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AfilCa [17]
2 years ago
13

Two equal forces are applied to a door. The first force is applied at the midpoint of the door; the second force is applied at t

he doorknob. Both forces are applied perpendicular to the door. Which force exerts the greater torque?a) the first at the midpoint.
b) the second at the doorknob.
c) both exert equal non-zero torques.
d) both exert zero torques.
e) Additional information is needed.
Physics
1 answer:
Olenka [21]2 years ago
5 0

Answer:

b) the second at the doorknob.

Explanation:

The torque applied by a force is given by the equation

\tau = F_p d

where

\tau is the torque

F_p is the component of the force perpendicular to the direction between the axis of rotation and the point of application of the force

d is the arm (the distance between the axis of rotation and the point of application of the force)

In this problem, we have two equal forces F both applied perpendicular to a door, so

F_p = F

The first force is applied at the midpoint of the door; if we call L the width of the door, then the arm of this force is d_1=\frac{L}{2}, so the torque applied by this first force is

\tau_1 = F\frac{L}{2}

The 2nd force instead is applied at the doorknob, so the arm in this case is L:

d = L

So, the torque exerted by the second force is

\tau_2 = FL

Therefore, we see that the 2nd force exerts a greater torque.

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Answer:

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Explanation:

From the above question, we are told that:

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Step 1

We convert 10.0 MeV to Joules

1 Mev = 1.602 × 10-13 Joules

10.0 MeV = 10.0 × 1.602 × 10^-13 Joules = 1.602 × 10^-12 J

Hence, the Kinectic energy of a proton = 1.602 × 10^-12 J

Step 2

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The formula for Kinectic Energy =

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Where

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v = speed of the proton

K.E of the proton = 1.602 × 10^-12 J

Mass of the proton = 1.6726219 × 10^-27 kilograms

Speed of the proton = ?

1.602 × 10^-12J = 1/2 × 1.6726219 × 10^-27 × v²

1.602 × 10^-12J = 8.3631095 ×10^-28 × v²

v² = 1.602 × 10^-12/8.3631095 ×10^-28

v = √(1.602 × 10^-12/8.3631095 ×10^-28)

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Step 3

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The formula for Magnetic Field =

B = m v / q r

We are told that the proton executes a circular orbit, hence,

mv = √2m(KE)

m = Mass of the proton = 1.6726219 × 10^-27 kg

K.E of the proton = 1.602 × 10^-12 J

v = speed of the proton = 4.4 × 10^7 m/s

q = Electric charge = 1.6 × 10^-19 C

r = radius of the orbit = 5.80Ã10^10 m

= 5.8 × 10^10m

Magnetic Field =

=√ (2 × 1.6726219 × 10^-27 kg × 1.602 × 10^-12 J) /( 1.6 × 10^-19 C × 5.80 × 10^10 m)

= 7.88 × 10^-12 T

The magnetic field in that region of space is approximately 7.88 × 10^-12 T

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Answer:

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Explanation:

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