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3 years ago

What term described the capability to prevent one participant in an electronic transaction from denying it performed an action?

Physics

1 answer:

vampirchik [111]3 years ago

7 0

Answer:

Plausible deniability.

Explanation:

Plausible deniability.

It is the ability of a person to refuse to accept knowledge of responsibility, if any secret(demeaning) action is done by others in an organizational hierarchy as there may be lack of evidence to confirm their involvement. Specially for high level official who can willfully deny any awareness of the to avoid themselves getting involved in blame game.

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What do you think causes such change in wave velocity?

saveliy_v [14]

Example: Increasing the tension in A string causes the speed of waves on the string to increase. Since the wavelengths of the standing waves remains constant, this results in a larger frequency of oscillations in the string, which we percieve as a higher pitch when the string vibrates the air.

8 0

2 years ago

Four uniform spheres, with masses mA = 200 kg, mB = 250 kg, mC = 1700 kg, and mD = 100 kg, have (x, y) coordinates of (0, 50 cm)

rusak2 [61]

Answer:

$F_2=2.43\times10^{-19}\ N$

$\theta=21.496^{\circ}$ in the anticlockwise direction from the positive x- axis.

Explanation:

Given are the masses of various spheres with their names in subscript:

$m_A=200\ kg$

$m_B=250\ kg$

$m_C=1700\ kg$

$m_D=100\ kg$

Now their respective coordinate positions (in cm) are as given below:

$P_A=(0,50)$

$P_B=(0,0)$

$P_C=(-80,0)$

$P_D=(40,0)$

Now force on B due to A:

$F_{BA}=G\times \frac{200\times 250}{50^2}$

$F_{BA}=20G\ N$

Now force on B due to C:

$F_{BC}=G\times \frac{1700\times 250}{80^2}$

$F_{BC}=66.406G\ N$

Now force on B due to D:

$F_{BD}=G\times \frac{100\times 250}{40^2}$

$F_{BD}=15.625G\ N$

We observe that the forces due to masses C&D act opposite in direction.

So, the net force in the x-direction:

$F_x=F_{BC}-F_{BD}$

$F_x=66.406G-15.625G$

$F_x=50.781G\ N$ in the positive x-direction

We have only one force in y-direction due to mass A.

So,

$F_y=20G\ N$ in the positive y-direction.

Now the net force:

$F_2=\sqrt{F_y^2+F_x^2}$

$F_2=\sqrt{(20G)^2+(50.781G)^2}$

$F_2=54.5776G^2\ N$

$F_2=2.43\times10^{-19}\ N$

Now the direction of this force with respect to x-axis:

$tan\ \theta=\frac{F_y}{F_x}$

$tan\ \theta=\frac{20G}{50.781G}$

$\theta=21.496^{\circ}$ in the anticlockwise direction from the positive x- axis.

3 0

3 years ago

Read 2 more answers

A 1000-kg car moving at 10 m/s brakes to a stop in 5 s. the average braking force is

Marina86 [1]

To calcculate the braking force of the car moving, we use Newton's second law of motion which relates the acceleration and the force of an object moving. The force of an object moving is directly proportional to its acceleration and the proportionality constant is the mass of the object. It is expressed as:

Force = ma

Acceleration is the rate of change of the velocity of a moving object. We calculate acceleration from the velocity and the time given above.

a = (10 m/s) / 5 s = 2 m/s^2

So,
Force = ma
Force = 1000 kg ( 2 m/s^2 )
Force = 2000 kg m/s^2 or 2000 N

4 0

3 years ago

A horizontal insulating rod of length 11.8-cm and charge 19 nC is in a plane with a long straight vertical uniform line charge.

SCORPION-xisa [38]

Answer:

11.962337 × 10^-4 N

Explanation:

Given the following :

Length L = 11.8

Charge = 29nC = 29 × 10^-9 C

Linear charge density λ = 1.4 × 10^-7 C/m

Radius (r) = 2cm = 2/100 = 0.02 m

Using the relation:

E = 2kλ/r ; F =qE

F = 2kλq/L × ∫dr/r

F = 2*k*q*λ/L × (In(0.02 + L) - In(0.02))

2*k*q*λ/L = [2 × (9 * 10^9) * (29 * 10^9) * (1.4 * 10^-7)]/ 0.118] = 6193.2203 × 10^(9 - 9 - 7) = 6193.2203 × 10^-7 = 6.1932203 × 10^-4

In(0.02 + 0.118) - In(0.02) = In(0.138) - In(0.02) = 1.9315214

Hence,

(6.1932203 × 10^-4) × 1.9315214 = 11.962337 × 10^-4 N

3 0

4 years ago

How do you calculate the braking distance

Anettt [7]

The braking distance is given by $s=\frac{-u^2}{2a}$

Explanation:

When the driver of a car hits the pedal of the brakes, the car starts decelerating until it stops. Assuming the deceleration is constant, then the motion is a uniformly accelerated motion, so we can use the following suvat equation:

$v^2-u^2=2as$