A parallel-plate capacitor is constructed from two aluminum foils of 1 square centimeter area each placedon both sides of a rubber square of the same size. The rubber dielectric is 2.5 mm thick, hasr2.5, andbreakdown field strength of 25 megavolts per meter. Find the voltage rating of the capacitor using a safetyfactor of 10.

Answers

Answer 1

Answer:

The voltage will be 0.0125V

Explanation:

See the picture attached

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How fast can the car take this curve this curve without skidding to the outside of the curve?

Answers

Lets write the data down. That will help us solve the problem later:

R = 36 m

θ = 18º

m = 1492 kg

μ = 0.67

g = 9.8 m/s²

Lets draw all the forces that act on the car:

In order to the car won't skidding to the outside of the curve we must have the centripetal force equals the friction force:

$F_(cp)=f_a$

$(m.v^(2))/(R)=\mu.F_N$

A 75 kg man starts climbing a ladder that leans against a wall. If the weight of the ladder is negligible, determine how far up the ladder the man can climb before the ladders starts to slip. The coefficient of friction on both surfaces is μS=0.25

Answers

The man can climb $\bold { X (max) = 0.25* L* tan \alpha }$ , before the ladders starts to slip.

A - point at the top of the ladder

B - point at the bottom of the ladder

C - point where the man is positioned in the ladder

L- the length of the ladder

α - angle between ladder and ground

x - distance between C and B

The forces act on the ladder,

Horizontal reaction force (T) of the wall against the ladder

Vertical (upward) reaction force (R) of ground against the ladder.

Frictionalhorizontal ( to the left ) force (F)

Vertical( downwards) of the man,

mg = 75 Kg x 9.8 m/s² = 735 N

in static conditions,

∑Fx = T - F = 0 Since, T = F

∑Fy = mg - R = 0 Since, 735 - R = 0, R = 735

∑ Torques(b) = 0

In point B the torque produced by forces R and F is Zero

Then:

∑Torques(b) = 0

And the arm lever for each force,

mg = 735

Since, ∑Torques(b) = 0

$\bold {735* x* cos\alpha = F* L* sin\alpha }$ Since,T = F

$\bold {F = \frac {735* x* cos\apha }{L* sin\alpha }}$ $\bold { \frac {cos \alpha } { sin\alpha }= cot\alpha =\frac 1{tan\alpha}}$

$\bold {F = \frac {735* x* cos\apha }{L }}$

$\bold {F = 735* x* tan\alpha }}$

F < μR the ladder will starts slipping over the ground

μ(s) = 0.25

$\bold { X (max) = 0.25* L* tan \alpha }$

Therefore, the man can climb $\bold { X (max) = 0.25* L* tan \alpha }$ , before the ladders starts to slip. \

To know more about Torque,

brainly.com/question/6855614

Answer:

x (max) = 0,25*L*tanα

Explanation:

Letá call

A: point at the top of the ladder

B: the point at the bottom of the ladder

C: point where the man is up the ladder

L the length of the ladder

α angle between ladder and ground

"x" distance between C and B

Description

The following forces act on the ladder

Point A: horizontal (to the right) reaction (T) of the wall against the ladder

Point B : Vertical (upwards) reaction (R) of ground against the ladder

frictional horizontal ( to the left ) force (F)

Point C : Weight (vertical downwards)) of the man mg

mg = 75 Kg * 9,8 m/s² mg = 735 [N]

Then in static conditions:

∑Fx = T - F = 0 ⇒ T = F

∑Fy = mg - R = 0 ⇒ 735 - R = 0 ⇒ R = 735

∑Torques(b) = 0

Note: In point B the torque produced by forces R and F are equal to 0

Then:

∑Torques(b) = 0

And the arm lever for each force is:

mg = 735

d₁ for mg and d₂ for T

cos α = d₁/x then d₁ = x*cosα

sin α = d₂ / L then d₂ = L*sinα

Then:

∑Torques(b) = 0 ⇒ 735*x*cosα - T*L*sinα = 0

735*x*cosα = T*L*sinα

T = F then 735*x*cosα = F*L*sinα

F = (735)*x*cosα/L*sinα cos α / sinα = cotgα = 1/tanα

F = (735)*x*cotanα/L or F = (735)*x/L*tanα

When F < μ* R the ladder will stars slippering over the ground

μ(s) = 0,25 0,25*R = 735*x/L*tanα

x = 0,25*R*tanα*L/735

But R = mg = 735 then

0,25*L*tanα = x

Then x (max) = 0,25*L*tanα

What is the motion of the particles in this kind of wave? A hand holds the left end of a set of waves. The waves themselves make a larger set of waves in the same direction as that of the smaller waves. A label Wave motion is above the series of waves and an arrow next to the label points right. The particles will move up and down over large areas. The particles will move up and down over small areas. The particles will move side to side over small areas. The particles will move side to side over large areas.

Answers

Answer:

→A←

Explanation:

D its incorrect in edge

Answer:

Explanation:

The particles will move side to side over large areas

What happens to the pressure in all parts of a confined fluid if the pressure in one part is increased? The pressure in the other parts remains the same.The pressure everywhere increases by different amounts depending on the area of each part.
The pressure everywhere increases by the same amount.
The pressure everywhere decreases to conserve total pressure.

Answers

Answer:

option C

Explanation:

the correct answer is option C

When in a confined fluid the pressure is increased in one part than the pressure will equally distribute in the whole system.

According to Pascal's law when pressure is increased in the confined system then the pressure will equally transfer in the whole system.

This law's application is used in machines like hydraulic jacks.

A river 500 ft wide flows with a speed of 8 ft/s with respect to the earth. A woman swims with a speed of 4 ft/s with respect to the water.1) If the woman heads directly across the river, how far downstream is she swept when she reaches the opposite bank?
2) If she wants to be swept a smaller distance downstream, she heads a bit upstream. Suppose she orients her body in the water at an angle of 37° upstream (where 0° means heading straight accross, how far downstream is she swept before reaching the opposite bank?
3) For the conditions, how long does it take for her to reach the opposite bank?

Answers

Answer:

1) $\Delta s=1000\ ft$

2) $\Delta s'=998.11\ ft.s^(-1)$

3) $t\approx125\ s$

$t'\approx463.733\ s$

Explanation:

Given:

width of river, $w=500\ ft$

speed of stream with respect to the ground, $v_s=8\ ft.s^(-1)$

speed of the swimmer with respect to water, $v=4\ ft.s^(-1)$

Now the resultant of the two velocities perpendicular to each other:

$v_r=√(v^2+v_s^2)$

$v_r=√(4^2+8^2)$

$v_r=8.9442\ ft.s^(-1)$

Now the angle of the resultant velocity form the vertical:

$\tan\beta=(v_s)/(v)$

$\tan\beta=(8)/(4)$

$\beta=63.43^(\circ)$

Now the distance swam by the swimmer in this direction be d.

so,

$d.\cos\beta=w$

$d* \cos\ 63.43=500$

$d=1118.034\ ft$

Now the distance swept downward:

$\Delta s=√(d^2-w^2)$

$\Delta s=√(1118.034^2-500^2)$

$\Delta s=1000\ ft$

On swimming 37° upstream:

The velocity component of stream cancelled by the swimmer:

$v'=v.\cos37$

$v'=4* \cos37$

$v'=3.1945\ ft.s^(-1)$

Now the net effective speed of stream sweeping the swimmer:

$v_n=v_s-v'$

$v_n=8-3.1945$

$v_n=4.8055\ ft.s^(-1)$

The component of swimmer's velocity heading directly towards the opposite bank:

$v'_r=v.\sin37$

$v'_r=4\sin37$

$v'_r=2.4073\ ft.s^(-1)$

Now the angle of the resultant velocity of the swimmer from the normal to the stream:

$\tan\phi=(v_n)/(v'_r)$

$\tan\phi=(4.8055)/(2.4073)$

$\phi=63.39^(\circ)$

Now let the distance swam in this direction be d'.

$d'* \cos\phi=w$

$d'=(500)/(\cos63.39)$

$d'=1116.344\ ft$

Now the distance swept downstream:

$\Delta s'=√(d'^2-w^2)$

$\Delta s'=√(1116.344^2-500^2)$

$\Delta s'=998.11\ ft.s^(-1)$

Time taken in crossing the rive in case 1:

$t=(d)/(v_r)$

$t=(1118.034)/(8.9442)$

$t\approx125\ s$

Time taken in crossing the rive in case 2:

$t'=(d')/(v'_r)$

$t'=(1116.344)/(2.4073)$

$t'\approx463.733\ s$

A horizontal cylindrical tank 8.00 ft in diameter is half full of oil (60.0 Ib/ft3). Find the force on one end

Answers

Answer:

Assuming h as the height of the cylindrical tank

$F=480\pi h \,g\,\, (lb)/(ft)$

Explanation:

Assuming that the height is $h$ we can find the volume of the cylindrical tank, then:

$V=\pi*r^2*h$

The diameter is 8.00 ft then $r=4.00 ft$ the total volume of the tank is:

$V=\pi (4.00 ft)^2 h=16\pi h\,\, ft^2$

But the tank is half full of oil, then we need half of the volume. For that reason the volume of oil is:

$V_(oil)=(16\pi h)/(2)ft^2=8\pi h \,\,ft^2$

We know the density of the oil $\rho=60.0\,lb/ft^3$ , with this we can fing the mass of oil that we have because:

$\rho=(m)/(V)$ then $m=\rho V$

Then the mass of oil that we have is:

$m=(60.0(lb)/(ft^3))(8\pi h\,\,ft^2)$

$m=480\pi h (lb)/(ft)$

Note that with the value of h we have the mass in correct units.

Finally to find the force we now that $F=mg$ then we just need to multiply the mass by the gravity.

$F=480\pi h \,g\,\, (lb)/(ft)$

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