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Naddik [55]
3 years ago
7

Part a consider another special case in which the inclined plane is vertical (θ=π/2). in this case, for what value of m1 would t

he acceleration of the two blocks be equal to zero?
Physics
1 answer:
Aloiza [94]3 years ago
6 0

The answer would be m1 = m2

FBD of m1 is the sum of the forces in the y direction = 0 
0 = T - m1*g 
transposing:
m1*g = T 

FBD of m2 is the sum of the forces in the y direction = 0 
0 = T - m2*g 
transposing:
m2*g = T 

set them equal to each other and solve for m1 
m1*g = m2*g

= m1 = m2

 

Explanation:

The force on individual mass would be downwards or descending and equal to the tension shaped by the other mass, which would be upward or rising by the act of the seamless pulley, so the forces cancel on each mass.

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My buddy and I have just finished a dive to 15 metres/50 feet for 60 minutes. We want to return to the same site and depth and s
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Answer:

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Explanation:

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The centripetal force acting on the space shuttle
tamaranim1 [39]

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(4) weight

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r is the radius of its circular orbit

When the shuttle orbits the Earth, the centripetal force that keeps the shuttle in circular motion is given by the gravitational attraction between the shuttle and the Earth, which corresponds to the weight of the shuttle, and it is given by:

F=G\frac{Mm}{r^2}

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4 years ago
A 600-turn solenoid, 25 cm long, has a diameter of 2.5 cm. A 14-turn coil is wound tightly around the center of the solenoid. If
Delvig [45]

Answer:

The induced emf in the short coil during this time is 1.728 x 10⁻⁴ V

Explanation:

The magnetic field at the center of the solenoid is given by;

B = μ(N/L)I

Where;

μ is permeability of free space

N is the number of turn

L is the length of the solenoid

I is the current in the solenoid

The rate of change of the field is given by;

\frac{\delta B}{\delta t} = \frac{\mu N \frac{\delta i}{\delta t} }{L} \\\\\frac{\delta B}{\delta t} = \frac{4\pi *10^{-7} *600* \frac{5}{0.6} }{0.25}\\\\\frac{\delta B}{\delta t} =0.02514 \ T/s

The induced emf in the shorter coil is calculated as;

E = NA\frac{\delta B}{\delta t}

where;

N is the number of turns in the shorter coil

A is the area of the shorter coil

Area of the shorter coil = πr²

The radius of the coil = 2.5cm / 2 = 1.25 cm = 0.0125 m

Area of the shorter coil = πr² = π(0.0125)² = 0.000491 m²

E = NA\frac{\delta B}{\delta t}

E = 14 x 0.000491 x 0.02514

E = 1.728 x 10⁻⁴ V

Therefore, the induced emf in the short coil during this time is 1.728 x 10⁻⁴ V

8 0
3 years ago
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