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SSSSS [86.1K]
3 years ago
9

Alex is asked to move two boxes of books in contact with each other and resting on a rough floor. He decides to move them at the

same time by pushing on box A with a horizontal pushing force FP = 8.9 N. Here, A has a mass mA = 10.2 kg and B has a mass mB = 7.0 kg. The contact force between the two boxes is FC. The coefficient of kinetic friction between the boxes and the floor is 0.04. (Assume FP acts in the +x direction.)
Physics
1 answer:
Lilit [14]3 years ago
5 0

Answer:

Explanation:

The force of friction acting on the system

= .04 x 9.8 ( 10.2 + 7 )

= 6.74 N

Net force = 8.9 - 6.74

= 2.16 N

Acceleration in the system

= 2.16 /  ( 10.2 + 7 )

= .12558 m / s ²

Contact force between boxes = FP

Considering force on box A

Net force = 8.9 - FP

Applying Newton's law on box A

8.9 - FP = 10.2 x .12558

= 1.28

FP = 8.9 - 1.28

= 7.62 N

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Gold or copper than silver

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I think the acceleration is 12m/s

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3 years ago
Light is incident on the left face of an isosceles prism; with an apex angle of 49o, such that the light exiting the right face
Sunny_sXe [5.5K]

Answer:

\mu = 1.645

Explanation:

By Snell's law we know at the left surface

\theta_i = 19^o

\theta_r = ?

\mu_1 = 1

\mu_2 = \mu

now we have

1 sin19 = \mu sin\theta_r

0.33 = \mu sin\theta_r

now on the other surface we know that

angle of incidence = \theta_r'

\theta_e = 90

so again we have

\mu sin\theta_r' = 1 sin90

so we have

\theta_r = sin^{-1}\frac{0.33}{\mu}

\theta_r' = sin^{-1}\frac{1}{\mu}

also we know that

\theta_r + \theta_r' = 49

sin^{-1}\frac{0.33}{\mu} + sin^{-1}\frac{1}{\mu} = 49

By solving above equation we have

\mu = 1.645

3 0
3 years ago
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I’ll give brainliest!! please help and answer correctly! plsss answer quick
Rashid [163]

Answer: The motion of the object will remain the same

Explanation:

6 0
3 years ago
Two square air-filled parallel plates that are initially uncharged are separated by 1.2 mm, and each of them has an area of 190
julia-pushkina [17]

Answer:

5.5\cdot 10^{-11} C

Explanation:

The capacitance of the parallel-plate capacitor is given by:

C=\epsilon_0 \frac{A}{d}

where

\epsilon_0 = 8.85\cdot 10^{-12} F/m is the vacuum permittivity

A=190 mm^2 = 190 \cdot 10^{-6} m^2 is the area of the plates

d=1.2 mm = 0.0012 m is the separation between the plates

Substituting,

C=(8.85\cdot 10^{-12}) \frac{190 \cdot 10^{-6}}{0.0012}=1.4\cdot 10^{-12}F

The energy stored in the capacitor is given by

U=\frac{Q^2}{2C}

Since we know the energy

U=1.1 nJ = 1.1 \cdot 10^{-9} J

we can re-arrange the formula to find the charge, Q:

Q=\sqrt{2UC}=\sqrt{2(1.1\cdot 10^{-9} J)(1.4\cdot 10^{-12}F )}=5.5\cdot 10^{-11} C

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3 years ago
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