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3 years ago

15

When you place charge on a conducting doorknob by touching it after walking across a carpet, what happens to the charge on the d

oorknob?

1 answer:

earnstyle [38]3 years ago

7 0

Answer:

Your body generates a static energy which gives it a net positive Charge as such the charge spreads uniformly over the surface of the doorknob

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During normal beating, a heart creates a maximum 3.95-mV potential across 0.305 m of a person’s chest, creating a 0.75-Hz electr

WINSTONCH [101]

Answer:

E = 0.0130 V/m.

Explanation:

The electric field is related to the potential difference as follows:

$E = \frac{\Delta V}{d}$

<u>Where:</u>

E: is electric field

ΔV: is the potential difference = 3.95 mV

d: is the distance of a person's chest = 0.305 m

Then, the electric field is:

$E = \frac{\Delta V}{d} = \frac{3.95 \cdot 10^{-3} V}{0.305 m} = 0.0130 V/m$

Therefore, the maximum electric field created is 0.0130 V/m.

I hope it helps you!

7 0

2 years ago

Gala2k [10]

Acceleration increases.

4 0

2 years ago

Read 2 more answers

A mass m at the end of a spring oscillates with a frequency of 0.89 hz. when an additional 603 g mass is added to m, the frequen

MrMuchimi

The solution for this problem:

Given:

f1 = 0.89 Hz

f2 = 0.63 Hz

Δm = m2 - m1 = 0.603 kg

The frequency of mass-spring oscillation is:
f = (1/2π)√(k/m)
k = m(2πf)²

Then we know that k is constant for both trials, we have:
k = k

m1(2πf1)² = m2(2πf2)²

m1 = m2(f2/f1)²

m1 = (m1+Δm)(f2/f1)²

m1 = Δm/((f1/f2)²-1)

m 1 = 0.603/ (0.89/0.63)^2 – 1

= 0.609 kg or 0.61kg or 610 g

5 0

3 years ago

What type of pressure is transmitted by fluids

Evgesh-ka [11]

Cerebrospinal fluid or Hydrocephalus

5 0

3 years ago

A projectile is shot from the edge of a cliff above the ground level with initial velocity of at an angle with the horizontal. (

Xelga [282]

Answer:

t = √2y/g

Explanation:

This is a projectile launch exercise

a) The vertical velocity in the initial instants ( $v_{oy}$ = 0) zero, so let's use the equation

y = $v_{oy}$ t -1/2 g t²

y= - ½ g t²

t = √2y/g

b) Let's use this time and the horizontal displacement equation, because the constant horizontal velocity

x = vox t

x = v₀ₓ √2y/g

c) Speeds before touching the ground

vₓ = vox = constant

$v_{y}$ = $v_{oy}$ - gt

$v_{y}$ = 0 - g √2y/g

$v_{y}$ = - √2gy

tan θ = Vy / vx

θ = tan⁻¹ (vy / vx)

θ = tan⁻¹ (√2gy / vox)

d) The projectile is higher than the cliff because it is a horizontal launch

6 0

3 years ago