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4 years ago

13

***PLEASE HELP WITH ANSWER AND EXPLANATION: Imagine the current in a current-carrying wire is flowing into the screen. What is t

he direction of the magnetic field around this current?
The magnetic field is clockwise.
The magnetic field is counterclockwise.
The magnetic field is moving out of the screen.
The magnetic field is moving into the screen.

2 answers:

DerKrebs [107]4 years ago

7 0

Answer:

the magnetic field is clockwise

Explanation:

aalyn [17]4 years ago

5 0

Magnetic field direction is given by right hand thumb rule.

If we put our thumb in the direction of current then curl of fingers will show the magnetic field direction around the wire.

Now here since current is going into the screen so we will put our thumb into the screen and then the curl of fingers is clockwise around it.

The magnetic field is clockwise.

So this would be the direction of magnetic field

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I need answers and solvings to these questions

den301095 [7]

1) The period of a simple pendulum depends on B) III. only (the length of the pendulum)

2) The angular acceleration is C) $15.7 rad/s^2$

3) The frequency of the oscillation is C) 1.6 Hz

4) The period of vibration is B) 0.6 s

5) The diameter of the nozzle is A) 5.0 mm

6) The force that must be applied is B) 266.7 N

Explanation:

1)

The period of a simple pendulum is given by

$T=2\pi \sqrt{\frac{L}{g}}$

where

T is the period

L is the length of the pendulum

g is the acceleration of gravity

From the equation, we see that the period of the pendulum depends only on its length and on the acceleration of gravity, while there is no dependence on the mass of the pendulum or on the amplitude of oscillation. Therefore, the correct option is

B) III. only (the length of the pendulum)

2)

The angular acceleration of the rotating disc is given by the equation

$\alpha = \frac{\omega_f - \omega_i}{t}$

where

$\omega_f$ is the final angular velocity

$\omega_i$ is the initial angular velocity

t is the time elapsed

For the compact disc in this problem we have:

$\omega_i = 0$ (since it starts from rest)

$\omega_f = 300 rpm \cdot \frac{2\pi rad/rev}{60 s/min}=31.4 rad/s$ is the final angular velocity

t = 2 s

Substituting, we find

$\alpha = \frac{31.4-0}{2}=15.7 rad/s^2$

3)

For a simple harmonic oscillator, the acceleration and the displacement of the system are related by the equation

$a=-\omega^2 x$

where

a is the acceleration

x is the displacement

$\omega$ is the angular frequency of the system

For the oscillator in this problem, we have the following relationship

$a=-100 x$

which implies that

$\omega^2 = 100$

And so

$\omega = \sqrt{100}=10 rad/s$

Also, the angular frequency is related to the frequency $f$ by

$f=\frac{\omega}{2\pi}$

Therefore, the frequency of this simple harmonic oscillator is

$f=\frac{10}{2\pi}=1.6 Hz$

4)

When the mass is hanging on the sping, the weight of the mass is equal to the restoring force on the spring, so we can write

$mg=kx$

where

m is the mass

$g=9.8 m/s^2$ is the acceleration of gravity

k is the spring constant

x = 8.0 cm = 0.08 m is the stretching of the spring

We can re-arrange the equation as

$\frac{k}{m}=\frac{g}{x}=\frac{9.8}{0.08}=122.5$

The angular frequency of the spring is given by

$\omega=\sqrt{\frac{k}{m}}=\sqrt{122.5}=11.1 Hz$

And therefore, its period is

$T=\frac{2\pi}{\omega}=\frac{2\pi}{11.1}=0.6 s$

5)

According to the equation of continuity, the volume flow rate must remain constant, so we can write

$A_1 v_1 = A_2 v_2$

where

$A_1 = \pi r_1^2$ is the cross-sectional area of the hose, with $r_1 = 5 mm$ being the radius of the hose

$v_1 = 4 m/s$ is the speed of the petrol in the hose

$A_2 = \pi r_2^2$ is the cross-sectional area of the nozzle, with $r_2$ being the radius of the nozzle

$v_2 = 16 m/s$ is the speed in the nozzle

Solving for $r_2$ , we find the radius of the nozzle:

$\pi r_1^2 v_1 = \pi r_2^2 v_2\\r_2 = r_1 \sqrt{\frac{v_1}{v_2}}=(5)\sqrt{\frac{4}{16}}=2.5 mm$

So, the diameter of the nozzle will be

$d_2 = 2r_2 = 2(2.5)=5.0 mm$

6)

According to the Pascal principle, the pressure on the two pistons is the same, so we can write

$\frac{F_1}{A_1}=\frac{F_2}{A_2}$

where

$F_1$ is the force that must be applied to the small piston

$A_1 = \pi r_1^2$ is the area of the first piston, with $r_1= 2 cm$ being its radius

$F_2 = mg = (1500 kg)(9.8 m/s^2)=14700 N$ is the force applied on the bigger piston (the weight of the car)

$A_2 = \pi r_2^2$ is the area of the bigger piston, with $r_2= 15 cm$ being its radius

Solving for $F_1$ , we find

$F_1 = \frac{F_2A_1}{A_2}=\frac{F_2 \pi r_1^2}{\pi r_2^2}=\frac{(14700)(2)^2}{(15)^2}=261 N$

So, the closest answer is B) 266.7 N.

Learn more about pressure:

brainly.com/question/4868239

brainly.com/question/2438000

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5 0

3 years ago

The surface charge density on an infinite charged plane is −1.70 x 10^−6 C/m^2 . A proton is shot straight away from the plane a

goldfiish [28.3K]

Answer:

Acceleration of proton will be $-9.197\times 10^{12}m/sec^2$

Explanation:

We have given surface charge density of an infinite charged plate $\sigma = -1.70\times 10^{-6}C/m^2$

Electric field due to infinite sheet charge is given by $E=\frac{\sigma }{2\epsilon _0}=\frac{-1.7\times 10^{-6}}{2\times 8.85\times10^{-12}}=-0.096\times 10^6=-9.6\times 10^4N/C$

Charge in proton is given by $e=1.6\times 10^{-19}C$

So force on proton $F=qE=1.6\times 10^{-19}\times -9.6\times 10^4=-15.36\times 10^{-15}N$

Mass of proton $m=1.67\times 10^{-27}kg$

According to newtons second law force F = mass × acceleration

So $-15.36\times10^{-15}=1.67\times 10^{-27}a$

$a=-9.197\times 10^{12}m/sec^2$

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3 years ago

All simple machines are variations of which two basic machines?

SCORPION-xisa [38]

L<span>ever; inclined plane</span>

5 0

3 years ago

How do the gravitational and electrical forces between a proton and neutron compare

uysha [10]

Assume distance of seperarion is 1m

F.elec = kQq/r^2

charge of a proton: 1.6×10^-19 C
charge of a neutron: 0 C

F.elec = 0 N

F.grav = GMm/r^2

mass of a proton: 1.672621898×10^-27 kg
mass of a neutron: 1.674927471×10^-27 kg

F.grav = (6.67408×10^-11)×(1.674927471×10^-27)×(1.672621898×10^-27)÷(1^2)

F.grav = 1.8699588×10^-64 N

4 0

4 years ago

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Feliz [49]

Http://www.sengpielaudio.com/calculator-ohm.htm

8 0

3 years ago