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iren [92.7K]
2 years ago
11

Using your hand to exert a horizontal force, you push a physics textbook across the floor at a steady pace. The frictional force

exerted by the floor on the book opposes its motion. Is the "push" force exerted by your hand greater than, less than, or equal to that frictional force?
Physics
1 answer:
bija089 [108]2 years ago
8 0

Answer:

It's equal to  the friction force.

Explanation:

  • When moving at a constant speed, the book must obey (as any body) Newton's 2nd Law.
  • As a consequence of its mathematical expression (F=m*a), if a=0, this means that the total force must be zero.
  • If we decompose this force in two components perpendicular each other (horizontal and vertical, for instance), both components must be zero too.
  • Taking the horizontal direction, there are only two forces acting on the book: the pushing force exerted by the hand, and the friction force oppposing to this force.
  • Both forces must be equal and opposite each other, in order to the total force be zero, and the book can move at a constant speed, as requested.
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Vector A⃗ has magnitude 8.00 m and is in the xy-plane at an angle of 127∘ counterclockwise from the +x–axis (37∘ past the +y-axi
WINSTONCH [101]

Answer:

293.7 degrees

Explanation:

A = - 8 sin (37) i + 8 cos (37) j

A + B = -12 j

B = a i+ b j , where and a and b are constants to be found

A + B = (a - 8 sin (37) ) i + ( 8cos(37) + b ) j

- 12 j = (a - 8 sin (37) ) i + ( 8cos(37) + b ) j

Comparing coefficients of i and j:

a = 8 sin (37) = 4.81452 m

b = -12 - 8cos(37) = -18.38908

Hence,

B = 4.81452 i  - 18.38908 j  ..... 4 th quadrant

Hence,

cos ( Q ) = 4.81452 / 12

Q = 66.346 degrees

360 - Q = 293.65 degrees from + x-axis in CCW direction

5 0
2 years ago
Suppose an isolated box of volume 2V is divided into two equal compartments. An ideal gas occupies half of the container and the
SpyIntel [72]

Answer:

A. the internal energy stays the same

Explanation:

From the first law of thermodynamics, "energy can neither be created nor destroyed but can be transformed from one form to another.

Based on this first law of thermodynamic, the new internal energy of the gas is the same as the internal energy of the original system.

Therefore, when the partition separating the two halves of the box is removed and the system reaches equilibrium again, the internal energy stays the same.

7 0
2 years ago
A student decides to give his bicycle a tune up. He flips it upside down (so there’s no friction with the ground) and applies a
Alinara [238K]

Answer:

Tangential velocity = 10.9 m/S

Explanation:

As per the data given in the question,

Force = 20 N

Time = 1.2 S

Length = 16.5 cm

Radius = 33.0 cm

Moment of inertia = 1200 kg.cm^2 = 1200 × 10^(-4) kg.m^2

= 1200 × 10^(-2) m^2

Revolution of the pedal ÷ revolution of wheel = 1

Torque on the pedal = Force × Length

= 20 × 16.5 10^(-2)

= 3.30 N m

So, Angular acceleration = Torque ÷ Moment of inertia

= 3.30 ÷ 12 × 10^(-2)

= 27.50 rad ÷ S^2

Since wheel started rotating from rest, so initial angular velocity = 0 rad/S

Now, Angular velocity = Initial angular velocity + Angular Acceleration × Time

= 0 + 27.50 × 1.2

= 33 rad/S

Hence, Tangential velocity = Angular velocity × Radius

= 33 × 33 × 10^(-2)

= 10.9 m/S

7 0
2 years ago
At time t, gives the position of a 3.0 kg particle relative to the origin of an xy coordinate system ( ModifyingAbove r With rig
Elena-2011 [213]

Complete Question

  The complete Question is shown on the first uploaded image

Answer:

a

The torque acting on the particle is  \tau = 48t \r k

b

The magnitude of the angular momentum increases relative to the origin

Explanation:

From the equation we are told that

      The position of the particle is   \= r = 4.0 t^2 \r i - (2.0 t - 6.0 t^2 ) \r j

       The mass of the particle is m = 3.0 kg

        The time is  t

   

The torque acting on  the particle is mathematically represented as

           \tau = \frac{ d \r l }{dt}

where \r l is change in angular momentum which is mathematically represented as

       \r l = m (\r r \ \ X  \ \ \r v)

Where X mean cross- product

   \r v is the velocity which is mathematically represented as

           \r v = \frac{d \r r }{dt}

Substituting for  \r r

           \r v = \frac{d }{dt} [ 4 t^2 \r i - (2t + 6t^2 ) \r j]

           \r v =  8t \r i - (2 + 12 t) \r j

Now the cross product of \r r \ and \ \r v is  mathematically evaluated as    

          \r r  \  \ X \ \ \r v = \left[\begin{array}{ccc}{\r i}&{\r j}&{\r k}\\{4t^2}&{-2t -6t^2}&0\\{8t}&{-2 -12t}&0\end{array}\right]

                       = 0 \r i + 0 \r j + (- 8t^2 -48t^3 + 16t^2 + 48t^3 ) \r k

                      \r r \ \  X \ \ \r v = 8t^2 \r k

So the angular momentum becomes

       \r l = m (8t^2 \r k)

Substituting for m

      \r l = 3 *  (8t^2 \r k)

      \r l =24t^2  \r k

Substituting into equation for torque

       \tau = \frac{d}{dt} [24t^2 \r k]

       \tau = 48t \r k

The magnitude of the angular momentum can be evaluated mathematically as

        |\r l| = \sqrt{(24 t^2) ^2}

        |\r l| = 24 t^2

From the is equation we see that the magnitude of the angular momentum is varies directly with square of the time so it would relative to the origin because at the origin t= 0s and we move out from origin t increases hence angular momentum increases also

4 0
2 years ago
Un pasager a intarziar la tren. Ajuns pe peron a observat trenul in miscare si a constatat ca unul dintre vagoane a trecut prin
ExtremeBDS [4]

b) intervalul de timp t3 in care prin fata sa trece vagonul urmator

Sper că am ajutat!

3 0
2 years ago
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