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MA_775_DIABLO [31]
3 years ago
5

In this problem, you will analyze a system composed of two blocks, 1 and 2, of respective masses m1 and m2. To simplify the anal

ysis, we will make several assumptions:
The blocks can move in only one dimension, namely, along the x axis.
The masses of the blocks remain constant.
The system is closed.
At time t, the x components of the velocity and the acceleration of block 1 are denoted by v1(t) and a1(t). Similarly, the x components of the velocity and acceleration of block 2 are denoted by v2(t)and a2(t). In this problem, you will show that the total momentum of the system is not changed by the presence of internal forces.
Find p(t), the x component of the total momentum of the system at time t. Express your answer in terms of m1, m2, v1(t), and v2(t).
Part B
Find the time derivative dp(t)/dt of the x component of the system's total momentum.
Express your answer in terms of a1(t), a2(t), m1, and m2.
Part C
The quantity ma (mass times acceleration) is dimensionally equivalent to which of the following?
momentum
energy
force
acceleration
inertia
Acceleration is due to which of the following physical quantities?
velocity
speed
energy
momentum
force
Since we have assumed that the system composed of blocks 1 and 2 is closed, what could be the reason for the acceleration of block 1?
the large mass of block 1
air resistance
Earth's gravitational attraction
a force exerted by block 2 on block 1
a force exerted by block 1 on block 2
Let us denote the x component of the force exerted by block 1 on block 2 by , and the x component of the force exerted by block 2 on block 1 by . Which of the following pairs equalities is a direct consequence of Newton's second law?
F12=m2a2 and F21=m1a1
F12=m1a1 and F21=m2a2
F12=m1a2 and F21=m2a1
F12=m2a1 and F21=m1a2
Let us recall that we have denoted the force exerted by block 1 on block 2 by , and the force exerted by block 2 on block 1 by . If we suppose that is greater than , which of the following statements about forces is true?
|F12|>|F21|
|F21|>|F12|
Both forces have equal magnitudes.
Now recall the expression for the time derivative of the x component of the system's total momentum: . Considering the information that you now have, choose the best alternative for an equivalent expression to .
0
nonzero constant
kt
kt2
Physics
1 answer:
Dafna1 [17]3 years ago
7 0

Answer:

a) p = m1 v1 + m2 v2 , b) dp / dt = m1 a1 + m2 a2 , c) It is equivalent to force

dp / dt = 0

Explanation:

In this problem we have two blocks and the system is formed by the two bodies.

Part A. Initially they ask us to find the moment of the whole system

    p = m1 v1 + m2 v2

Part B.

Find the derivative

     dp / dt = m1 dv1dt + m2 dv2 / dt

     dp / dt = m1 a1 + m2 a2

Part C.

Let's analyze the dimensions

     m a = [kg] [m / s2] = [N]

It is equivalent to force

Part d

Acceleration is due to a net force applied

Part e

The acceleration of block 1 is due to the force exerted by block 2 during the moment change

Part f

Force of block 1 on block 2

True f12 = m1a1        f21 = m2a2

Part g

By the law of action and reaction are equal magnitude F12 = f21

Part H

     dp / dt = 0

Isolated system F12 = F21 and the masses are constant. The total moment is only redistributed

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Two very large parallel sheets are 5.00 cm apart. sheet a carries a uniform surface charge density of -9.70 μc/m2 , and sheet b,
MAVERICK [17]
Question is missing. Found on internet the complete text of the problem:

"<span>Two very large parallel sheets are 5.00 cm apart. Sheet A carries a uniform surface charge density of −9.70µC/m2, and sheet B, which is to the right or A, carries a uniform charge density of −11.5 µC/m2. Assume the sheets are large enough to be treated as infinite. Find the magnitude and direction of the net electric field these sheets produce at a point (a) 4.00 cm to the right of sheet A; (b) 4.00 cm to the left of sheet A; (c) 4.00 cm to the right of sheet B."

Solution:

(a) The electric field produced by a uniformly charged sheet at any distance is given by
</span>E= \frac{\sigma}{2 \epsilon _0}
where \sigma is the charge density and \epsilon _0 = 8.85\cdot 10^{-12} F/m is the vacuum permittivity.

First of all, let's compute the fields generated by the two sheets separately. The two densities of charge are \sigma_A = -9.70 \mu C/m^2=-9.70\cdot 10^{-6}C/m^2 and \sigma_B = -11.5 \mu C/m^2 = -11.5\cdot 10^{-6} C/m^2.

Sheet a gives an electric field of
E_A= \frac{\sigma_A}{2\epsilon _0}= \frac{-9.7\cdot 10^{-6} C/m^2}{2\cdot 8.85\cdot 10^{-12} F/m} = -5.48\cdot 10^5 V/m
where the negative sign means the field points towards sheet A, in any point of the space.

The electric field produced by sheet B is given by:
E_B = \frac{\sigma_A}{2\epsilon _0}= \frac{-11.5\cdot 10^{-6} C/m^2} {2\cdot 8.85\cdot 10^{-12} F/m} =-6.50\cdot 10^{5} V/m
and again, the negative sign means that the field at any point of the space points towards sheet B.

The point at which we have to compute the total field is at 4.00 cm right of sheet A. Since the two sheets are 5.00cm far apart, it means that this point is between the two sheets. Therefore, in this point the two fields point into opposite directions. Therefore, the total field is
E=E_1-E_2= -5.48\cdot 10^5 V/m - (-6.50\cdot 10^{5} V/m)=1.02\cdot 10^5 V/m
And the direction is towards sheet B, since it has a field with stronger intensity.

(b) Field at 4.00 cm to the left of sheet A: in this point of the space, the two fields point towards same direction (on the right, towards both sheet A and sheet B). So, the total field is simply the sum of the two fields:
E=E_1+E_2=-11.98\cdot 10^5 V/m
towards right.

(c) Field at 4.00 cm to the right of sheet B. As before, the two fields in this point have same direction (both towards left, pointing towards both sheet A and sheet B). And so, the total field is simply the sum of the two fields:
E=E_1+E_2=11.98 \cdot 10^5 V/m
towards left.
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The return-air ventilation duct in a home has a cross-sectional area of 900 cm^2. The air in a room that has dimensions 7.0 m ××
koban [17]

Answer:

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Explanation:

A = cross-sectional area of the duct = 900 cm² = 900 x 10⁻⁴ m²

v = speed of air in the duct

t = time period of circulation = 40 min = 40 x 60 sec = 2400 sec

V = Volume of the air in the room = volume of room = 7 x 11 x 2.4 = 184.8 m³

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