10. Most reabsorption occurs in the

What name is given to the turning effect of a force?

Alex787 [66]

Answer:

The turning effect of a force is known as the moment. It is the product of the force multiplied by the perpendicular distance from the line of action of the force to the pivot or point where the object will turn. When undoing a nut fastened to a screw by hand one realises that the amount of force required is a lot greater than when undoing the same nut using a spanner.

Explanation:

6 0

3 years ago

At what latitudd is the coriolis effect the smallest

alexandr1967 [171]

The Coriolis effect is strongest at the north and south poles, and zero on the equator (zero latitude).

6 0

4 years ago

If a box is pushed across a floor with a force of 130N. The frictional force acting between the box and the floor is 30N, over a

Sidana [21]

The force acting in the front direction is the 130N.
The frictional force is acting backwards 30N.

1) The net force is 130N - 30N = 100N

2) s = ut + (1/2)at^2    u = 0, Start from rest, s = 25m t =5.

25 = 0*5 + (1/2)* a * 5^2.

25 = 0 + 25/2 * a.

25 = (25/2)a. Divide 25 from both sides.

1 = (1/2)* a. Cross multiply.

2 = a.

a = 2 m/s^2.

3) Mass of the box
Net Force, F = ma
   100 = m*2. Divide both sides by 2.

   100/2 = m
   50    = m.
      m = 50 kg.

4) Final velocity ,   v = u + at.
   v = 0 + 2*5 = 10 m/s.

   Kinetic Energy, K = (1/2) * mv^2.
      = 1/2 * 50 * 10 * 10.
      = 2500 J.

3 0

3 years ago

Instruments on board the trmm (tropical rainfall measuring mission) satellite show 3d images of very tall rain columns called __

Ket [755]

These columns are called hot towers

7 0

3 years ago

The 20-g bullet is travelling at 400 m/s when it becomes embedded in the 2-kg stationary block. The coefficient of kinetic frict

nikklg [1K]

Answer:

The distance the block will slide before it stops is 3.3343 m

Explanation:

Given;

mass of bullet, m₁ = 20-g = 0.02 kg

speed of the bullet, u₁ = 400 m/s

mass of block, m₂ = 2-kg

coefficient of kinetic friction, μk = 0.24

Step 1:

Determine the speed of the bullet-block system:

From the principle of conservation of linear momentum;

m₁u₁ + m₂u₂ = v(m₁ + m₂)

where;

v is the speed of the bullet-block system after collision

(0.02 x 400) + (2 x 0) = v (0.02 + 2)

8 = v (2.02)

v = 8/2.02

v = 3.9604 m/s

Step 2:

Determine the time required for the bullet-block system to stop

Apply the principle of conservation momentum of the system

$v(m_1+m_2) -F_kt = v_f(m_1 +m_2)\\\\v(m_1+m_2) -N \mu_kt = v_f(m_1 +m_2)\\\\v(m_1+m_2) -g(m_1 +m_2) \mu_kt = v_f(m_1 +m_2)\\\\3.9604(2.02)-9.8(2.02)0.24t = v_f(2.02)\\\\8 - 4.751t = 2.02v_f\\\\3.9604 - 2.352t = v_f$

when the system stops, vf = 0

3.9604 -2.352t = 0

2.352t = 3.9604

t = 3.9604/2.352

t = 1.684 s

Thus, time required for the system to stop is 1.684 s

Finally, determine the distance the block will slide before it stops

From kinematic, distance is the product of speed and time

$S = \int\limits {v} \, dt \\\\S = \int\limits^t_0 {(3.9604-2.352t)} \, dt\\\\ S = 3.9604t - 1.176t^2$

Now, recall that t = 1.684 s

S = 3.9604(1.684) - 1.176(1.684)²

S = 6.6693 - 3.3350

S = 3.3343 m

Thus, the distance the block will slide before it stops is 3.3343 m

3 0

4 years ago

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