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3 years ago

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A stack of books rests on a level frictionless surface. A force F acts on the stack, and it accelerates at 3.0 m/s2. A 1.0 kg bo

ok is then added to the stack. The same force is applied, and now the stack accelerates at 2.0 m/s2.

1 answer:

Oxana [17]3 years ago

4 0

Answer:

m1 = 2 kg

m2 = 3kg

Explanation:

The force can be getting by

F = m * a

F1 = F2

a1 = 3.0 m/s^2

a2 = 2.0 m/s^2

The force F1=F2 because the force is applied so get the a2 acceleration

m1 * a1 = m2 * a2

m2 = m1 + 1kg

m1 *(3.0 m/s^2) = m2* (2.0 m/s^2)

m1 *(3.0 m/s^2) = (m1 + 1kg) * (2.0 m/s^2)

m1*(3.0m/s^2-2.0m/s^2)=2 kg*m/s^2

Solve to find the mass

m1 m/s^2= 2 kg*m/s^2

m1 = 2 kg

m2 = 3kg

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A 20 cm-radius ball is uniformly charged to 71 nC.

artcher [175]

Answer:

Part a)

$\rho = 2.12\mu C/m^3$

Part b)

$q_1 = 1.11 nC$

$q_2 = 8.88 nC$

$q_3 = 71 nC$

Part c)

$E_1 = 3996 N/C$

$E_2 = 7992 N/C$

$E_3 = 15975 N/C$

Explanation:

Part a)

As we know that charge density is the ratio of total charge and total volume

So here the volume of the charge ball is given as

$V = \frac{4}{3}\pi R^3$

$V = \frac{4}{3}\pi(0.20)^3$

$V = 0.0335 m^3$

now the charge density of the ball is given as

$\rho = \frac{71 nC}{0.0335} = 2.12\mu C/m^3$

Part b)

Now the charge enclosed by the surface is given as

$q = \rho V$

at radius of 5 cm

$q = (2.12 \mu C/m^3)(\frac{4}{3}\pi(0.05)^3$

$q = 1.11 nC$

at radius of 10 cm

$q = (2.12 \mu C/m^3)(\frac{4}{3}\pi(0.10)^3$

$q = 8.88 nC$

at radius of 20 cm

$q = 71 nC$

Part c)

As we know that electric field is given as

$E = \frac{kq}{r^2}$

so we have electric field at r = 5 cm

$E_1 = \frac{(9\times 10^9)(1.11 nC)}{0.05^2}$

$E_1 = 3996 N/C$

electric field at r = 10 cm

$E_2 = \frac{(9\times 10^9)(8.88 nC)}{0.10^2}$

$E_2 = 7992 N/C$

electric field at r = 20 cm

$E_3 = \frac{(9\times 10^9)(71 nC)}{0.20^2}$

$E_3 = 15975 N/C$

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