F x wavelength λ

i.e V = F λ

In the first case:

Wavespeed = 30 m/s

Frequency of sound = 6Hz

Wavelength = 5m

In the second case:

Wavespeed = ?

Frequency of sound = (2x 6Hz = 12Hz)

Wavelength = 5m (remains constant)

Apply V = F λ

Wavespeed = 12 Hz x 5m

Wavespeed = 60m/s

Therefore, when frequency is doubled, the speed is also doubled. Thus, the new speed of the wave is 60m/s

8 0

3 years ago