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2 years ago

8

A proton is located at the point (x = 1.0 nm, y = 0.0 nm) and an electron is located at the point (x = 0.0 nm, y = 4.0 nm). Find

the magnitude of the electrostatic force that each one exerts on the other. (k = 1/4πε0 = 9.0 × 109 N ∙ m2/C2, e = 1.6 × 10-19 C) A) 5.3 × 10-18 NB) 5.3 × 108 N C) 1.4 × 10-11 N D) 5.9 × 10-15 N

1 answer:

OLga [1]2 years ago

7 0

Answer:

C : 1.4*10^(-11) N.

Explanation:

q_1 = q_2 = 1.6 * 10^(-19)

R^2 = (1)^2 + (4)^2 = 1.7 * 10^(-17) m^2

The coulomb's law is as follows:

F_e = k*q_1*q_2 / R^2

F_e = k*q^2 / R^2

F_e = (9.0*10^9) * (1.6 * 10^(-19))^2 / 1.7 * 10^(-17)

F_e = 1.35 * 10^(-11) N

Hence, answer closes to obtained is option C : 1.4*10^(-11) N.

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Suppose a capacitor is fully charged by a battery and then disconnected from the battery. The positive plate has a charge +q and

dimaraw [331]

Answer:-q

Explanation:

Given

Capacitor is charged to a battery and capacitor acquired a charge of q i.e.

+q on Positive Plate and -q on negative Plate.

If the plate area is doubled and the plate separation is reduced to half its initial separation then capacitor becomes four times of initial value because capacitor is given by

$c=\epsilon_0 \cdot \frac{A}{d}$

where A=area of capacitor plate

d=Separation between plates

This change in capacitance changes the Potential such that new charge on the negative plate will remain same -q

8 0

3 years ago

A .5 kg air puck moves to the right at 3 m/s, colliding with a 1.5kg air puck that is moving to the left at 1.5 m/s.

arlik [135]

Answer:

part (a) v = 1.7 m/s towards right direction

part (b) Not an elastic collision

part (c) F = -228.6 N towards left.

Explanation:

Given,

Mass of the first puck = $m_1\ =\ 5\ kg$
Mass of the second puck = $m_2\ =\ 3\ kg$
initial velocity of the first puck = $u_1\ =\ 3\ m/s.$
Initial velocity of the second puck = $u_2\ =\ -1.5\ m/s.$

Part (a)

Pucks are stick together after the collision, therefore the final velocities of the pucks are same as v.

From the conservation of linear momentum,

$m_1u_1\ +\ m_2u_2\ =\ (m_1\ +\ m_2)v\\\Rightarrow v\ =\ \dfrac{m_1u_1\ +\ m_2u_2}{m_1\ +\ m_2}\\\Rightarrow v\ =\ \dfrac{5\times 3\ -\ 1.5\times 1.5}{5\ +\ 1.5}\\\Rightarrow v\ =\ 1.7\ m/s.$

Direction of the velocity is towards right due to positive velocity.

part (b)

Given,

Final velocity of the second puck = $v_2\ =\ 2.31\ m/s.$

Let $v_1$ be the final velocity of first puck after the collision.

From the conservation of linear momentum,

$m_1u_1\ +\ m_2u_2\ +\ m_1v_1\ +\ m_2v_2\\\Rightarrow v_1\ =\ \dfrac{m_1u_1\ +\ m_2u_2\ -\ m_2v_2}{m_1}\\\Rightarrow v_1\ =\ \dfrac{5\times 3\ -\ 1.5\times 1.5\ -\ 1.5\times 2.31}{5}\\\Rightarrow v_1\ =\ 1.857\ m/s.$

For elastic collision, the coefficient of restitution should be 1.

From the equation of the restitution,

$v_1\ -\ v_2\ =\ e(u_2\ -\ u_1)\\\Rightarrow e\ =\ \dfrac{v_1\ -\ v_2}{u_2\ -\ u_1}\\\Rightarrow e\ =\ \dfrac{1.857\ -\ 2.31}{-1.5\ -\ 3}\\\Rightarrow e\ =\ 0.1\\$

Therefore the collision is not elastic collision.

part (c)

Given,

Time of impact = t = $25\times 10^{-3}\ sec$

we know that the impulse on an object due to a force is equal to the change in momentum of the object due to the collision,

$\therefore I\ =\ \ m_1v_1\ -\ m_1u_1\\\Rightarrow F\times t\ =\ m_1(v_1\ -\ u_1)\\\Rightarrow F\ =\ \dfrac{m_1(v_1\ -\ u_1)}{t}\\\Rightarrow F\ =\ \dfrac{5\times (1.857\ -\ 3)}{25\times 10^{-3}}\\\Rightarrow F\ =\ -228.6\ N$

Negative sign indicates that the force is towards in the left side of the movement of the first puck.

3 0

3 years ago

Which of the following is right?

barxatty [35]

Scientific theory because it's a theory it's already an answer but it might change depending on the condition.

7 0

3 years ago

What property of matter determines how much inertia an object has

tester [92]

The mass of an object determines the object's inertia.

4 0

2 years ago

Consider a simple tension member that carries an axial load of P=22.44N. Find the total elongation in the member due to the load

rodikova [14]

Answer:

The total elongation for the tension member is of 0.25mm

Explanation:

Assuming that material is under a linear deformation then the relation between the stress and the specific elongation is given as:

$\sigma=E*\epsilon$ (1)

Where E is the modulus of elasticity, σ the stress and ε the specific deformation. Also, the total longitudinal elongation can be expressed as:

$\delta L=L*\epsilon$ (2)

Here L is the member extension and δL the change total longitudinal elongation.

Now if the stress is found then the deformation can be calculated by solving the stress-deformation equation (1). The stress applied sigama is computed dividing the axial load P by the cross-sectional area A:

$\sigma=P/A$

$\sigma=22.44N / 1290 mm^2$

$\sigma=0.0174 N/mm^2$

Solving for epsilon and replacing the calculated value for the stress and the value for the modulus of elasticity:

$=\sigma=E*\epsilon$

$\epsilon=\sigma/E$

$\epsilon=0.0174 \frac{N}{mm^2}/\ 204 \frac{N}{mm^2}$

$\epsilon=8.53*10^-{5}$

Finally introducing the specific deformation and the longitudinal extension in the equation of total elongation (2):

$\delta L=3048 mm * 8.53*10^{-5}$

$\delta L= 0.25 mm$

4 0

2 years ago