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Sedaia [141]
3 years ago
7

18 kilogram Mass Blokus addressed a level surface if the coefficient of static friction between the Block in the surface is 0.6

what is horizontal force is required to take gravity 10m/s2
​
Physics
1 answer:
Tasya [4]3 years ago
4 0

Question: 18 kilogram Mass Block rest on level surface if the coefficient of static friction between the Block and the surface is 0.6 what  horizontal force is required to just move the blcok ( take gravity as 10m/s2 )

Answer:

108 N

Explanation:

From the question,

Applying

F' = mgμ................ Equation 1

Where F' = Frictional force = horizontal  force required to just move the block,  m = mass of the block, g = acceleration due to gravity, μ = coefficient of static friction.

From the question,

Given: m = 18 kg, μ = 0.6, g = 10 m/s²

Substitute these values into equation 1

F' = 18×0.6×10

F' = 108 N

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The frictional force is 218.6 N

Explanation:

The block in the problem is at rest along the inclined surface: this means that the net force acting along the direction parallel to the incline must be zero.

There are two forces acting along this direction:

- The component of the weight parallel to the incline, downward along the plane, of magnitude

mg sin \theta

where

m = 46 kg is the mass

g=9.8 m/s^2 is the acceleration of gravity

\theta=29^{\circ} is the angle of the incline

- The (static) frictional force, acting upward, of magnitude F_f

Since the block is in equilibrium, we can write

mg sin \theta - F_f = 0

And substituting, we find the force of friction:

F_f = mg sin \theta = (46)(9.8)(sin 29^{\circ})=218.6 N

Learn more about frictional force along an inclined plane:

brainly.com/question/5884009

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3 years ago
A man is sitting in the outfield bleachers watching a baseball game through binoculars. Which is MOST LIKELY true?
Anestetic [448]
Choice-A is the correct one. It doesn't say it, but it means he'll see the ball reach the catcher's mitt first BEFORE HE HEARS IT slap the mitt.
7 0
4 years ago
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Pages 64 to 65 of your reading material trace an extended problem that involes lifting a bag of sugar up to a shelf at first it
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Answer:

d) the amount of work is the same whether the bag is moved all at once or in two stages, provided the total height lifted is the same in either case.

Explanation:

While moving the bag to the shelf in one shot we can say that the total work done is given as

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here we know that

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now when we raise the bag to first shelf and then move it to next shelf

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W = W_1 + W_2[tex][tex]W = mgH + mgH

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so the correct answer will be

d) the amount of work is the same whether the bag is moved all at once or in two stages, provided the total height lifted is the same in either case.

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3 years ago
A professor determined the relationship between the time spent studying (in hours) and performance on an exam.
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Lets write again formula for determening Ann's performance.

P = 70.443 + 4,885*t

where t is in hours. This is equation with P(t) which means that P only depends on variable t. If we express t=2.6 in formula we will find her expected performance.

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3 years ago
A 10-turn coil of wire having a diameter of 1.0 cm and a resistance of 0.50 Ω is in a 1.0 mT magnetic field, with the coil orien
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Answer:

The voltage across the capacitor is 1.57 V.

Explanation:

Given that,

Number of turns = 10

Diameter = 1.0 cm

Resistance = 0.50 Ω

Capacitor = 1.0μ F

Magnetic field = 1.0 mT

We need to calculate the flux

Using formula of flux

\phi=NBA

Put the value into the formula

\phi=10\times1.0\times10^{-3}\times\pi\times(0.5\times10^{-2})^2

\phi=7.85\times10^{-7}\ Tm^2

We need to calculate the induced emf

Using formula of induced emf

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Put the value into the formula

\epsilon=\dfrac{7.85\times10^{-7}}{dt}

Put the value of emf from ohm's law

\epsilon =IR

IR=\dfrac{7.85\times10^{-7}}{dt}

Idt=\dfrac{7.85\times10^{-7}}{R}

Idt=\dfrac{7.85\times10^{-7}}{0.50}

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We know that,

Idt=dq

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We need to calculate the voltage across the capacitor

Using formula of charge

dq=C dV

dV=\dfrac{dq}{C}

Put the value into the formula

dV=\dfrac{1.57\times10^{-6}}{1.0\times10^{-6}}

dV=1.57\ V

Hence, The voltage across the capacitor is 1.57 V.

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