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3 years ago

How would amperage and voltage affect the power of the fence

1 answer:

8 0

So it's easier to break amps and volts down as if they were water through a pipe. Voltage is the pressure at which the water is forced through so you can survive high voltage. While Amperes would be the amount or volume of water that flows out over a period of time it doesn't take much amps to kill you as goes the saying "amps give the cramps and volts give the jolts"

You might be interested in

What is the meaning of smooth?

damaskus [11]

A surface in which is flat or very soft to the touch and reduces splinters or anything sticking out, having an surface which does not have lumps, or indentations.

5 0

3 years ago

Robert has just bought a new model rocket, and is trying to measure its flight characteristics. The rocket engine package claims

777dan777 [17]

Answer:

The work done by the drag force is given by 29.96 J

Explanation:

Given :

Thrust force $F = 12.3$ N

Displacement $d = 10.2$ m

Mass of rocket $m = 0.663$ Kg

From work energy theorem,

$W = \Delta K$

$W_{t} - Wd - W_{g} = KE$

Where $W_{t} =$ thrust work $W_{g} =$ gravitational work

$KE = 12.3 \times 10.2 -Wd - 0.663 \times 9.8 \times 10.2$

$KE = 59.2 -Wd$

After cutoff kinetic energy is converted into potential energy,

$KE = Wd' + mg\Delta h$

Put value of KE

$59.2 -Wd = Wd' + 0.663 \times 9.8 \times 4.5$

Work done by drag force is given by,

$Wd'+Wd = 59.2 -29.23$

$= 29.96$ J

Therefore, the work done by the drag force is given by 29.96 J

5 0

3 years ago

How many valence electrons do most stable atoms have?

IceJOKER [234]

Answer:

D) 8

Explanation:

Due to the octet rule the most stable atoms will have 8 valence electrons.

8 0

2 years ago

A 30-g bullet is fired with a horizontal velocity of 450 m/s and becomes embedded in block B, which has a mass of 3 kg. After th

Zolol [24]

Answer:

(a) The velocity of the bullet and B after the first impact is 4.4554 m/s.

(b) The velocity of the carrier is 0.40872 m/s.

Explanation:

(a) To solve the question, we apply the principle of conservation of linear momentum as follows.

we note that the distance between B and C is 0.5 m

Then we have

Sum of initial momentum = Sum of final momentum

0.03 kg × 450 m/s = (0.03 kg + 3 kg) × v₂

Therefore v₂ = (13.5 kg·m/s)÷(3.03 kg) = 4.4554 m/s

The velocity of the bullet and B after the first impact = 4.4554 m/s

(b) The velocity of the carrier is given as follows

Therefore from the conservation of linear momentum we also have

(m₁ + m₂)×v₂ = (m₁ + m₂ + m₃)×v₃

Where:

m₃ = Mass of the carrier = 30 kg

Therefore

(3.03 kg)×(4.4554 m/s) = (3.03 kg+30 kg) × v₃

v₃ = (13.5 kg·m/s)÷ (33.03 kg) = 0.40872 m/s

The velocity of the carrier = 0.40872 m/s.

3 0

3 years ago

If the magnitude of the initial velocity of the ball v0 = 7.94 ± 0.03, and the "gun" is tilted 31 ± 0.4º upwards, what is the un

ycow [4]

Answer:

0.05

Explanation:

Given a variable C which is the product of two variables A, B:

$C=A\cdot B$

Then the absolute error on C is given by:

$\frac{\sigma_C}{C}=\frac{\sigma_A}{A}+\frac{\sigma_B}{B}$

where $\sigma_A, \sigma_B, \sigma_C$ are the uncertanties on the measure of A, B and C, respectively.

In this problem, the horizontal component of the velocity $v_x$ is given by

$v_x = v_0 cos \theta$

Therefore, the uncertainty on vx is given by:

$\frac{\sigma_{v_x}}{v_x}=\frac{\sigma_{v_0}}{v_0}+\frac{\sigma_{cos \theta}}{cos \theta}$ (1)

where we have:

$v_0 = 7.94$

$\sigma_{v_0}=0.03$

$\theta=31^{\circ}$ , so

$cos \theta=cos 31^{\circ}=0.857$

The uncertainty on $cos \theta$ is given by:

$\sigma_{cos \theta}=|sin \theta|\sigma_\theta$

where:

$|sin \theta| = |sin 31^{\circ}|=0.515$

and

$\sigma_\theta=0.4^{\circ}=0.007 rad$

$\sigma_{cos \theta}=|0.515|\cdot 0.007 = 0.0036$

Also,

$v_x = v_0 cos \theta = (7.94)(cos 31^{\circ})=6.81 m/s$

So, combininb everything into (1), we find:

$\sigma_{v_x}=(\frac{\sigma_{v_0}}{v_0}+\frac{\sigma_{cos \theta}}{cos \theta})v_x=(\frac{0.03}{7.94}+\frac{0.0036}{0.857})(6.80)=0.05$

4 0

3 years ago