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AfilCa [17]
3 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]3 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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Fudgin [204]

Answer:

g'_h=1.096\times 10^{-5}\ m.s^{-2}

Explanation:

We know that the gravity on the surface of the moon is,

  • g'=\frac{g}{6}
  • g'=1.63\ m.s^{-2}

<u>Gravity at a height h above the surface of the moon will be given as:</u>

g'_h=\frac{G.m}{(r+h)^2} ..........................(1)

where:

G = universal gravitational constant

m = mass of the moon

r = radius of moon

We have:

  • G=6.67\times 10^{-11}\ m^3.s^{-2}.kg^{-1}
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  • r=1.74\times 10^6\ m
  • h=384.4\times 10^6\ m is the distance between the surface of the earth and the moon.

Now put the respective values in eq. (1)

g'_h=\frac{6.67\times 10^{-11}\times 7.35\times 10^{22}}{(1.74\times 10^6+384.4\times 10^6)^2}

g'_h=1.096\times 10^{-5}\ m.s^{-2} is the gravity on the moon the earth-surface.

4 0
3 years ago
A jet plane flying 600 m/s experiences an acceleration of 4g when pulling out of the dive. What is the radius of curvature of th
Gnoma [55]

Answer:

91.84 m/s²

Explanation:

velocity, v = 600 m/s

acceleration, a = 4 g = 4 x 9.8 = 39.2 m/s^2

Let the radius of the loop is r.

he experiences a centripetal force.

centripetal acceleration,

a = v² / r

39.2 x r = 600 x 600

r = 3600 / 39.2

r = 91.84 m/s²

Thus, the radius of the loop is 91.84 m/s².

8 0
3 years ago
when the heat is added to the material, which factors are important in order to determine how much the material will rise in tem
posledela

The heat Q transferred to cause a temperature change depends on the magnitude of the temperature change, the mass of the system, and the substance and phase involved.

Explanation:

https://courses.lumenlearning.com/physics/chapter/14-2-temperature-change-and-heat-capacity/

5 0
3 years ago
The weight of a block on the inclined plane is 500 N and the angle of incline is 30 degrees. What is the magnitude of the force
yanalaym [24]

Answer:

250 N

433 N

Explanation:

N = Normal force by the surface of the inclined plane

W = Weight of the block = 500 N

f = static frictional force acting on the block

Parallel to incline, force equation is given as

f = W Sin30

f = (500) Sin30

f = 250 N

Perpendicular to incline force equation is given  

N = W Cos30

N = (500) Cos30

N = 433 N

3 0
3 years ago
Initially a car accelerates at 2 m/s2 for x seconds. The car then travels at a velocity of -6 m/s for x seconds. If the car disp
Luda [366]

Answer:

The time travel is

t=8 s

Explanation:

a= 2 \frac{m}{s^{2} } \\v=-6 \frac{m}{s} \\x=16m

x_{f}=x_{o}+v_{o}*t+\frac{1}{2} *a*t^{2}

x_{f}=0-6*t+\frac{1}{2} *2*t^{2}

t^{2}-6*t-16=0\\ using :\\\frac{-b+/-\sqrt{b^{2}-4*c*a } }{2} \\\frac{-(-6)+/-\sqrt{(-6)^{2}-4*(-16)*(1) } }{2}=\frac{3}{2} +/- \frac{10}{2} \\t_{1} = 2s \\t_{2} = 8s

Check

t_{2}=8s

x_{f}=x_{o}+v_{o}*t+\frac{1}{2} *a*t^{2}

x_{f}=0-6*+\frac{1}{2} *2*8^{2}

x_{f}=-48+64\\x_{f}=16

5 0
3 years ago
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