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Maksim231197 [3]
3 years ago
12

A square metal plate of edge length 12 cm and negligible thickness has a total charge of 5.6 × 10-6 C. (a) Estimate the magnitud

e E of the electric field just off the center of the plate (at, say, a distance of 0.49 mm from the center) by assuming that the charge is spread uniformly over the two faces of the plate. (b) Estimate E at a distance of 29 m (large relative to the plate size) by assuming that the plate is a charged particle.
Physics
1 answer:
hodyreva [135]3 years ago
8 0

Answer:

a) 2.2*10^7 N/C

b) 60 N/C

Explanation:

To start with, we say that the

Area charge of the plate, σ = q/2A

σ = 5.6*10^-6 / 2(12*10^-2)²

σ = 5.6*10^-6 / 0.0288

σ = 1.944*10^-4 C²/m

Next, we find the electric field

Electric field, E = σ/Eo

Electric field, E = 1.944*10^-4 / 8.85*10^-12

Electric field, E = 2.2*10^7 N/C

b)

E = kq/r²

E = [8.99*10^9 * 5.6*10^-6] / 29²

E = 50344 / 841

E = 60 N/C

This means that the E at a distance of 29 m is 60 N/C

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A mass of 0.250 kg is attached to a spring and undergoes simple harmonic oscillations with a period of 0.640 s. What is the forc
Paha777 [63]

The force constant of the spring is approximately 24.038 newtons per meter.

As we are talking about Simple Harmonic Motion. In this exercise we need to determine the Spring Constant (k), in newtons per meter, from the equation of the Period (T), in seconds, which is described below:

T = 2\pi\cdot \sqrt{\frac{m}{k} } (1)

Where m is the mass of the moving element, in kilograms.

If we know that T = 0.640\,s and m = 0.250\,kg, then the spring constant of the spring is:

0.640 = 2\pi\cdot \sqrt{\frac{0.250}{k} }

\sqrt{\frac{0.250}{k} } \approx 0.102

\frac{0.250}{k} \approx 0.0104

k \approx 24.038\,\frac{N}{m}

The force constant of the spring is approximately 24.038 newtons per meter.

Please see this question related to Simple Harmonic Motion for further details: brainly.com/question/17315536

6 0
3 years ago
Read 2 more answers
A body with mass of 200 g is attached to the end of a spring that is stretched 20 cm by a force of 9 N. At time tequals0 the bod
11Alexandr11 [23.1K]

Answer:

X(t) = 13/13 cos(12t+α)

C =13/13

π/6 s

Explanation:

(A) A body with mass 200 g is attached to the end of a spring that is stretched 20 cm by a force of 9 N. At time t = 0 the body is pulled 1 m in to the right, stretching the spring, 3 set in motion with an initial velocity of 5 m/s to the left.  

(a) Find X(t) in the form c • cos(w_o*t— α)  

(b) Find the amplitude 3 Period of motion of the body 1  

mass: m = 200g =  0.200 kg  

displacement: ΔX = 20 cm =  0.20 m

Spring Constant: K =  9/0.20 = 45 N/m

IV:   X(0) = 1m V(0) = -5 m/s

Simple Harmonic Motion: c•cos(cosw_t— α) = X(t)  

Circular Frequency: w_o = √k/m= √36/(0.20) = 13 rad/s

X(0) = 1m =c_1

X'(0) = V(0) = c_2*w_o/w_o

        = -5/12 =   c_2

"radians Technically Unitless"  

Amplitude: c = √ci^2 + c^2 ==> √1^2 + (-5/12)^2 = 1 m =13/13 = c

X(t) = 13/13 cos(12t+α)

since, C>0 : damped forced vibration c_1>0, c_2>0

phase angle 2π+tan^-1(c_2/c_1)

                        =2π+tan^-1(-5/12/1)= 5.884

period: T =2π/w_o

                =π/6 s

6 0
4 years ago
A rock climber wears a 8.1 kg backpack while scaling a cliff. After 28.2 min, the climber is 9.4 m above the starting point. How
mart [117]

Answer:

Explanation:

mass of backpack, m = 8.1 kg

weight of climber, W = 656 N

height raised, h = 9.4 m

time, t = 28.2 min = 28.2 x 60 = 1692 second

weight of backpack, w = m x g = 8.1 x 9.8 = 79.38 N

Work done by the climber on the backpack = mg x h = 79.38 x 9.4 = 746.17 J

Wok done in lifting herself + backpack = (W + w) x h

                                                                 = (656 + 79.38) x 9.4 = 6912.57 J

Power developed by the climber,P = Total work / time

P = 6912.57 / 1692 = 4.09 W

3 0
4 years ago
If the direction of the position is north and the direction of the velocity is up, then what is the direction of the angular mom
Kay [80]

Answer:

the direction of angular momentum = EAST

Explanation:

given

Direction of position = r = north

Direction of velocity = v = up

angular momentum = L = m(r x v)

where m is the mass, r is the radius, v is the velocity

utilizing the right hand rule, the right finger heading towards the course of position vector and curl them toward direction of velocity, at that point stretch thumb will show the bearing of the angular momentum.

then L = north x up = East

6 0
3 years ago
I NEED THE ANSWER TO PASS SO I CAN GET MY PHONE BACK HELPPPPPPPPPPPPPP
lianna [129]

Answer:

it would be c.

Explanation:

starts at zero and isn't a straight shot up

6 0
3 years ago
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