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lidiya [134]
4 years ago
9

If it takes 600 N to move a box 5 meters, how much work is done on the box?

Physics
1 answer:
Arada [10]4 years ago
6 0

Answer:3000joules

Explanation: work=force×distance

Work=600×5

Work=3000joules

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How does changing the number of neutrons affect an atom?
katrin2010 [14]

Answer:

If you change the number of neutrons in an atom, you get an isotope of the same element. The addition of a neutron can make an atom radioactive. When you change the number of protons in an atom, you will change the atom from one element to a different element.

Explanation:

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3 years ago
An object exerts a reaction force when it
r-ruslan [8.4K]

Answer:

all the forces occur in pairs that if one object exerts a force on another object, then the second object exerts an equal and opposite reaction force on the first.

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3 years ago
Describe 4 common workplace practices that are risk factors for injury
Aleonysh [2.5K]
High task repitition, forceful exertions, repetitive or sustained awkward posture
7 0
4 years ago
If the moment acting on the cross section is M=630N⋅m, determine the maximum bending stress in the beam. Express your answer to
-BARSIC- [3]

Answer:

2.17 Mpa

Explanation:

The location of neutral axis from the top will be

\bar y=\frac {(240\times 25)\times \frac {25}{2}+2\times (20\times 150)\times (25+(\frac {150}{2}))}{(240\times 25)+2\times (20\times 150)}=56.25 mm

Moment of inertia from neutral axis will be given by \frac {bd^{3}}{12}+ ay^{2}

Therefore, moment of inertia will be

\frac {240\times 25^{3}}{12}+(240\times 25)\times (56.25-25/2)^{2}+2\times [\frac {20\times 150^{3}}{12}+(20\times 150)\times ((25+150/2)-56.25)^{2}]=34.5313\times 10^{6} mm^{4}}

Bending stress at top= \frac {630\times 10^{3}\times (175-56.25)}{34.5313\times 10^{6}}=2.1665127\approx 2.17 Mpa

Bending stress at bottom=\frac {630\times 10^{3}\times 56.25}{34.5313\times 10^{6}}=1.026242858\approx 1.03 Mpa

Comparing the two stresses, the maximum stress occurs at the bottom and is 2.17 Mpa

8 0
4 years ago
Three point charges are located on the x-axis. The first charge, q1 = 10 μC, is at x = -1.0 m. The second charge, q2 = 20 μC, is
victus00 [196]

Answer:

<em>3.15 N towards the positive x-axis</em>

<em></em>

Explanation:

first charge has charge q1 = 10 μC = 10 x 10^-6 C

second charge has charge q2 = 20 μC = 20 x 10^-6 C

third charge has charge q3 = -30 μC = -30 x 20^-6 C

According to coulomb's law, force between two charged particle is given as

F = \frac{-kQq}{r^2}

Where

F is the force between the charges

k is Coulomb's constant = 9 x 10^9 kg⋅m^3⋅s^−2⋅C^−2.

Q is the magnitude of one charge

q is the magnitude of the other charge

is the distance between these two charges

For the force on q2 due to q1,

distance r between them = 0 - (-1.0) = 1 m

F = \frac{-9*10^{9}*10*10^{-6}*20*10^{-6}}{1^2} = -1.8 N (the negative sign indicates a repulsion on q2 towards the positive  x-axis)

For the force on q2 due to q3,

distance between them = 2.0 - 0 = 2 m

F = \frac{-9*10^{9}*20*10^{-6}*(-30*10^{-6})}{2^2} = 1.35 N (the positive sign indicates an attraction on q2 towards the positive x-axis)

Resultant force on q2 = 1.8 N + 1.35 N = <em>3.15 N towards the positive x-axis</em>

3 0
3 years ago
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