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soldi70 [24.7K]
3 years ago
5

Where are auroras frequently seen

Physics
2 answers:
Fynjy0 [20]3 years ago
7 0
Auroras as in northern lights? Usually around northern Canada, Greenland, Iceland, Russia or Alaska
zysi [14]3 years ago
4 0
Around the person....is where they are seen...
You might be interested in
What can you infer about a wave with a short wavelength?
raketka [301]

Answer:

- It can be infer that it has a lower frequency.

<em>In the case of electromagnetic waves.</em>

- A short wavelength means a lower energy,

Explanation:

The wavelength is the distance between two consecutive crests or valleys while the frequency is the number of crests that pass for a specific point in an interval of time.

For example, a person makes laundry once a weak.

In this example, the event is represented by the laundry and the interval of time is once a weak

The velocity of a wave is defined as:

v = \nu \cdot \lambda  (1)

Where nu is the frequency and \lambda is the wavelenth

\lambda =  \frac{v}{\nu}  (2)

Notice from equation 2 that the wavelength is inversely proportional to the frequency (when the wavelength increases the frequency decreases).

In the case of electromagnetic waves, a short wavelength means a lower energy, as it can be seen in equation 4 (inversely proportional).

E = h\nu  (3)

E = \frac{hc}{\lambda} (4)

8 0
2 years ago
Read 2 more answers
a bal is launched upward with a velocity of v0 from the edge of a cliff of height D. it reaches a maximum height of H above its
lilavasa [31]

Answer:

D/H =15

Explanation:

  • We can find first the peak height H, taking into consideration, that at the maximum height, the ball will reach momentarily to a stop.
  • At this point, we can find the value of H, applying the following kinematic equation:

       v_{f} ^{2} -v_{0} ^{2} = 2* g* H (1)

  • If vf=0, if we assume that the positive direction is upwards, we can find the value of H as follows:

       H = \frac{v_{0} ^{2} }{2*g} (2)

  • We can use the same equation, to find the value of D, as follows:

        v_{f} ^{2} -v_{1} ^{2} = 2* g* D (3)

  • In order to find v₁, we can use the same kinematic equation that we used to get H, but now, we know that v₀ = 0.
  • When we replace these values in (1), we find that  v₁ = -v₀.
  • Replacing in (3), we have:

        (4*v_{0})^{2} - (-v_{0}) ^{2}  = 2* g* D\\ \\ 15*v_{0}^{2}  = 2*g*D

  • Solving for  D:

       D = \frac{15*v_{0} ^{2} }{2*g}

  • From (2) we know that H can be expressed as follows:

       H = \frac{v_{0} ^{2} }{2*g}

  • ⇒ D = 15 * H

        \frac{D}{H} = 15

3 0
2 years ago
A cart is pulled by a force of 250 N at an angle of 35° above the horizontal. The cart accelerates at 1.4 m/s2. The free-body di
Pachacha [2.7K]

Answer:

m=146.277kg which is rounded to 146kg

Explanation:

Remember that F=ma

But F represents not 250N, but 250cos(35)N since the force is being pulled above the horizontal.

So 250cos(35)=204.7880111 approximately, and since a=1.4m/s^2, we have 204.7880111=m(1.4m/s^2). Then we divide both sides by the acceleration to get the mass. So m=146.2771508kg which the nearest number is 146kg

Mass is always in kg, unless stated otherwise.

4 0
2 years ago
Read 2 more answers
How does the thermoflask prevent heat loss​
riadik2000 [5.3K]

Answer:

The way that the flask is built it has 3 protective layers.... the inside layer to keep the heat in, the outside layer to reflective the cold, and a vacuum layer, which is an empty layer that limits conduction and convection

Explanation:

3 0
3 years ago
A 2.50-m segment of wire carries 1000 A current and feels a 4.00-N repulsive force from a parallel wire 5.00 cm away. What is th
Stolb23 [73]

Answer:

The current is  I_b  =  400 \ A

Explanation:

From the question we are told that

    The  length of the segment is  l  =  2.50  \  m

     The current is  I_a  =  1000 \ A

     The force felt is  F  =  4.0 \  N

        The distance of the second wire is  d =  5.0 \ cm  = 0.05 \  m

Generally the current on the second wire is mathematically represented as

        I_b  =  \frac{2 \pi * r * F }{ l *  \mu_o  *  I_a }

Here  \mu_o is the permeability of free space with value  \mu_o =  4 \pi * 10^{-7} \ N/A^2

=>      I_b  =  \frac{2 * 3.142  *  0.05 *  4 }{ 2.50  *  4\pi *10^{-7}  * 1000 }

=>      I_b  =  400 \ A

4 0
2 years ago
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