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Eddi Din [679]
2 years ago
8

If you pick up and carry a piece of heavy furniture which type of friction do you have with the floor

Physics
1 answer:
Phoenix [80]2 years ago
7 0
Sliding Friction hope its right
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He distance between two consecutive crests is 2.5 meters. Which characteristic of the wave does this distance represent? amplitu
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Frequency is measured in units of reciprocal time.
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It must be either amplitude or wavelength.

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The question is talking about points on consecutive waves.

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If youre walking from point a to b, the magnitude of your displacement will always be equal or less than or greater than your di
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The magnitude of your displacement can be equal to the distance you covered, or it can be less than the distance you covered.  But it can never be greater than the distance you covered.

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Why would knowing the characteristics of circuits be important in designing electrical circuits?
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A single-turn circular loop of wire of radius 65 mm lies in a plane perpendicular to a spatially uniform magnetic field. During
BabaBlast [244]

Answer:

Explanation:

change in flux = no of turns x area of loop x change in magnetic field

= 1 x π 65² x 10⁻⁶ x ( 650 - 350 ) x 10⁻³

= 3.9 x 10⁻³ weber .

rate of change of flux  = change of flux / time

= 3.9 x 10⁻³ / .10

= 39 x 10⁻³ V

= 39 mV .

Since the magnetic flux is directed outside page and it is increasing , induced current will be clockwise so that magnetic field is produced in opposite direction to reduce it , as per Lenz's law.

5 0
3 years ago
A box rests on top of a flat bed truck. The box has a mass of m = 16.0 kg. The coefficient of static friction between the box an
3241004551 [841]

Answer:

1) 1.31 m/s2

2) 20.92 N

3) 8.53 m/s2

4) 1.76 m/s2

5) -8.53 m/s2

Explanation:

1) As the box does not slide, the acceleration of the box (relative to ground) is the same as acceleration of the truck, which goes from 0 to 17m/s in 13 s

a = \frac{\Delta v}{\Delta t} = \frac{17 - 0}{13} = 1.31 m/s2

2)According to Newton 2nd law, the static frictional force that acting on the box (so it goes along with the truck), is the product of its mass and acceleration

F_s = am = 1.31*16 = 20.92 N

3) Let g = 9.81 m/s2. The maximum static friction that can hold the box is the product of its static coefficient and the normal force.

F_{\mu_s} = \mu_sN = mg\mu_s = 16*9.81*0.87 = 136.6N

So the maximum acceleration on the block is

a_{max} = F_{\mu_s} / m = 136.6 / 16 = 8.53 m/s^2

4)As the box slides, it is now subjected to kinetic friction, which is

F_{\mu_s} = mg\mu_k = 16*9.81*0.69 = 108.3 N

So if the acceleration of the truck it at the point where the box starts to slide, the force that acting on it must be at 136.6 N too. So the horizontal net force would be 136.6 - 108.3 = 28.25N. And the acceleration is

28.25 / 16 = 1.76 m/s2

5) Same as number 3), the maximum deceleration the truck can have without the box sliding is -8.53 m/s2

3 0
3 years ago
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