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3 years ago

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Coc and ball torture (CBT), ponies torture or dic torture is a sexual activity involving application of pain or constriction to

the ponies or testcool. This may involve directly painful activities, such as genital piercing, wax play, genital spanking, squeezing, ball-busting, genital flogging, urethral play, tickle torture, erotic electrostimulation or even kicking. The recipient of such activities may receive direct physical pleasure via masochism, or emotional pleasure through erotic humiliation, or knowledge that the play is pleasing to a sadistic dominant. Many of these practices carry significant health risks

2 answers:

zhenek [66]3 years ago

8 0

Answer:

this is very valuable information

Explanation:

thanks for letting me know ahead of time

maw [93]3 years ago

6 0

Answer:

CABT

Explanation:

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What is the principle of potentiometer?

kupik [55]

Answer:

The principle of a potentiometer is that the potential dropped across a segment of a wire of uniform cross-section carrying a constant current is directly proportional to its length. The potentiometer is a simple device used to measure the electrical potentials (or compare the e.m.f of a cell).

Explanation:

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6 0

3 years ago

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What is the acceleration of a car that travels in a straight line?

Sphinxa [80]

Acceleration occurs when there is a change in speed or direction. If it travels in a straight line, there is no speed or change in direction as it is constant througout, hence 0 acceleration.

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6 0

3 years ago

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In the "before" part of Fig. 9-60, car A (mass 1100 kg) is stopped at a traffic light when it is rear-ended by car B (mass 1400

liq [111]

Complete Question:

In the "before" part of Fig. 9-60, car A (mass 1100 kg) is stopped at a traffic light when it is rear-ended by car B (mass 1400 kg). Both cars then slide with locked wheels until the frictional force from the slick road (with a low ?k of 0.15) stops them, at distances dA = 6.1 m and dB = 4.4 m. What are the speeds of (a) car A and (b) car B at the start of the sliding, just after the collision? (c) Assuming that linear momentum is conserved during the collision, find the speed of car B just before the collision.

Answer:

a) Speed of car A at the start of sliding = 4.23 m/s

b) speed of car B at the start of sliding = 3.957 m/s

c) Speed of car B before the collision = 7.28 m/s

Explanation:

NB: The figure is not provided but all the parameters needed to solve the question have been given.

Let the frictional force acting on car A, $f_{ra} = \mu mg\\$ ............(1)

Since frictional force is a type of force, we are safe to say $f_{ra} = ma$ .......(2)

Equating (1) and (2)

$ma = \mu mg\\a = \mu g\\\mu = 0.15\\a = 0.15 * 9.8 = 1.47 m/s^{2}$

a) Speed of A at the start of the sliding

$d_{A} = 6.1 m\\Speed of A at the start of sliding, v_{A} = \sqrt{2ad_{A} }\\ v_{A} = \sqrt{2*1.47*6.1 } \\v_{A} = \sqrt{17.934 } \\v_{A} = 4.23 m/s$

b) Speed of B at the start of the sliding

$d_{A} = 4.4 m\\Speed of A at the start of sliding, v_{B} = \sqrt{2ad_{B} }\\ v_{B} = \sqrt{2*1.47*4.4 } \\v_{B} = \sqrt{12.936 } \\v_{B} = 3.957 m/s$

Let the speed of car B before collision = $v_{B1}$

Momentum of car B before collision = $m_{B} v_{B1}$

Momentum after collision = $m_{A} v_{A} + m_{B} v_{B2}$

Applying the law of conservation of momentum:

$m_{B} v_{B1} = m_{A} v_{A} +m_{B} v_{B2}$

$m_{A} = 1100 kg\\m_{B} = 1400 kg$

$(1400*v_{B1} ) = (1100 * 4.23) + ( 1400 * 3.957)\\(1400*v_{B1} ) = 10192.8\\v_{B1} = 10192.8/1400\\v_{B1 = 7.28 m/s$

3 0

4 years ago

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If 0.035pC of charge is transferred via the movement of Al3+ ions, how's many of these must be transferred in total? Please add

mr Goodwill [35]

Each $Al^+^3$ ion contains three extra protons. Hence, the extra charge on each $Al^+^3$ = $3 \times 1.6 \times 10^-^1^9$ C

Total charge = 0.035 pC

Total charge (Q) = $0.035 \times 10^-^1^2$ C

Let the number of $Al^+^3$ ions be n.

According to question:

$n \times 3 \times 1.6 \times 10^-^1^9 =0.035 \times 10^-^1^2$

$n = \frac{0.035 \times 10^-^1^2}{3 \times 1.6 \times 10^-^1^9}$

$n = 7.29167 \times 10^4$

n = 72917

Hence, the total number of ions needed to be transferred is 72917

3 0

3 years ago

What is the weight of a dog with a mass of 12 kg? Explain how you obtained your answer.

butalik [34]

Answer: 117.60N

Explanation:

Weight is a force. Therefore, we can use the force formula to find weight.

$Formula: W=m*g$

W = weight

m = mass

g = acceleration due to gravity ( $9.8m/s^2$ )

$W=(12kg)(9.8m/s^2)\\W=117.60N$

5 0

4 years ago

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