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Dovator [93]
3 years ago
14

In fig. 30.11, suppose that and with switch open, switch is left closed until a constant current is established. then is closed

and opened, taking the battery out of the circuit. (a what is the initial current in the resistor, just after is closed and is opened? (b what is the current in the resistor at (c what is the potential difference between points b and c at which point is at a higher potential? (d how long does it take the current to decrease to half its initial value?
Physics
1 answer:
Gnom [1K]3 years ago
3 0
In a particle accelerator a positron (C= +1.6 x 10-19) travels through a perpendicular magnet field with a magnitude of 3.1 x 10-2 T. At what speed must the positron travel in order for it to experience a force of 4.75 x 10-14 N? 28. An alpha particle (2 protons and 2 neutrons) experiences a downward force of 2.9 x 10-14 N while traveling in a magnetic field with a strength of 5.1 x 10-19 T pointing to the north. Find the speed of the particle and the direction it must be traveling in. 29. Find the length of a wire if it experiences a .63N force when it travels through a magnetic field with a strength of 0.85T whilst carrying 5.0 amps of current. 30. A coil with 462 turns of wire, a total resistance of 36Ω , and a cross-sectional area of 0.25 m2 is positioned with its plane perpendicular to the field of a powerful electromagnet. What average current is induced in the coil during the 0.37s that the magnetic field drops from 3.1 T to 0.0 T? 31. A step-up transformer has a potential difference across the primary of 28 V and a potential difference across the secondary of 3.0 × 104 V. There are 28 turns in the primary coil. How many turns are in the secondary? 32. A step-up transformer is used to create a potential difference of 1.6872 × 105 V across the secondary. The potentia
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The amount of energy carried by a wave and the waves maximum displacement from the rest position describe what?
Mademuasel [1]
What you are describing is Amplitude.
6 0
3 years ago
A ball is kicked horizontally at 4.6 m/s off of a cliff 13.4 m high. How far from the cliff will it land.
Mekhanik [1.2K]
<h3>Answer:</h3>

7.53 m

<h3>Explanation:</h3>

<u>We are given:</u>

Initial Horizontal Velocity of the Ball = 4.6 m/s

Initial Vertical Velocity of the Ball = 0 m/s

Height from which ball is kicked = 13.4 m

<u>Time taken by the ball to reach the ground:</u>

The ball has an initial vertical velocity of 0 m/s

it also has a downward acceleration of 10 m/s² due to gravity

<u>Solving for the time taken:</u>

s = ut + 1/2(at²)                 [second equation of motion]

replacing the values

13.4 = (0)(t) + 1/2 (10)(t²)

13.4 = 5t²

t² = 13.4/5                  [dividing both sides by 5]

t² = 2.68

t = 1.637 seconds     [taking the square root of both sides]

<u>Horizontal distance covered by the ball:</u>

Since there are no horizontal opposing forces on the ball,

the ball will more horizontally at a velocity of 4.6 m/s until it hits the ground

We calculated that the ball will hit the ground in 1.637 seconds

<u>Distance covered:</u>

s = ut + 1/2 (at²)                            [seconds equation of motion]

s = ut                                            [since a = 0m/s² in the horizontal plane]

replacing the values

s = 4.6 * 1.637

s = 7.53 m

Hence, the ball landed 7.53 m from the cliff

5 0
3 years ago
una bala de 20 g choca con un fango como se muestra en la figura y penetra una distancia de 6 cm antes de detenerse. calcule la
aleksklad [387]

Answer:

A 20g bullet collides with a mud as shown in the figure and penetrates a distance of 6cm before stopping. calculate the braking force f if the input speed was 80m/s

Explanation:

Given that,

A bullet of mass

M = 20g = 0.02kg

The bullet is fired into a mud initially at rest

The bullet penetrate a distance of 6cm in the mud

S = 6cm = 0.06m

The input velocity into the mud is 80m/s, this implies that, the initial velocity is 80m/s

U = 80m/s

So, the bullet stop after a distance of 6cm, so the final velocity is 0m/s

V = 0m/s

So, we need to find the braking force

Breaking force (F) is given as

F = ma

Where m is mass and 'a' is deceleration

So, we need to find the deceleration, using equation of motion

V² = U² + 2as

0² = 80² + 2 × a × 0.06

0 = 6400 + 0.12a

0.12a = -6400

a = -6400 / 0.12

a = - 53,333.33 m/s²

So, the negative sign shows that the bullet is decelerating

So, a = 53,333.33 m/s²

So, breaking force

F = ma

F = 0.02 × 53,333.33

F = 1066.67 N

The breaking force is 1066.67 N

In spanish

Dado que,

Una bala de masa

M = 20 g = 0.02 kg

La bala se dispara en un lodo inicialmente en reposo

La bala penetra una distancia de 6 cm en el barro.

S = 6 cm = 0.06 m

La velocidad de entrada al lodo es de 80 m / s, esto implica que la velocidad inicial es de 80 m / s

U = 80 m / s

Entonces, la bala se detiene después de una distancia de 6 cm, por lo que la velocidad final es de 0 m / s

V = 0 m / s

Entonces, necesitamos encontrar la fuerza de frenado

La fuerza de ruptura (F) se da como

F = ma

Donde m es masa y 'a' es desaceleración

Entonces, necesitamos encontrar la desaceleración, usando la ecuación de movimiento

V² = U² + 2as

0² = 80² + 2 × a × 0.06

0 = 6400 + 0.12a

0.12a = -6400

a = -6400 / 0.12

a = - 53,333.33 m / s²

Entonces, el signo negativo muestra que la bala se está desacelerando

Entonces, a = 53,333.33 m / s²

Entonces, fuerza de ruptura

F = ma

F = 0.02 × 53,333.33

F = 1066.67 N

La fuerza de ruptura es 1066.67 N

7 0
3 years ago
When the ambulance passes the person (switching from moving towards him to moving away from him), the perceived frequency of the
dlinn [17]

To develop this problem we will apply the considerations made through the concept of Doppler effect. The Doppler effect is the change in the perceived frequency of any wave movement when the emitter, or focus of waves, and the receiver, or observer, move relative to each other. At first the source is moving towards the observer. Than the perceived frequency at first

F_1 = F \frac{{343}}{(343-V)}

Where F is the actual frequency and v is the velocity of the ambulance

Now the source is moving away from the observer.

F_2 = F\frac{343}{(343+V)}

We are also so told the perceived frequency decreases by 11.9%

F_2 = F_1 - 9.27\% \text{ of } F_1

F_2 = F_1-0.0927F_1

F_2 = 0.9073F_1

Equating,

F\frac{343}{(343+V)}= 0.9073(F\frac{343}{(343-V)})

\frac{1}{(343+V)}= 0.9073\frac{1}{(343-V)}

0.9073(343+V) = 343-V

(0.9073)(343)+(0.9073)V = 343-V

V+0.9073V = 343-(0.9073)(343)

Solving for V,

V = 16.67 m/s

5 0
3 years ago
On which factor hydraulic pressure depend?​
Andreas93 [3]
Amount of gas, temperature, and volume of container.
6 0
3 years ago
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