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prisoha [69]
3 years ago
14

A negative charge is moving across a rook from south to north. A magnetic field runs from East to west. In what direction is the

magnetic force on the moving negative charge?
A. Toward the East
B. Toward the west
C. Toward the floor
D. Toward the ceiling
Physics
1 answer:
Zinaida [17]3 years ago
5 0

Answer:

The magnetic force on the moving negative charge is towards the ceiling.

Explanation:

The magnetic force acting on a wire is given by the relation as follows :

F=q(v\times B)

q is charged particle

v is velocity of charged particle

B is magnetic field

In this case, a negative charge is moving across a rook from south to north. A magnetic field runs from East to west.

q = -q

v =+j

B = -i

So,

F=-q(vj\times B(-i))

Since, j\times i=-k

So,

F=q(v\times B)(-k)

So, the magnetic force on the moving negative charge is towards the ceiling.

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In a Little League baseball game, the 145 g ball enters the strike zone with a speed of 17.0 m/s . The batter hits the ball, and
VMariaS [17]

Hi there!

Impulse = Change in momentum

I = Δp = mΔv = m(vf - vi)

Where:

m = mass of object (kg)

vf = final velocity (m/s)

vi = initial velocity (m/s)

Begin by converting grams to kilograms:

1 kg = 1000g ⇒ 145g = .145kg

Now, plug in the given values. Remember to assign directions since velocity is a vector. Let the initial direction be positive and the opposite be negative.

I = (.145)(-20 - 17) = -5.365 Ns

The magnitude is the absolute value, so:

|-5.365| = 5.365 Ns

4 0
2 years ago
I have a combination of myopia and presbyopia—overall, the power of my visual system is too large, but I also have a very limite
e-lub [12.9K]

Answer:

The range of powers is    - 5 \ D \le P \le - 2.667\  D

Explanation:

From the question we are told that

       The far point of the left eye is n_f = 20 cm

       The near point of the left eye is  n =  15cm

       The near point with the glasses on is n_g =25 \ cm

     

From these parameter we can see that with the glass on that for near point the

         Object distance would be u = -25 \ cm

          Image distance would be  v =  -15 \ cm

To obtain the focal length we would apply the lens formula which is mathematically represented as

              \frac{1}{f} =  \frac{1}{v}  -  \frac{1}{u}

substituting values

              \frac{1}{f} =  \frac{1}{-15}  -  \frac{1}{-25}

               f =  - \frac{75}{2} cm

           converting to  meters

               f =  - \frac{75}{2} * \frac{1}{100}

               f =  - \frac{75}{200} \ m

   Generally the power of the lens is mathematically represented as

                P  = \frac{1}{f}

Substituting values

                 P = -  \frac{200}{75}  m

                 P = - 2.667 \ D

   

From these parameter we can see that with the glass on that for far  point the

         Object distance would be u_f = - \infty \ cm

          Image distance would be  v_f =  -20  \ cm

To obtain the focal length of the lens we would apply the lens formula which is mathematically represented as

                    \frac{1}{f_f} =  \frac{1}{v_f}  -  \frac{1}{u_f}

substituting values

                  \frac{1}{f} =  \frac{1}{-20}  -  \frac{1}{- \infty}

                 \frac{1}{f} =  \frac{1}{-20}  -  0      

                  f_f =  \frac{20}{1}  \ cm

converting to  meters

                f_f =  - \frac{20}{1}  * \frac{1}{100}

               

Generally the power of the lens is mathematically represented as

                P  = \frac{1}{f_f}

Substituting values

                 P = -  \frac{100}{20}  m

                 P = - 5 \ D

This implies that the range of powers of the lens in his glass is

                  - 5 \ D \le P \le - 2.667\  D

   

               

               

           

3 0
4 years ago
Elliptical galaxies are frequently found a) Inside the Milky Way b) In galaxy clusters c) In the Galactic bulge d) In the Local
sergejj [24]

Answer:

The correct option is b) In galaxy clusters

Explanation:

A type of galaxy that appear elliptical in shape and have an almost featureless and smooth image is known as the elliptical galaxy.

An elliptical galaxy is three dimensional and consists of more than one hundred trillion stars which are present in random orbits around the centre.

Elliptical galaxy is generally found in the galaxy clusters.

5 0
3 years ago
Se lanza verticalmente una esfera con una rapidez de 30m/se. Determinar la rapidez de la esfera a una altura de 40m (g=10m/s2)
sergeinik [125]

v^2-{v_0}^2=2a(x-x_0)

dónde v es la velocidad de la esfera, v_0 es suya velocidad inicial, a=-g la aceleración debida a la gravedad, x la posición, y x_0 la posición inicial. Tomamos x_0=0\,\mathrm m a referirse a la posición de la esfera en el momento que la esfera fue lanzada.

Entonces

v^2-\left(30\,\dfrac{\mathrm m}{\mathrm s}\right)^2=2\left(-10\,\dfrac{\mathrm m}{\mathrm s^2}\right)(40\,\mathrm m)

\implies v^2=100\,\dfrac{\mathrm m^2}{\mathrm s^2}\implies v=\pm10\,\dfrac{\mathrm m}{\mathrm s}

Esto nos dice que la esfera alcanza una altura de 40 m en dos momentos - una vez hacia arriba y una vez hacia abajo. Sin embargo, independientemente del signo de la velocidad, sabemos que suya magnitud es 10 m/s, y así tenemos una rapidez de 10 m/s también en ambos momentos.

4 0
3 years ago
A cylinder is fitted with a piston, beneath which is a spring, as in the drawing. The cylinder is open to the air at the top. Fr
Mazyrski [523]

Answer:

x = 0.0537 m or 5.37 cm

Explanation:

Given:

spring constant'k'= 4900 N/m

radius 'r' =0.029 m

Area 'A' =r²π = 0.029²π => 2.6 x 10^{-3} m²

Here, Pressure 'P' is given by,

Pressure = Force / Area

And we know that, for a spring :

F = kx, where k is the spring constant and x is the change in length.

P = kx/A

As P = 101325 Pa

101325 = 4900x / ( 2.6 x 10^{-3})

x = 0.0537 m or 5.37 cm

6 0
3 years ago
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