What type of energy to high frequency waves have?

When a certain air-filled parallel-plate capacitor is connected across a battery, it acquires a charge of magnitude 172 μC on ea

crimeas [40]

Answer:

k = 2.279

Explanation:

Given:

Magnitude of charge on each plate, Q = 172 μC

Now,

the capacitance, C of a capacitor is given as:

C = Q/V

where,

V is the potential difference

Thus, the capacitance due to the charge of 172 μC will be

C = $\frac{(172\ \mu C)}{V}$

Now, when the when the additional charge is accumulated

the capacitance (C') will be

C' = $\frac{(172+220)\ \mu C)}{V}$

or

C' = $\frac{(392)\ \mu C)}{V}$

now the dielectric constant (k) is given as:

$k=\frac{C'}{C}$

substituting the values, we get

$k=\frac{\frac{(392\ \mu C)}{V}}{\frac{(172)\ \mu C)}{V}}$

or

k = 2.279

6 0

3 years ago

A 49 kg bear slides, from rest, 11 m down a lodgepole pine tree, moving with a speed of 3.3 m/s just before hitting the ground.

hichkok12 [17]

Answer:

a) $\Delta U_g=-5.3kJ$

b) $K=0.27kJ$

c) $F_f=0.45kN$

Explanation:

the gravitational potential energy is given by:

$U_g=m.g.h\\$

$\Delta U_g=m.g.h_f-m.g.h_i\\\Delta U_g=49kg*9.8m/s^2*(0m-11m)\\\Delta U_g=-5.3kJ$

The kinetic energy is given by:

$K=\frac{1}{2}m.v^2\\$

the initial kinetic energy is zero because the motion started from rest, so:

$K=\frac{1}{2}*49kg*(3.3m/s^2)^2\\K=0.27kJ$

applying the conservation of energy theorem:

$U_g-W_f=K_f\\W_f=-(\Delta K+\Delta U)\\W_F=5.3kJ-0.27kJ\\W_F=-5.0kJ$

The work done by the friction force is given by:

$W_f=F_f.h.cos(\theta)\\$

the angle of the force is 180 degrees because it's against the movement:

$F_f=\frac{W_f}{h.cos(\theta)}\\\\F_f=\frac{-5.0kJ}{11m.cos(180^o)}\\\\F_f=0.45kN$

8 0

3 years ago

The velocity of a body is given by the equation v= a + bx, where 'x' is displacement. The unit of b is .......

valina [46]

Answer:

s^ -1 ( or 1/sec)

Explanation:

Velocity is given in units of displacement / sec

like feet /sec or m/sec

so b would have units of s^-1

(or perhaps a more general term would be time^-1)

8 0

1 year ago

Blake and Sandra are having a rummage sale. Blake drags 3 boxes a distance of 10 meters each. He exerts a force of 20 newtons on

vampirchik [111]

For Blake:

3 boxes at a distance of 10 meters each, each box weighs 20 N

Work done by Blake = 3 * 10m * 20N
= 600 J
Power = 600 J/ 2 min
= 300 J/min

For Sandra:

4 boxes, 15 N each at a distance of 12 meters each.

Work done by Sandra = 4 * 15 N *12m
= 720 J
Power = 720 J/ 4 min
= 180 J/min

Blake does less work than Sandra.
Blake's power is more than Sandra's.

5 0

3 years ago

Read 2 more answers

The end diastolic volume of a heart is 140 mL Assume that it is a sphere. At end diastole, the intraventricular pressure is 7mmI

Vera_Pavlovna [14]

Answer:

Explanation:

We know that, V = 140 mL = 0.00014 m3

Assume that it is a sphere. so, we have

V = (4/3) \pir3

r3 = (0.00014 m3) (3) / (4) (3.14)

r = \sqrt[3]{}\sqrt[3]{}3\sqrt{}3.34 x 10-5 m3

r = 1.93 x 10-7 m

(a) The wall tension at end diastole will be given as :

using a formula, we have

T = P r / 2 H

where, P = intraventricular pressure at end diastole = 7 mmHg = 933.2 Pa

H = wall thickness at this time = 0.011 m

then, we get

T = (933.2 Pa) (1.93 x 10-7 m) / 2 (0.011 m)

T = 8.18 x 10-3 N

(b) The wall tension at the end of isovolumetric contraction will be given as :

using a formula, we have

T = P r / 2 H

where, P = intraventricular pressure at end of isovolumetric contraction = 80 mmHg = 10665.7 Pa

H = wall thickness at this time = 0.011 m

then, we get

T = (10665.7 Pa) (1.93 x 10-7 m) / 2 (0.011 m)

T = 9.35 x 10-2 N

(d) The wall stress from A and B which will be given as :

we know that, \sigma = T / w

For part A, we have

\sigmaA = (8.18 x 10-3 N) / (0.011 m)

\sigmaA = 0.743 N/m

For part B, we have

\sigmaB = (9.35 x 10-2 N) / (0.011 m)

\sigmaB = 8.5 N/m

4 0

2 years ago