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Reil [10]
3 years ago
5

In these formulas, it is useful to understand which variables are parameters that specify the nature of the wave. The variables

E0E0E_0 and B0B0B_0 are the __________ of the electric and magnetic fields. Choose the best answer to fill in the blank.
Physics
1 answer:
Juli2301 [7.4K]3 years ago
7 0
They made me do it I don’t even know what to say I’m so sorry
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How would you find the speed of a person who walked
Alex Ar [27]

Answer:

v = 1.25 m/s

Explanation:

We have,

Distance covered by a person is 10 meters

Time taken by him to cover that distance is 8 seconds.

If we want to find the speed of a person, we must know distance covered by it and taken. In this case, we know both distance and time. His speed is given by :

v=\dfrac{d}{t}\\\\v=\dfrac{10\ m}{8\ s}\\\\v=1.25\ m/s

So, the speed of the person is 1.25 m/s.

4 0
3 years ago
What quantity is multiplied by the hydraulic lift system of a dump truck?
kogti [31]

funny, actually, i'm a hydraulic, and pneumatic cylinder, pump, and line system mechanic, and the answer is pressure.... pressure builds up from the hydraulic pump, and then transfers through hoses to extend, and contract the cylinders by filling them with hydraulic fluid, and vice versa for contracting them.

hope this helps!

4 0
3 years ago
Read 2 more answers
if the Horse and Rider have a combined mass of 572 kg what force would be required to accelerate them 5 kph per second
Vlad [161]

Answer:

   Force required to accelerate = 794.44 N

Explanation:

 Force required = Mass of horse x Acceleration of horse

 Mass of horse and rider, m=   572 kg

 Acceleration of horse and rider, a = 5 kph per second

                                      =\frac{5*1000}{60*60} =1.39 m/s^2

  Force required = ma

                             = 572 x 1.39 = 794.44 N

  Force required to accelerate = 794.44 N

8 0
3 years ago
A pilot is upside down at the top of an inverted loop of radius 3.20 x 103 m. At the top of the loop his normal force is only on
n200080 [17]

Answer:

6858.5712 m/s

Explanation:

Given that:

Radius, r

R = 3.20 * 10^3.

Normal force = 0.5 * normal weight

Normal force = Fn ; Normal weight = Fg

Fn = 0.5Fg

Recall:

mv² / R = Fn + Fg

Fn = 0.5Fg

mv² / R = 0.5Fg + Fg

mv² /R = 1.5Fg

mv² = 1.5Fg * R

F = mg

mv² = 1.5* mg * R

v² = 1.5gR

v = sqrt(1.5gR)

V = sqrt(1.5 * 9.8 * 3.2 * 10^3)

V = sqrt(47.04^3)

V = 6858.5712 m/s

6 0
3 years ago
With the same block-spring system from above, imagine doubling the displacement of the block to start the motion. By what factor
Fofino [41]

Answer:

A)     K / K₀ = 4   b)     v / v₀ = 4

Explanation:

A) For this exercise we can use the conservation of mechanical energy

in the problem it indicates that the displacement was doubled (x = 2xo)

starting point. At the position of maximum displacement

      Em₀ = Ke = ½ k (2x₀)²

final point. In the equilibrium position

      Em_{f} = K = ½ m v²

        Em₀ = Em_{f}

        ½ k 4 x₀² = K

        (½ K x₀²) = K₀

         K = 4 K₀

          K / K₀ = 4

B) the speed value

          ½ k 4 x₀² = ½ m v²

          v = 4 (k / m) x₀

if we call

           v₀ = k / m x₀

          v = 4 v₀

         v / v₀ = 4

3 0
3 years ago
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