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Nina [5.8K]
3 years ago
6

Name two ways to reduce friction and two ways to increase it.

Physics
1 answer:
Maurinko [17]3 years ago
4 0
Ways to increase friction 

<span>- increase the roughness of the contact materials </span>
<span>- increase the pressure on the contact </span>


<span>Ways to decrease friction </span>

<span>- float the moving body on air </span>
<span>- suck out any air </span>
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A 50.0 ohm and a 30.0 ohm resistor are connected in parallel. What is their equivalent resistance? Unit=Ohms
Alchen [17]

R(parallel) = product/ sum

50×30/50+30

1500/80

18,75 ohms

4 0
3 years ago
Which of these will always produce a magnetic field?
vovikov84 [41]

Answer:

Technically everything has somewhat of a magnetic field. I guess

7 0
3 years ago
A garage door opener has a power rating of 350 watts. If the door is in operation for 30 seconds, how many joules of energy are
Ksju [112]
Because of the hint we can conclude what equation we need to solve this problem. We have power and duration that means that we need to express energy:

1 joule = 1watt * 1 second
or
E (energy) = P (power) * t (time duration)
E = 350 * 30 = 10500 joules.
7 0
3 years ago
One complete expression of a waveform beginning at a certain point, progressing through the zero line to the wave’s highest (cre
elixir [45]

Answer:

wavelength.

Explanation:

One complete expression of a waveform beginning at a certain point, progressing through the zero line to the wave’s highest (crest) and lowest (trough) points, and returning to the same value as the starting point is called a is called wavelength. Its can be also defined as the distance between two successive crests or trough points in wave form.

4 0
2 years ago
) Force F = − + ( 8.00 N i 6.00 N j ) ( ) acts on a particle with position vector r = + (3.00 m i 4.00 m j ) ( ) . What are (a)
natali 33 [55]

To develop this problem it is necessary to apply the concepts related to the Cross Product of two vectors as well as to obtain the angle through the magnitude of the angles.

The vector product between the Force and the radius allows us to obtain the torque, in this way,

\tau = \vec{F} \times \vec{r}

\tau = (8i+6j)\times(-3i+4j)

\tau = (8*4)(i\times j)+(6*-3)(j\times i)

\tau = 32k +18k

\tau = 50 k

Therefore the torque on the particle about the origen is 50k

PART B) To find the angle between two vectors we apply the definition of the dot product based on the vector quantities, that is,

cos\theta = \frac{r\cdot F}{|\vec{r}|*|\vec{F}|}

cos\theta = \frac{(8*-3)+(4*3)}{\sqrt{(-3)^2+4^2}*\sqrt{8^2+6^2}}

cos\theta = -0.24

\theta = cos^{-1} (-0.24)

\theta = 103.88\°

Therefore the angle between the ratio and the force is 103.88°

5 0
3 years ago
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