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telo118 [61]
3 years ago
10

What is the relationship between the poles of a magnet? which direction does the magnet field move?

Physics
1 answer:
Assoli18 [71]3 years ago
4 0
A magnet has a South Pole and a North Pole. South Pole and South Pole can't connect to her other, same as North and North. The same poles push each other away.
South Pole and North Pole connect.
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Help Please
densk [106]

Answer:

Energy=3.1times 10^-17 J

Rest mass: 6.2 kg

Speed: 47.5 m/s

Wavelength: 2.659 times 10^-6

Momentum: 67.3 kg(m/s)

Explanation:

3 0
3 years ago
Explain why people have different skin colors?
stiks02 [169]

I think it because of UV rays ultra violet ray which can make their colors different

5 0
3 years ago
What types of electromagnetic radiation does the sun emit?
Ksju [112]

Answer:

https://gml.noaa.gov/education/info_activities/pdfs/LA_radiation.pdf

Explanation:

5 0
2 years ago
A traffic light is weighing 200N hangs from a vertical cable tied to two other cables that are fastened to a support. The upper
Levart [38]

Answer:

T₁ = 93.6 N , T₂ = 155.6 N , T₃ = 200 N

Explanation:

This is a balance exercise where we must apply the expressions for translational balance in the two axes

     ∑  F = 0

Suppose that cable t1 goes to the left and the angles are 41º with respect to the horizontal and cable t2 goes to the right with angles of 63º

decompose the tension of the two upper cables

          cos 41 = T₁ₓ / T1

          sin 41 = T₁y / T1

          T₁ₓ = T₁  cos 41

          T₁y= T₁  sin 41

for cable gold

           cos 63 = T₂ / T₂

           sin 63 = T_{2y} / T₂

We apply the two-point equilibrium equation: The junction point of the three cables and the point where the traffic light joins the vertical cable.

Let's start by analyzing the point where the traffic light meets the vertical cable

              T₃ - W = 0

              T₃ = W

              T₃ = 200 N

now let's write the equations for the single point of the three wires

X axis

   - T₁ₓ + T₂ₓ = 0

  T₁ₓ = T₂ₓ

   T1 cos 41 = T2 cos 63

   T1 = T2 cos 63 / cos 41                (1)

y Axis

      T_{1y} + T_{2y} - T3 = 0

       T₁ sin 41 + T₂ sin 63 = T₃          (2)

to solve the system we substitute equation 1 in 2

        T₂ cos 63 / cos 41 sin 41 + T₂ sin 63 = W

         T₂ (cos 63 tan 41 + sin 63) = W

         T₂ = W / (cos 63 tan 41 + sin 63)

We calculate

          T₂ = 200 / (cos 63 tan 41 + sin 63)

          T₂ = 200 / 1,2856

           T₂ = 155.6 N

we substitute in 1

            T₁ = T₂ cos 63 / cos 41

             T₁ = 155.6 cos63 / cos 41

             T₁ = 93.6 N

therefore the tension in each cable is

            T₁ = 93.6 N

             T₂ = 155.6 N

             T₃ = 200 N

6 0
3 years ago
A car of mass m accelerates from speed v_1 to speed v_2 while going up a slope that makes an angle theta with the horizontal. Th
Karo-lina-s [1.5K]

Answer:

Work done by external force is given as

Work_{external} = mgLsin\theta + \mu mgLcos(\theta) + \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2

Explanation:

As per work energy Theorem we can say that work done by all force on the car is equal to change in kinetic energy of the car

so we will have

Work_{external} + Work_{gravity} + Work_{friction} = \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2

now we have

W_{gravity} = -mg(Lsin\theta)

W_{friction} = -\mu mgcos(\theta) L

so from above equation

Work_{external} - mgLsin\theta - \mu mgLcos(\theta) = \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2

so from above equation work done by external force is given as

Work_{external} = mgLsin\theta + \mu mgLcos(\theta) + \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2

8 0
3 years ago
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