Resolve the weight of the box to find the component of the weight acting parallel to the slope.

A professor designing a class demonstration connects a parallel-plate capacitor to a battery, so that the potential difference b

Lesechka [4]

Answer:

a) Q = 397.57 pC , Q = 3.18 104 pC , b) C = 1.157 10⁻¹⁰ F , V = 3.4375 V ,

c) U = 54.7 nJ , d) ΔU = 54 nJ,

Explanation:

a) The capacity of a capacitor is defined

C = Q / V

Q = C V

can also be calculated using geometry consideration

C = e or A / d

we reduce to the SI system

A = 25.0 cm² (1 m / 10² cm) 2 = 25.0 10⁻⁴ m²

d = 1.53 cm = 1.53 10⁻² m

we substitute

Q = eo A / d V

Q = 8.85 10⁻¹² 25 10⁻⁴ / 1.53 10⁻² 275

Q = 3.9757 10⁻¹⁰ C

let's reduce to pC

Q = 3.9757 10⁻¹⁰ C (10¹² pC / 1 C)

Q = 397.57 pC

when the capacitor is introduced into the water the dielectric constant is different

Q = k Q₀

Q = 80 397.57

Q = 3.18 104 pC

b) Find capacitance and voltage after submerged in water

C = k C₀

C = 80 8.85 10⁻¹² 25 10⁻⁴ / 1.53 10⁻²

C = 1.157 10⁻¹⁰ F

V = Vo / k

V = 275/80

V = 3.4375 V

c) The stored energy is

U = ½ C V²

U = ½, 85 10⁻¹² 25 10⁻⁴ / 1.53 10⁻² 275²

U = 5.47 10⁻⁸ J

let's reduce to nJ

109 nJ = 1 J

U = 54.7 nJ

d) energy after submerging

U = ½ (kCo) (Vo / k) 2

U = ½ Co Vo2 / k

U = U₀ / k

U = 54.7 / 80 nJ

U = 0.68375 nJ

the energy change is

ΔU = U₀ -U

ΔU = 54.7 - 0.687375

6 0

3 years ago

The variation in the pressure of helium gas, measured from its equilibrium value, is given by ΔP = 2.9 × 10−5 cos (6.20x − 3 000

nadya68 [22]

Answer:

The wavelength of this wave is 1.01 meters.

Explanation:

The variation in the pressure of helium gas, measured from its equilibrium value, is given by :

$\Delta P=2.9\times 10^{-5}\ cos(6.2x-3000t)$ ..............(1)

The general equation is given by :

$\Delat P=P_o\ cos(kx-\omega t)$ ...........(2)

On comparing equation (1) and (2) :

$k=6.2$

Since, $k=\dfrac{2\pi}{\lambda}$

$\dfrac{2\pi}{\lambda}=6.2$

$\lambda=1.01\ m$

So, the wavelength of this wave is 1.01 meters. Hence, this is the required solution.

6 0

3 years ago

PLEASE HELP!

Kobotan [32]

Answer:

No

Explanation:

Cause a monster truck don

3 0

3 years ago

A puck of mass 0.110 kg slides across ice in the positive x-direction with a kinetic friction coefficient between the ice and pu

lara [203]

Answer:

a) Ffr = -0.18 N

b) a= -1.64 m/s2

c) t = 9.2 s

d) x = 68.7 m.

e) W= -12.4 J

f) Pavg = -1.35 W

g) Pinst = -0.72 W

Explanation:

a)

While the puck slides across ice, the only force acting in the horizontal direction, is the force of kinetic friction.
This force is the horizontal component of the contact force, and opposes to the relative movement between the puck and the ice surface, causing it to slow down until it finally comes to a complete stop.
So, this force can be written as follows, indicating with the (-) that opposes to the movement of the object.

$F_{frk} = -\mu_{k} * F_{n} (1)$

where μk is the kinetic friction coefficient, and Fn is the normal force.

Since the puck is not accelerated in the vertical direction, and there are only two forces acting on it vertically (the normal force Fn, upward, and the weight Fg, downward), we conclude that both must be equal and opposite each other:

$F_{n} = F_{g} = m*g (2)$

We can replace (2) in (1), and substituting μk by its value, to find the value of the kinetic friction force, as follows:

$F_{frk} = -\mu_{k} * F_{n} = -0.167*9.8m/s2*0.11kg = -0.18 N (3)$

b)

According Newton's 2nd Law, the net force acting on the object is equal to its mass times the acceleration.
In this case, this net force is the friction force which we have already found in a).
Since mass is an scalar, the acceleration must have the same direction as the force, i.e., points to the left.
We can write the expression for a as follows:

$a= \frac{F_{frk}}{m} = \frac{-0.18N}{0.11kg} = -1.64 m/s2 (4)$

c)

Applying the definition of acceleration, choosing t₀ =0, and that the puck comes to rest, so vf=0, we can write the following equation:

$a = \frac{-v_{o} }{t} (5)$

Replacing by the values of v₀ = 15 m/s, and a = -1.64 m/s2, we can solve for t, as follows:

$t =\frac{-15m/s}{-1.64m/s2} = 9.2 s (6)$

d)

From (1), (2), and (3) we can conclude that the friction force is constant, which it means that the acceleration is constant too.
So, we can use the following kinematic equation in order to find the displacement before coming to rest:

$v_{f} ^{2} - v_{o} ^{2} = 2*a*\Delta x (7)$

Since the puck comes to a stop, vf =0.
Replacing in (7) the values of v₀ = 15 m/s, and a= -1.64 m/s2, we can solve for the displacement Δx, as follows:

$\Delta x = \frac{-v_{o}^{2}}{2*a} =\frac{-(15.0m/s)^{2}}{2*(-1.64m/s2} = 68.7 m (8)$

e)

The total work done by the friction force on the object , can be obtained in several ways.
One of them is just applying the work-energy theorem, that says that the net work done on the object is equal to the change in the kinetic energy of the same object.
Since the final kinetic energy is zero (the object stops), the total work done by friction (which is the only force that does work, because the weight and the normal force are perpendicular to the displacement) can be written as follows:

$W_{frk} = \Delta K = K_{f} -K_{o} = 0 -\frac{1}{2}*m*v_{o}^{2} =-0.5*0.11*(15.0m/s)^{2} = -12.4 J (9)$

f)

By definition, the average power is the rate of change of the energy delivered to an object (in J) with respect to time.
$P_{Avg} = \frac{\Delta E}{\Delta t} (10)$
If we choose t₀=0, replacing (9) as ΔE, and (6) as Δt, and we can write the following equation:

$P_{Avg} = \frac{\Delta E}{\Delta t} = \frac{-12.4J}{9.2s} = -1.35 W (11)$

g)

The instantaneous power can be deducted from (10) as W= F*Δx, so we can write P= F*(Δx/Δt) = F*v (dot product)
Since F is constant, the instantaneous power when v=4.0 m/s, can be written as follows:

$P_{inst} =- 0.18 N * 4.0m/s = -0.72 W (12)$

7 0

3 years ago

Which type of plate boundary or zone would be most likely to lead to above-ground volcanic activity?

Anna007 [38]

Answer:

Convergent plate boundary

Explanation:

The convergent plate boundary refers to the type of boundary where two plates move towards each other. Due to this type of motion, there forms a subduction zone, where the denser plate subducts below the lighter plate. This zone of subduction is commonly identified by the presence of a deep and narrow V-shaped depression which is commonly known as the oceanic trench.

When the subducting plate enters into the region of the asthenosphere, the rocks melt and mix with the magma. This magma is then pushed upward due to the force exerted by the convection current that forms in the mantle, and further reaches the over-riding plate and eventually give rise to the formation of volcanoes and volcanic/island arcs.

Thus, this type of plate boundary is responsible for the formation of above-ground volcanic activities.

6 0

3 years ago