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liraira [26]
3 years ago
11

Two charged objects separated by some distance attract each other. If the charges on both objects are doubled with no change in

distance between them, what happens to the magnitude of the force between the objects?
a. the force between them quadruples
b. the between them doubles
c. the between them is halved
d. the force between them is quartered
e. the force between them is unchanged
Physics
1 answer:
beks73 [17]3 years ago
4 0

Answer:

a. the force between them quadruples

Explanation:

The electrostatic force between two charges is given by

F=k\frac{q_1 q_2}{r^2}

where

k is the Coulomb's constant

q1 and q2 are the two charges

r is the separation between the two charges

In this problem, the charges on both objects are doubled, so

q_1' = 2q_1\\q_2' = 2q_2

While the distance does not change, so the new force will be

F'= k \frac{(2q_1)(2q_2)}{r^2}=4 (k\frac{q_1 q_2}{r^2})=4 F

so, the force will quadruple.

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3 years ago
26. A 40 kg boy jumps from a height of 4m onto a plate-form mounted on springs. As the
denpristay [2]

Answer:

c. 1600J

Explanation:

The loss in potential energy of the boy is given by:

U=mg \Delta h

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Substituting, we find

\Delta U = (40 kg)(9.8 m/s^2)(4.02 m) = 1577 J \sim 1600 J

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3 years ago
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5 0
3 years ago
What derived unit is used to measure the slope of the line in this graph?
Alexxx [7]

Answer:

g/cm³

Explanation:

From the question given above,

The y-axis is representing mass (g)

The x-axis is representing volume (cm³)

Unit of slope =?

Slope of a graph is simply defined as the change in y-axis divided by the change in x-axis. Mathematically it is expressed as:

Slope = change in y-axis (Δy)/change in x-axis (Δx)

Slope = Δy/Δx

Thus, with the above formula, we can obtain the unit used for measuring the slope as follow:

y-axis = mass (g)

x-axis = volume (cm³)

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5 0
3 years ago
ASTM A229 oil-tempered carbon steel is used for a helical coil spring. The spring is wound with D = 50 mm d = 10.0 mm, and a pit
horrorfan [7]

Answer

given,

D = 50 mm = 0.05 m

d = 10 mm = 0.01 m

Force to compress the spring

F = \dfrac{d^4G\delta}{8D^3N}

\dfrac{\delta}{N} = p - d = 14 - 10 = 4 mm

F = \dfrac{d^4G}{8D^3}\times 0.004

F = \dfrac{0.1^4\times 79\times 10^9}{8\times 0.05^3}\times 0.004

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stress correction factor from stress correction curve is equal to 1.1

now, calculation of corrected stress

\tau = \dfrac{8FDk_s}{\pi d^3}

\tau = \dfrac{8\times 3160 \times 0.05 \times 1.1}{\pi \times 0.01^3}

              = 442.6 Mpa

The tensile strength of the steel material of  ASTM A229 is equal to 1300 Mpa

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since corrected stress is less than the \tau_s

hence, spring will return to its original shape.

6 0
3 years ago
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