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julsineya [31]
3 years ago
15

If an electron is released at the equator and falls toward earth under the influence of gravity, the magnetic force on the elect

ron will be toward the
Physics
1 answer:
Alborosie3 years ago
4 0
Using Fleming’s left hand rule,

The magnetic force on the electron will be west. Example: If it is dropped above the equator in Africa, the electron will head towards North America.
You might be interested in
Jupiter and Saturn each have the same three basic cloud layers, but the spacing of the layers differs on the two planets. Why is
-Dominant- [34]

Answer:

b. Jupiter’s greater gravity has compressed the layers, so they are closer together there.

Explanation:

The value for Jupiter mass is 1.8981×10²⁷kg, while the mass of Saturn is 5.6832×10²⁶kg, so the different layers of clouds in Jupiter will be submitted to a greater gravitational pull because it has a bigger mass, as is established in the law of universal gravitation:

F = G\frac{m1m2}{r^{2}}   (1)

Where m1 and m2 are the masses of two objects, G is the gravitational constant and r is the distance between the two objects.

As it can be seen in equation 1, the gravitational force is directly proportional to the product of the masses of the objects, so if the mass increase the gravitational force will do it too.  

For the case of Saturn, it has a lower mass so its layers of clouds will suffer a weaker gravitational pull. That leads to the three clouds being more spacing that the ones of Jupiter.    

3 0
3 years ago
How many times louder is the intensity of sound at a rock concert in comparison with that of a whisper, if the two intensity lev
Phoenix [80]

Answer:

 5 × 10¹⁰ or 5 billion times louder is the intensity of sound at a rock concert in comparison with that of a whisper

Explanation:

Given that at a rock concert;

the intensity of sound is 110 dB compared to a whisper of 3 dB

Now; how many times louder will that of the whisper be compared to the sound of the rock concert

Using the formula for calculating decibels (dB);

For 110 dB

dB = 10log₁₀(W)

110 dB =  10^{(110/10)}

110 dB = 10¹¹

For 3dB

dB = 10log₁₀(W)

3 dB = 10^{(3/10)}

3 dB = 2

Now:

(110/3)dB = 10¹¹/2

(110/3)dB =  5 × 10¹⁰ or 5 billion times louder is the intensity of sound at a rock concert in comparison with that of a whisper

7 0
3 years ago
Which can do work faster-a 700 watt gasoline engine or a 300 watt electric motor
iogann1982 [59]

"700 watts" means 700 joules of work per second.

"300 watts" means 300 joules of work per second.

If the labels on both machines are true, and both machines
are loaded to their full capacity, then the 700-watt engine
is doing work faster than the 300-watt one.
3 0
3 years ago
Question Part Points Submissions Used A pitcher throws a 0.200 kg ball so that its speed is 19.0 m/s and angle is 40.0° below th
postnew [5]

Answer:

The impulse is (10.88 i^ + 7.04 j^) N s

maximum force on the ball is  (4.53 10 2 i^ + 2.93 102 j ^) N  

Explanation:

In a problem of impulse and shocks we must use the impulse equation

       I = dp = pf-p₀         (1)

       p = m V

With we have vector quantities, let's decompose the velocities on the x and y axes

      V₀ = -19 m / s

      θ₀ = 40.0º  

      Vf = 46.0 m / s

      θf = 30.0º

Note that since the positive direction of the x-axis is from the batter to the pitcher, the initial velocity is negative and the angle of 40º is measured from the axis so it is in the third quadrant

      Vcx = Vo cos θ

      Voy = Vo sin θ

      Vox= -19 cos (40) = -14.6 m/s

      Voy = -19 sin (40) =  -12.2 m/s

      Vfx = 46 cos 30 = 39.8 m/s

      Vfy = 46 sin 30 =  23.0 m/s

   a) We already have all the data, substitute and calculate the impulse for each axis

      Ix = pfx -pfy

      Ix = m ( vfx -Vox)

      Ix = 0.200 ( 39.8 – (-14.6))

      Ix = 10.88 N s

      Iy = m (Vfy -Voy)

      Iy = 0.200 ( 23.0- (-12.2))

      Iy=  7.04 N s

In vector form it remains

       I =  (10.88 i^ + 7.04 j^) N s

   b) As we have the value of the impulse in each axis we can use the expression that relates the impulse to the average force and your application time, so we must calculate the average force in each interval.

         I = Fpro Δt

In the first interval

        Fpro = (Fm + Fo) / 2

With the Fpro the average value of the force, Fm the maximum value and Fo the minimum value, which in this case is zero

         Fpro = (Fm +0) / 2

In the second interval the force is constant

          Fpro = Fm

In the third interval

         Fpro = (0 + Fm) / 2

Let's replace and calculate

         I =  Fpro1 t1 +Fpro2 t2  +Fpro3 t3

         I = Fm/2 4 10⁻³ + Fm 20 10⁻³+ Fm/2 4 10⁻³  

         I = Fm  24 10⁻³ N s

         Fm = I / 24 10⁻³

         Fm = (10.88 i^ + 7.04 j^) / 24 10⁻³

         Fm = (4.53 10² i^ + 2.93 10² j ^) N

maximum force on the ball is  (4.53 10 2 i^ + 2.93 102 j ^) N  

3 0
3 years ago
What happens once a scientific theory is accepted
nordsb [41]
You does any experience
8 0
3 years ago
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