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3 years ago

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Two charged spheres are 20 cm apart and exert an attractive force of 8 x 10-9 n on each other. What will the force of attraction

be when the spheres are moved to 10 cm apart?

1 answer:

Pavel [41]3 years ago

6 0

Answer:

$3.2\cdot 10^{-8} N$

Explanation:

The inital electrostatic force between the two spheres is given by:

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

where

$F=8\cdot 10^{-9} N$ is the initial force

k is the Coulomb's constant

q1 and q2 are the charges on the two spheres

r is the distance between the two spheres

The problem tells us that the two spheres are moved from a distance of r=20 cm to a distance of r'=10 cm. So we have

$r'=\frac{r}{2}$

Therefore, the new electrostatic force will be

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

So the force has increased by a factor 4. By using $F=8\cdot 10^{-9} N$ , we find

$F'=4(8\cdot 10^{-9} N)=3.2\cdot 10^{-8} N$

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Answer:

a) $v=13.2171\,m.s^{-1}$

b) $H=8.9605\,m$

Explanation:

Given:

mass of bullet, $m=4.97\times 10^{-3}\,kg$

compression of the spring, $\Delta x=0.0476\,m$

force required for the given compression, $F=9.12 \,N$

(a)

We know

$F=m.a$

where:

a= acceleration

$9.12=4.97\times 10^{-3}\times a$

$a\approx 1835\,m.s^{-2}\\$

we have:

initial velocity, $u=0\,m.s^{-1}$

Using the eq. of motion:

$v^2=u^2+2a.\Delta x$

where:

v= final velocity after the separation of spring with the bullet.

$v^2= 0^2+2\times 1835\times 0.0476$

$v=13.2171\,m.s^{-1}$

(b)

Now, in vertical direction we take the above velocity as the initial velocity "u"

so,

$u=13.2171\,m.s^{-1}$

∵At maximum height the final velocity will be zero

$v=0\,m.s^{-1}$

Using the equation of motion:

$v^2=u^2-2g.h$

where:

h= height

g= acceleration due to gravity

$0^2=13.2171^2-2\times 9.8\times h$

$h=8.9129\,m$

is the height from the release position of the spring.

So, the height from the latched position be:

$H=h+\Delta x$

$H=8.9129+0.0476$

$H=8.9605\,m$

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3 years ago

A student weighing 120 lbs climbs a 12 ft flight of stairs in 9 seconds. how much power did the student create?

Alex Ar [27]

Power can be calculate through the equation,
Power = Force x velocity

It should be noted that velocity is calculated by dividing displacement by time. Thus, from the given in this item we can calculate for the power.
Power = (120 lb) x (12 ft/9 s)
<em> </em><span><em>Power = 160 lb.ft/s</em></span>

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3 years ago

Figure 1 shows the motion of three balls. The curved paths followed by balls B and Care examples of

nlexa [21]

Answer:

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3 years ago

The energy of an object can be converted to heat due to the friction of the car on the hill. The difference between the potentia

MAVERICK [17]

Answer:

Energy Lost for group A's car = 0.687 J

Energy Lost for group B's car = 0.55 J

Explanation:

The exact question is as follows :

Given - The energy of an object can be converted to heat due to the friction of the car on the hill. The difference between the potential energy of the car and its kinetic energy at the bottom of the hill equals the energy lost due to friction.

To find - How much energy is lost due to heat for group A's car ?

How much for Group B's car ?

Solution -

We know that,

GPE = 1 Joule (Potential Energy)

Now,

For Group A -

Energy Lost = GPE - KE

= 1 J - 0.313 J

= 0.687 J

So,

Energy Lost for group A's car = 0.687 J

Now,

For Group B -

Energy Lost = GPE - KE

= 1 J - 0.45 J

= 0.55 J

So,

Energy Lost for group B's car = 0.55 J

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3 years ago

A person is standing on a raft; their

krok68 [10]

Answer:

The volume of water displaced by the raft is 0.233 m³

Explanation:

The question relates to Archimedes' principle which states that the buoyant force experienced by an object immersed in a fluid is equal to the weight of (the force of gravity on) the displaced fluid

The given parameters are;

The combined mass of the person and the raft, m = 233 kg

The liquid on which the raft is located = Water

The density of water, $\rho _{water}$ = 1000 kg/m³

Weight = Mass, m × g

Where;

m = The mass of the object

g = The acceleration due to gravity = 9.8 m/s²

Given that the raft is on the surface of the water (floating), the buoyant force is equal to the combined weight of the person and the raft = 233 kg

The combined weight of the person and the raft, $W_{combined}$ = 233 kg × 9.8 m/s² = 2,283.4 N

Therefore;

The buoyant force = 2,283.4 N = The weight of the water displaced

The mass of the water displaced, $m_{water}$ , = 2,283.4 N/(9.8 m/s²) = 233 kg

Density = Mass/Volume

The volume of water displaced by the raft = The mass of the water displaced/(The density of the water) = 233 kg/(1,000 kg/m³) = 0.233 m³.

3 0

2 years ago

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