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2 years ago

10

Identify and explain one way in which we try to deal with infiltration? What are the benefits and cost of that particular soluti

on?

1 answer:

tigry1 [53]2 years ago

5 0

Answer:

i vigilate

Explanation:

You might be interested in

You have an electrically neutral toy that you divide into two pieces. You notice that at least one of those pieces has an electr

den301095 [7]

Answer:

The pieces will attract one another

Explanation:

From the law of conservation of energy, we know that energy can neither be created nor destroyed, but transformed. If one piece of the toy that was neutral ends up having an electric charge (positive or negative), from the conservation of energy, the other piece must have a charge opposite to that on the other charged piece but equal in magnitude. These two pieces which are oppositely charged attracts each other, this shows that electric charge is conserved.

8 0

3 years ago

Find the energy in joules required to lift a 55.0 megagram object a distance of 500cm

pantera1 [17]

1,000 grams = 1 kilogram
so 55 megagrams = 55,000 kilograms

100 cm = 1 meter
so 500 cm = 5 meters

Acceleration of gravity on Earth = 9.8 m/s²

Weight = (mass) x (gravity)

========================================

Work = increase in potential energy =

   (weight) x (height) =

   (mass) x (gravity) x (height) =

   (55,000 kg) x (9.8 m/s²) x (5 m) =

   2,695,000 joules .

5 0

3 years ago

An Atwood machine consists of two masses, mA = 6.8 kg and mB = 8.0 kg , connected by a cord that passes over a pulley free to ro

Lisa [10]

To solve this problem it is necessary to apply the concewptos related to Torque, kinetic movement and Newton's second Law.

By definition Newton's second law is described as

F= ma

Where,

m= mass

a = Acceleration

Part A) According to the information (and as can be seen in the attached graph) a sum of forces is carried out in mass B, it is obtained that,

$\sum F = m_b a$

$m_Bg-T_B = m_Ba$

$T_B = m_Bg-m_Ba$

In the case of mass A,

$\sum F = m_A a$

$T_A = m_Ag-m_Aa$

Making summation of Torques in the Pulley we have to

$\sum\tau = I\alpha$

$T_BR_0-T_AR_0=I\alpha$

$T_B-T_A=I\frac{a}{R^2_0}$

Replacing the values previously found,

$(m_Bg-m_Ba )-(m_Ag-m_Aa )=I\frac{a}{R^2_0}$

$(m_B-m_A)g-(m_B+m_A)a=I\frac{a}{R^2_0}$

$a = \frac{(m_B-m_A)g}{\frac{I}{R_0^2}+(m_B+m_A)}$

$a = \frac{(m_B-m_A)g}{\frac{MR^2_0^2/2}{R_0^2}+(m_B+m_A)}$

$a =\frac{(m_B-m_A)g}{\frac{M}{2}+(m_B+m_A)}$

Replacing with our values

$a =\frac{(8-6.8)(9.8)}{\frac{0.8}{2}+(8+6.8)}$

$a=0.7736m/s^2$

PART B) Ignoring the moment of inertia the acceleration would be given by

$a' =\frac{(m_B-m_A)g}{(m_B+m_A)}$

$a' =\frac{(8-6.8)(9.8)}{(8+6.8)}$

$a' = 0.7945$

Therefore the error would be,

$\%error = \frac{a'-a}{a}*100$

$\%error = \frac{0.7945-0.7736}{0.7736}*100$

$\%error = 2.7%$

8 0

2 years ago

1. Consider the point exactly halfway between the two wires. Can you adjust the current so that, with current passing through ea

chubhunter [2.5K]

Answer:

hello your question is incomplete attached below is missing part of the question

answer:

1 ) Magnetic field due to long current carrying wire : $B = \frac{U_{0} I}{2\pi d}$

Therefore the net magnetic field due the both wires ; B = B $_{1}$ + B $_{2}$ . when we adjust the current I $_{1}$ = I $_{2}$ then the Netfield (B ) = zero

2) The distance between the field lines are not equally spaced and this is because the separation between field lines increases with the increase in the distance between the wires

3) Increase in current through the wire will lead to increase in force and this can be explained via this equation

$F = \frac{U_{0}I_{1}I_{2} }{2\pi d }$

Explanation:

1 ) Magnetic field due to long current carrying wire : $B = \frac{U_{0} I}{2\pi d}$

Therefore the net magnetic field due the both wires ; B = B $_{1}$ + B $_{2}$ . when we adjust the current I $_{1}$ = I $_{2}$ then the Netfield (B ) = zero

2) The distance between the field lines are not equally spaced and this is because the separation between field lines increases with the increase in the distance between the wires

3) Increase in current through the wire will lead to increase in force and this can be explained via this equation

$F = \frac{U_{0}I_{1}I_{2} }{2\pi d }$

8 0

2 years ago

Find the product of 30−−√ and 610−−√. Express it in standard form (i.

Helen [10]

The product of √30 and √610 is 10√183.

√30 = √(2×3×5)
and √610 = √(2×5×61

Since 61 can't be factorised further.
Therefore, the value of √30×√610 is
= √(2×3×5×2×5×61)
= 2×5×√(3×61)
=10√183

4 0

3 years ago

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