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Lelechka [254]
3 years ago
11

A mass of 0.40 kg is suspended on a spring which then stretches 10 cm. The mass is then removed and a second mass is placed on t

he spring which stretches the spring and does 20 J of work. How far is the spring stretched as a result of the work done by the second mass?
Physics
1 answer:
Vanyuwa [196]3 years ago
7 0

Answer:

 x' = 1.01 m

Explanation:

given,

mass suspended on the spring, m = 0.40 Kg

stretches to distance, x = 10 cm  = 0. 1 m

now,

we know

m g = k x

where k is spring constant

0.4 x 9.8 = k x 0.1

  k = 39.2 N/m

now, when second mass is attached to the spring work is equal to 20 J

work done by the spring is equal to

W = \dfrac{1}{2}kx'^2

20= \dfrac{1}{2}\times 39.2\times x'^2

 x'² = 1.0204

 x' = 1.01 m

hence, the spring is stretched to 1.01 m from the second mass.

 

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Which component of magma includes mobile ions of silicon?
Lemur [1.5K]

Answer:

Melt.

Explanation:

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When hydrogen is fused into helium, energy is released from Choose one: A. the increase in pressure. B. the decrease in the grav
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The nuclear fusion of hydrogen atoms releases a huge amount of energy. So the correct choice is C. Conversion of mass to energy.

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8 0
2 years ago
A third point charge q3 is now positioned halfway between q1 and q2. The net force on q2 now has a magnitude of F2,net = 4.444 N
Dmitriy789 [7]

Answer:

The value of third charge is 0.8μC.

Explanation:

Given that.

Magnitude of net force=4.444 N

According to figure,

Suppose, First charge = 2.4 μC

Second charge = 6.2 μC

Distance r₁ = 9.8 cm

Distance r₂ = 2.1 cm

We need to calculate the value of r

Using Pythagorean theorem

r=\sqrt{(r_{1})^2+(r_{2})^2}

Put the value into the formula

r=\sqrt{(9.8)^2+(2.1)^2}

r=10.02\ cm

We need to calculate the force

Using formula of force

F_{12}=\dfrac{kq_{1}q_{2}}{(r)^2}

Force F₁₂,

F_{12}=\dfrac{9\times10^{9}\times2.4\times10^{-6}\times6.2\times10^{-6}}{(10.02\times10^{-2})^2}

F_{12}=13.33\ N

F_{21}=-13.33\ N

Force F₂₃,

F_{23}=\dfrac{9\times10^{9}\times6.2\times10^{-6}\times q_{3}}{(10.02)^2}

We need to calculate the value of third charge

F_{net}=F_{21}+F_{23}

4.444=-13.33+\dfrac{9\times10^{9}\times6.2\times10^{-6}\times q_{3}}{(5.01)^2}

q_{3}=\dfrac{(4.444+13.33)\times(5.01\times10^{-2})^2}{9\times10^{9}\times6.2\times10^{-6}}

q_{3}=7.99\times10^{-7}\ C

q_{3}=0.8\times10^{-6}\ C

Hence, The value of third charge is 0.8μC.

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