Compare these two collisions of a PE student with a wall.

If you secure a refrigerator magnet about 2mmfrom the metallic surface of a refrigerator door and then move the magnet sideways,

Yuliya22 [3333]

Response:

(A) 4* 6 ^ ⁻6 T m² (B) 2 * 10 ^ ⁻6 v

Clarification:

Solution

Given that:

A refrigerator magnet with a depth of approximately 2 mm

The estimated magnetic field strength of the magnet is = 5 m T

The Area = 8 cm²

Now,

(A) The magnetic flux ΦB = BA

Therefore,

ΦB = (5 * 10^⁻ 3) ( 4 * 10 ^⁻2) * ( 2 * 10^ ⁻2) Tm²

Thus,

ΦB = 4* 6 ^ ⁻6 T m²

(B) By employing Faraday's Law, the subsequent equation applies:

Ε = Bℓυ

Where,

ℓ = 2 cm equals 2 * 10 ^⁻2 m

B = 5 m T = 5 * 10 ^ ⁻3 T

υ = 2 cm/s = 2 * 10 ^ ⁻2 m/s

Therefore,

Ε = (5 * 10 ^ ⁻3 T) * (2 * 10 ^ ⁻2) (2 * 10 ^ ⁻2) v

E =2 * 10 ^ ⁻6 v

7 0

1 month ago

A resistor with resistance R and an air-gap capacitor of capacitance C are connected in series to a battery (whose strength is "

kicyunya [3294]

Answer:

a) Q = C*emf

b) Decrease in electric field strength and electric potential

c) Initial current through the resistor = emf/R

d) The final charge = K*C*emf

Explanation:

a) The resistors and capacitors are linked in series with the battery

According to Kirchoff's voltage law, the total voltage in the circuit must equal zero

Let $V_{R}$ represent the Voltage across the Resistor

$V_{c}$ and

represent the Voltage across the capacitor

Implementing KVL;

$emf - V_{R} - V_{c} = 0\\$

.........................(1)

Since this is a series connection, the same current traverses through the circuit

$V_{R} = IR\\Q = CV_{c} \\V_{c} = Q/C$

Integrating $V_{c}$ and $V_{R}$ into equation (1)

$emf - IR - Q/C = 0$

Initially, as the capacitor reaches full charge, the current will drop to zero because of equilibrium

$I = 0A\\emf = Q/C\\Q = C* emf$

b) When the plastic sheet is inserted between the plates, current begins to flow because both the electric field intensity and electric potential decrease. As a result, charge diminishes, leading to current flow

c) The current through the resistor equates to the total current within the circuit (given the series connection)

$I = I_{o} \exp(\frac{-t}{RC} )\\At time the initial time, t\\t = 0\\ I_{o} = \frac{emf}{R} \\$

Substituting the values of t and I₀ into the aforementioned formula for I

$I = \frac{emf}{R} \exp(0)\\I = \frac{emf}{R}$

d) Note: The initial charge on the capacitor equals C * emf

Following the insertion of the plastic, the new charge will be:

$Q = K* Q_{initial} \\Q_{initial} = C *emf\\Q_{final} = KCemf$

4 0

1 month ago

If a person weighs 882 N on the surface of the earth, at what altitude above the earth’s surface must they be for their weight t

Sav [3153]

Para calcular el peso utilizamos la fórmula:
$F=m*g=882N$

Cuando tenemos dos cuerpos, empleamos la fórmula de la gravedad general:
$F=G* \frac{ M_{p}*M_{E} }{ r^{2} }$
Donde:
G = constante de gravedad = $6.67* 10^{-11} m^{3} kg^{-1} s^{-2}$
Mp = masa de la persona = 882 / 9.81= 89.9kg
ME = masa de la Tierra = $5.97* 10^{24} kg$
r = distancia entre la Tierra y la persona

A partir de estas dos fórmulas, deducimos que los lados izquierdos son equivalentes, por lo tanto, los lados derechos también deben serlo.
$m*g=G* \frac{ M_{p}*M_{E} }{ r^{2} }$

Resolviendo esta ecuación para r:
$r= \sqrt{G* \frac{ M_{p}*M_{E}}{M_{p}*g}$
r=6371116m = 6371.116km

Esta es la distancia desde el centro de la Tierra. El radio de la Tierra es 6370km y la altura sobre la superficie es 6371.116 - 6370 = 1.116km o 1116m.

3 0

2 months ago

Can pockets of vacuum persist in an ideal gas? Assume that a room is filled with air at 20∘C and that somehow a small spherical

kicyunya [3294]

Answer:

The time required is 20 μs

Explanation:

Here is the data provided:

temperature = 20°C = 293 K

radius = 1 cm

atomic mass of air = 29 u

To determine

the duration for air to refill the vacuum space

solution:

We calculate the root mean square velocity of air particles. This can be expressed as:

$\frac{1}{2}mv^2 = \frac{3}{2}RT$

where m indicates mass, t is temperature, v is speed, and R is the ideal gas constant, which is approximately 8.3145 (kg·m²/s²)/K·mol.

v = $\sqrt{\frac{3RT}{M} }$ ............................1

v = $\sqrt{\frac{3(8.314)293}{29*10{-3}kg} }$

Resulting in v = 501.99 m/s.

<pNow, to cover the distance of 1 cm,<pThe duration needed for air is calculated as:

time taken = $\frac{r}{v}$

which gives us:

time taken = $\frac{1*10^{-2}m}{501.99}$

so, time taken = 19.92 × $10^{6}$ seconds = 20μs.

Thus, the required time is 20 μs.

3 0

3 months ago

a force of 6lbs acts on an object with a weight of 35 lbs on earth. determine the objects acceleration. final answer must be 5.5

kicyunya [3294]

Weight of the object = 35 lbs
F = ma
m = F/a = 35/32 (with acceleration of 32 ft/s²)
m= 1.09

Again applying the same formula,
a = F/m
a= 6/1.09
a= 5.489

Thus, the acceleration is approximately 5.5 ft/s²!!

5 0

2 months ago

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