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aev [14]
3 years ago
11

The magnetic field at the equator points north. If you throw a positively charged object (for example, a baseball with some elec

trons removed) to the east, what is the direction of the magnetic force on the object? 1. Toward the east 2. Upward 3. Toward the west 4. Downward
Physics
1 answer:
PilotLPTM [1.2K]3 years ago
5 0

Answer:

The magnetic force points in the positive z-direction, which corresponds to the upward direction.

Option 2 is correct, the force points in the upwards direction.

Explanation:

The magnetic force on any charge is given as the cross product of qv and B

F = qv × B

where q = charge on the ball thrown = +q (Since it is positively charged)

v = velocity of the charged ball = (+vî) (velocity is in the eastern direction)

B = Magnetic field = (+Bj) (Magnetic field is in the northern direction; pointing forward)

F = qv × B = (+qvî) × (Bj)

F =

| î j k |

| qv 0 0|

| 0 B 0

F = i(0 - 0) - j(0 - 0) + k(qvB - 0)

F = (qvB)k N

The force is in the z-direction.

We could also use the right hand rule; if we point the index finger east (direction of the velocity), the middle finger northwards (direction of the magnetic field), the thumb points in the upward direction (direction of the magnetic force). Hence, the magnetic force is acting upwards, in the positive z-direction too.

Hope this Helps!!!

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Oh football player kicks a football from the height of 4 feet with an initial vertical velocity of 64 ft./s use the vertical mot
Stolb23 [73]

Answer:

4.1 seconds

Explanation:

The height of the football is given by the equation:

H = -16t^2 + V*t + S

Using the inicial position S = 4 and the inicial velocity V = 64, we can find the time when the football hits the ground (H = 0):

0 = -16t^2 + 64*t + 4

4t^2 - 16t - 1 = 0

Using Bhaskara's formula, we have:

\Delta = b^2 - 4ac = (-16)^2 - 4*4*(-1) = 272

t_1 = (-b + \sqrt{\Delta})/2a

t_1 = (16 + 16.49)/8 = 4.06\ seconds

t_2 = (-b - \sqrt{\Delta})/2a

t_2 = (16 - 16.49)/8 = -0.06\ seconds

A negative time is not a valid result for this problem, so the amount of time the football is in the air before hitting the ground is 4.1 seconds.

7 0
3 years ago
Read 2 more answers
A student hears a police siren. What would change the frequency that the student hears? check all that apply
andriy [413]

Considering the Doppler efect, the frequency heard by the student would change if:

  • if the student walked toward the police car.
  • if the student walked away from the police car.
  • if the police car moved toward the student.
  • if the police car moved away from the student.

<h3>Doppler effect</h3>

The Doppler effect is defined as the change in the apparent frequency of a wave produced by the relative motion of the source with respect to its observer. In other words, this effect is the change in the perceived frequency of any wave motion when the sender and receiver, or observer, move relative to each other.

The following expression is considered the general case of the Doppler effect:

f'=f\frac{v+-vR}{v-+vE}

Where:

  • f', f: Frequency perceived by the receiver and frequency emitted by the transmitter, respectively. Its unit of measurement in the International System (S.I.) is the hertz (Hz), which is the inverse unit of the second (1 Hz = 1 s⁻¹)
  • v: Wave propagation speed in the medium. It is constant and depends on the characteristics of the medium. In this case, the speed of sound in air is considered to be 343 m/s.
  • vR, vE: Receiver and transmitter speed respectively. Its unit of measure in the S.I. is the m/s
  • ±, ∓:
  • We will use the + sign:
  1. In the numerator if the receiver approaches the sender
  2. In the denominator if the sender moves away from the receiver
  • We will use the - sign:
  1. In the numerator if the receiver moves away from the sender
  2. In the denominator if the sender approaches the receiver

In summary, the Doppler Effect is an alteration of the observed frequency of a sound due to the movement of the source or the observer, that is, they are changes in the frequency and wavelength of a wave due to the relative movement between the wave source and the observer.

<h3>Changes on the frequency </h3>

In this case, considering the Doppler effect, the frequency heard by the student would change if:

  • if the student walked toward the police car.
  • if the student walked away from the police car.
  • if the police car moved toward the student.
  • if the police car moved away from the student.

Learn more about Doppler effect:

brainly.com/question/15307081

brainly.com/question/4052291

brainly.com/question/15097772

brainly.com/question/3841958

#SPJ12

7 0
2 years ago
Older televisions display a picture using a device called a cathode ray tube, where electrons are emitted at high speed and coll
gregori [183]

Answer:

a)  t = 3.027 10⁻⁹ s ,  b)   y = 2.25 10⁻² m

Explanation:

We can solve this problem using the kinematic relations

a) as on the x-axis there is no relationship

          vₓ = x / t

          t = x / vₓ

We reduce the magnitudes to the SI system

          x = 5.6 cm (1m / 100 vm) = 0.056 m

we calculate

          t = 0.056 / 1.85 10⁷

          t = 3.027 10⁻⁹ s

b) the time is the same for the two movements, on the y axis

         y = v₀t + ½ a t²

         

as the beam leaves horizontal there is no initial vertical velocity

         y = ½ a t²

         

let's calculate

         y = ½  5.45 10¹⁵ (3.027 10⁻⁹)²

         y = 2.25 10⁻² m

6 0
3 years ago
A boy pushes a 5 kg box across the floor with a force of 25 N. There is a 5 N frictional force opposing him. What is the net for
11Alexandr11 [23.1K]

Answer:net force:20N, accel. :4m/s2

Explanation:

8 0
3 years ago
What is the final velocity of a car that is originally traveling 12 m/s and then undergoes an acceleration of 2.3m/s squared for
Katarina [22]

To solve this problem we use the general kinetic equations.

We need to know the time it takes for the car to reach 130 meters.

In this way we have to:

x(t) = x_0 + v_0t + 0.5at ^ 2

Where

x_0 = initial position

v_0 = initial velocity

a = acceleration

t = time

x(t) = position as a function of time

130 = 0 + 12(t) + 0.5(2.3)t ^ 2

1.15t ^ 2 + 12t - 130.

We use the quadratic formula to solve the equation.

t = \frac{-12 \± \sqrt {(12) ^ 2-4(1.15)(- 130)}}{2 (1.15)}

t = 6.63 s and t = -17.1 s

We take the positive solution. This means that the car takes 6.63 s to reach 130 meters.

Then we use the following equation to find the final velocity:

v_f = v_0 + at

Where:

v_f = final speed

v_f = 12 +2.23(6.63)

The final speed of the car is 27.25 m/s

3 0
3 years ago
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