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denpristay [2]
3 years ago
9

Two boxes sit side by side on a smooth horizontal surface. The lighter box 5.2 kg, the heavier box has a mass of 7.4kg (a) find

the contact force between these boxes when a horizontal force of 5.0 N is applied to the light box. (Hint: you need to calculate acceleration first)
Physics
1 answer:
Paladinen [302]3 years ago
8 0

OK.  So you're pushing on the small box, and on the other side of it, the small
box is pushing on the big box. So you're actually pushing both of them.

-- The total mass that you're pushing is (5.2 + 7.4) = 12.6 kg.

-- You're pushing it with 5.0N of force.

-- Acceleration of the whole thing = (force)/(mass) = 5/12.6 = <em>0.397 m/s²</em> (rounded)

-- Both boxes accelerate at the same rate. So the box farther away from you ...
the big one, with 7.4 kg of mass, accelerates at the same rate.

The force on it to make it accelerate is (mass) x (acceleration) =

                                                              (7.4 kg) x (5/12.6 m/s²) =  <em>2.936 N.</em>

The only force on the big box comes from the small box, pushing it from behind. 
So that same  <em>2.936N</em>  must be the contact force between the boxes.

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Which of the following is same dimension quantity?
hram777 [196]

Answer:

Find the dimension of each and every quantity in all the options to check whether they are the same or not. We can use any one formula of each identity to find its dimension.

Complete step by step solution:

To find the dimension of a quantity, we can use any formula related to that quantity but we will use the easiest ones to save time.

Force-

from Newton’s law of motion,

F=maF=ma

Dimension of force =[M][LT−2]=[MLT−2]=[M][LT−2]=[MLT−2]

Work done-

W=F×sW=F×s

Dimension of work=[MLT−2][L]=[ML2T−2]=[MLT−2][L]=[ML2T−2]

Momentum-

p=mvp=mv

Dimension of momentum=[M][LT1]=[MLT−1]=[M][LT1]=[MLT−1]

Impulse-

I=F×tI=F×t

Dimension of impulse=[MLT−2][T]=[

3 0
3 years ago
Red Riding Hood sat down to rest and placed her 1.20-kg basket of goodies beside her. A wolf came along, spotted the basket, and
anygoal [31]

Answer:

  117.6°

Explanation:

The vertical component of a force directed at some angle α from the vertical is ...

  F·cos(α)

We want the vertical components of the wolf's force (Fw) and Red's force (Fr) to total zero. So for some angle from vertical α, Red's force will satisfy ...

  Fw·cos(25°) + Fr·cos(α) = 0

  cos(α) = -Fw/Fr·cos(25°) ≈ -(6.4 N)/(12.5 N)·0.906308 ≈ -0.464030

  α ≈ arccos(-0.464030) ≈ 117.6°

Red was pulling at an angle of about 117.6° from the vertical.

_____

<em>Additional comment</em>

That's about 27.6° below the horizontal.

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Read 2 more answers
Two stationary point charges of 3.00 nC and 2.00 nC are separated by a distance of 50.0 cm. An electron is released from rest at
andrew-mc [135]

Answer:

1. the electric potential energy of the electron when it is  at the midpoint is - 2.9 x 10^{-17} J

2. the electric potential energy of the electron when it is 10.0 cm from the 3.00 nC charge is - 5.04 x  10^{-17} J

Explanation:

given information:

q_{1} =  3 nC = 3 x 10^{-9} C

q_{2} =  2 nC = 2 x 10^{-9} C

r = 50 cm = 0.5 m

the electric potential energy of the electron when it is  at the midpoint

potential energy of the charge, F

F = k \frac{q_{e}q}{r}

where

k = constant (8.99 x 10^{9} Nm^{2} /C^{2})

electron charge, q_{e} = - 1.6 x 10^{-19} C

since it is measured at the midpoint,

r = \frac{0.5}{2}

  = 0.25 m

thus,

F = F_{1}+ F_{2}

  = k\frac{q_{e} q_{1} }{r} + k\frac{q_{e} q_{2} }{r}

  = \frac{kq_{e} }{r} (q_{1} +q_{2})

  = (8.99 x 10^{9})( - 1.6 x 10^{-19} )(3 x 10^{-9} +2 x 10^{-9})/0.25

  = - 2.9 x 10^{-17} J

the electric potential energy of the electron when it is 10.0 cm from the 3.00 nC charge

r_{1} = 10 cm = 0.1 m

r_{2} = 0.5 - 0.1 = 0.4 m

F = k\frac{q_{e} q_{1} }{r} + k\frac{q_{e} q_{2} }{r}

  = kq_{e}(\frac{q_{1} }{r_{1} }+\frac{q_{2} }{r_{2} })

  = (8.99 x 10^{9})( - 1.6 x 10^{-19} )(3 x 10^{-9} /0.1+2 x 10^{-9}/0.4)

  = - 5.04 x  10^{-17} J

3 0
3 years ago
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